Let $H$$H$HH$H$ be a subspace of ${\mathbb{R}}^4$${\mathbb{R}}^4$R^(4)\mathbb{R}^{4}${\mathbb{R}}^4$ spanned by the vectors $v_1=(1,-2,5,-3)$$v_1=(1,-2,5,-3)$v_(1)=(1,-2,5,-3)v_{1}=(1,-2,5,-3)$v_1=(1,-2,5,-3)$, $v_2=(2,3,1,-4),v_3=(3,8,-3,-5)$$v_2=(2,3,1,-4),v_3=(3,8,-3,-5)$v_(2)=(2,3,1,-4),v_(3)=(3,8,-3,-5)v_{2}=(2,3,1,-4), v_{3}=(3,8,-3,-5)$v_2=(2,3,1,-4),v_3=(3,8,-3,-5)$. Then find a basis and dimension of $H$$H$HH$H$, and extend the basis of $H$$H$HH$H$ to a basis of ${\mathbb{R}}^4$${\mathbb{R}}^4$R^(4)\mathbb{R}^{4}${\mathbb{R}}^4$.

Answer:

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Step 1: Find the Basis and Dimension of $H$$H$HH$H$

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✅ Conclusion:

• Dimension of $H$$H$HH$H$ = 2
• Basis of $H$$H$HH$H$ = Any two linearly independent vectors. We can take:
  $$\boxed{{\displaystyle \{v_1=(1,-2,5,-3),v_2=(2,3,1,-4)\}}}$$$$\boxed{{\displaystyle \left\{v_1=(1,-2,5,-3),v_2=(2,3,1,-4)\right\}}}$${v_(1)=(1,-2,5,-3),quadv_(2)=(2,3,1,-4)}\boxed{\left\{v_1 = (1,-2,5,-3),\quad v_2 = (2,3,1,-4)\right\}}$$\boxed{{\displaystyle \{v_1=(1,-2,5,-3),v_2=(2,3,1,-4)\}}}$$

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Step 2: Extend this Basis to a Basis for ${\mathbb{R}}^4$${\mathbb{R}}^4$R^(4)\mathbb{R}^4${\mathbb{R}}^4$

• $e_1$$e_1$e_(1)e_1$e_1$ is not in the span of $v_1$$v_1$v_(1)v_1$v_1$ and $v_2$$v_2$v_(2)v_2$v_2$ (since $a{v}_1+b{v}_2=e_1$$a{v}_1+b{v}_2=e_1$av_(1)+bv_(2)=e_(1)a v_1+b v_2=e_1$a{v}_1+b{v}_2=e_1$ leads to a contradiction).
• The set $\{v_1,v_2,e_1\}$$\left\{v_1,v_2,e_1\right\}${v_(1),v_(2),e_(1)}\left\{v_1, v_2, e_1\right\}$\{v_1,v_2,e_1\}$ is linearly independent.
• $e_2$$e_2$e_(2)e_2$e_2$ is not in the span of $\{v_1,v_2,e_1\}$$\left\{v_1,v_2,e_1\right\}${v_(1),v_(2),e_(1)}\left\{v_1, v_2, e_1\right\}$\{v_1,v_2,e_1\}$ (since $a{v}_1+b{v}_2+c{e}_1=e_2$$a{v}_1+b{v}_2+c{e}_1=e_2$av_(1)+bv_(2)+ce_(1)=e_(2)a v_1+b v_2+c e_1=e_2$a{v}_1+b{v}_2+c{e}_1=e_2$ leads to a contradiction).
• The set $\{v_1,v_2,e_1,e_2\}$$\left\{v_1,v_2,e_1,e_2\right\}${v_(1),v_(2),e_(1),e_(2)}\left\{v_1, v_2, e_1, e_2\right\}$\{v_1,v_2,e_1,e_2\}$ is linearly independent.

Let $T:{\mathbb{R}}^3\to {\mathbb{R}}^3$$T:{\mathbb{R}}^3\to {\mathbb{R}}^3$T:R^(3)rarrR^(3)T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$T:{\mathbb{R}}^3\to {\mathbb{R}}^3$ be a linear operator and $B=\{v_1,v_2,v_3\}$$B=\left\{v_1,v_2,v_3\right\}$B={v_(1),v_(2),v_(3)}B=\left\{v_{1}, v_{2}, v_{3}\right\}$B=\{v_1,v_2,v_3\}$ be a basis of ${\mathbb{R}}^3$${\mathbb{R}}^3$R^(3)\mathbb{R}^{3}${\mathbb{R}}^3$ over $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$. Suppose that $T{v}_1=(1,1,0),T{v}_2=(1,0,-1),T{v}_3=(2,1,-1)$$T{v}_1=(1,1,0),T{v}_2=(1,0,-1),T{v}_3=(2,1,-1)$Tv_(1)=(1,1,0),Tv_(2)=(1,0,-1),Tv_(3)=(2,1,-1)T v_{1}=(1,1,0),\; T v_{2}=(1,0,-1),\; T v_{3}=(2,1,-1)$T{v}_1=(1,1,0),T{v}_2=(1,0,-1),T{v}_3=(2,1,-1)$. Find a basis for the range space and null space of $T$$T$TT$T$.

Answer:

Step 1: Find the Matrix Representation of $T$$T$TT$T$

• $T{v}_1=(1,1,0)$$T{v}_1=(1,1,0)$Tv_(1)=(1,1,0)T v_1 = (1, 1, 0)$T{v}_1=(1,1,0)$
• $T{v}_2=(1,0,-1)$$T{v}_2=(1,0,-1)$Tv_(2)=(1,0,-1)T v_2 = (1, 0, -1)$T{v}_2=(1,0,-1)$
• $T{v}_3=(2,1,-1)$$T{v}_3=(2,1,-1)$Tv_(3)=(2,1,-1)T v_3 = (2, 1, -1)$T{v}_3=(2,1,-1)$

Step 2: Basis for the Range Space

• R2 = R2 – R1:

• R3 = R3 – R2 (not necessary, as R3 = R2), set R3 = 0:

• R2 = -R2:

• R1 = R1 – R2:

• Column 3: $(2,1,-1)$$(2,1,-1)$(2,1,-1)(2, 1, -1)$(2,1,-1)$
• Columns 1 and 2: $(1,1,0)$$(1,1,0)$(1,1,0)(1, 1, 0)$(1,1,0)$, $(1,0,-1)$$(1,0,-1)$(1,0,-1)(1, 0, -1)$(1,0,-1)$

Step 3: Basis for the Null Space

• From the third equation: $x_2=-x_3$$x_2=-x_3$x_(2)=-x_(3)x_2 = -x_3$x_2=-x_3$.
• From the second: $x_1=-x_3$$x_1=-x_3$x_(1)=-x_(3)x_1 = -x_3$x_1=-x_3$.
• Substitute into the first: $(-x_3)+(-x_3)+2x_3=-x_3-x_3+2x_3=0$$(-x_3)+(-x_3)+2x_3=-x_3-x_3+2x_3=0$(-x_(3))+(-x_(3))+2x_(3)=-x_(3)-x_(3)+2x_(3)=0(-x_3) + (-x_3) + 2 x_3 = -x_3 – x_3 + 2 x_3 = 0$(-x_3)+(-x_3)+2x_3=-x_3-x_3+2x_3=0$, which is satisfied.

Step 4: Express Null Space Basis in Standard Coordinates (if needed)

Final Answer

• Basis for the range space: $\{(1,1,0),(1,0,-1)\}$$\{(1,1,0),(1,0,-1)\}${(1,1,0),(1,0,-1)}\{ (1, 1, 0), (1, 0, -1) \}$\{(1,1,0),(1,0,-1)\}$
• Basis for the null space: $\{-v_1-v_2+v_3\}$$\{-v_1-v_2+v_3\}${-v_(1)-v_(2)+v_(3)}\{ -v_1 – v_2 + v_3 \}$\{-v_1-v_2+v_3\}$

Discuss the continuity of the function

Answer:

1. Continuity for $x\ne 0$$x\ne 0$x!=0x \neq 0$x\ne 0$:

• $e^{-1/x}$$e^{-1/x}$e^(-1//x)e^{-1/x}$e^{-1/x}$ is continuous for all $x\ne 0$$x\ne 0$x!=0x \neq 0$x\ne 0$.
• The denominator $1-e^{-1/x}$$1-e^{-1/x}$1-e^(-1//x)1 – e^{-1/x}$1-e^{-1/x}$ is continuous and non-zero for $x\ne 0$$x\ne 0$x!=0x \neq 0$x\ne 0$, since $e^{-1/x}\ne 1$$e^{-1/x}\ne 1$e^(-1//x)!=1e^{-1/x} \neq 1$e^{-1/x}\ne 1$ when $x\ne 0$$x\ne 0$x!=0x \neq 0$x\ne 0$.

2. Continuity at $x=0$$x=0$x=0x = 0$x=0$:

Case 1: $x\to 0^{+}$$x\to 0^{+}$x rarr0^(+)x \to 0^+$x\to 0^{+}$ (Right-Hand Limit)

• As $x\to 0^{+}$$x\to 0^{+}$x rarr0^(+)x \to 0^+$x\to 0^{+}$, $\frac{1}{x}\to +\mathrm{\infty }$$\frac{1}{x}\to +\mathrm{\infty }$(1)/(x)rarr+oo\frac{1}{x} \to +\infty$\frac{1}{x}\to +\mathrm{\infty }$, so $e^{-1/x}\to 0$$e^{-1/x}\to 0$e^(-1//x)rarr0e^{-1/x} \to 0$e^{-1/x}\to 0$.
• Thus:$$\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{+}}{lim}\frac{1}{1-e^{-1/x}}=\frac{1}{1-0}=1.$$$$\underset{x\to 0^{+}}{lim} f(x)=\underset{x\to 0^{+}}{lim} \frac{1}{1-e^{-1/x}}=\frac{1}{1-0}=1.$$lim_(x rarr0^(+))f(x)=lim_(x rarr0^(+))(1)/(1-e^(-1//x))=(1)/(1-0)=1.\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{1}{1 – e^{-1/x}} = \frac{1}{1 – 0} = 1.$$\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{+}}{lim}\frac{1}{1-e^{-1/x}}=\frac{1}{1-0}=1.$$

Case 2: $x\to 0^{-}$$x\to 0^{-}$x rarr0^(-)x \to 0^-$x\to 0^{-}$ (Left-Hand Limit)

• As $x\to 0^{-}$$x\to 0^{-}$x rarr0^(-)x \to 0^-$x\to 0^{-}$, $\frac{1}{x}\to -\mathrm{\infty }$$\frac{1}{x}\to -\mathrm{\infty }$(1)/(x)rarr-oo\frac{1}{x} \to -\infty$\frac{1}{x}\to -\mathrm{\infty }$, so $e^{-1/x}\to +\mathrm{\infty }$$e^{-1/x}\to +\mathrm{\infty }$e^(-1//x)rarr+ooe^{-1/x} \to +\infty$e^{-1/x}\to +\mathrm{\infty }$.
• Thus:$$\underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{-}}{lim}\frac{1}{1-e^{-1/x}}=\frac{1}{1-\mathrm{\infty }}=0.$$$$\underset{x\to 0^{-}}{lim} f(x)=\underset{x\to 0^{-}}{lim} \frac{1}{1-e^{-1/x}}=\frac{1}{1-\mathrm{\infty }}=0.$$lim_(x rarr0^(-))f(x)=lim_(x rarr0^(-))(1)/(1-e^(-1//x))=(1)/(1-oo)=0.\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1}{1 – e^{-1/x}} = \frac{1}{1 – \infty} = 0.$$\underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{-}}{lim}\frac{1}{1-e^{-1/x}}=\frac{1}{1-\mathrm{\infty }}=0.$$

Conclusion at $x=0$$x=0$x=0x = 0$x=0$:

3. Summary of Continuity:

• Continuous for all $x\ne 0$$x\ne 0$x!=0x \neq 0$x\ne 0$.
• Discontinuous at $x=0$$x=0$x=0x = 0$x=0$ because the limit does not exist (left and right limits disagree).

Final Answer:

Expand $\ln (x)$$\ln (x)$ln(x)\ln(x)$\ln (x)$ in powers of $(x-1)$$(x-1)$(x-1)(x-1)$(x-1)$ by Taylor’s theorem and hence find the value of $\ln (1\cdot 1)$$\ln (1\cdot 1)$ln(1*1)\ln (1 \cdot 1)$\ln (1\cdot 1)$ correct up to four decimal places.

Answer:

Step 1: Understand Taylor’s Theorem

Step 2: Compute the Function and Its Derivatives at $x=1$$x=1$x=1x = 1$x=1$

• Function value:
  $$f(x)=\ln (x),f(1)=\ln (1)=0$$$$f(x)=\ln (x),f(1)=\ln (1)=0$$f(x)=ln(x),quad f(1)=ln(1)=0f(x) = \ln(x), \quad f(1) = \ln(1) = 0$$f(x)=\ln (x),f(1)=\ln (1)=0$$
• First derivative:
  $$f^{\prime }(x)=\frac{d}{dx}\ln (x)=\frac{1}{x},f^{\prime }(1)=\frac{1}{1}=1$$$$f^{\prime }(x)=\frac{d}{dx}\ln (x)=\frac{1}{x},f^{\prime }(1)=\frac{1}{1}=1$$f^(‘)(x)=(d)/(dx)ln(x)=(1)/(x),quadf^(‘)(1)=(1)/(1)=1f'(x) = \frac{d}{dx} \ln(x) = \frac{1}{x}, \quad f'(1) = \frac{1}{1} = 1$$f^{\prime }(x)=\frac{d}{dx}\ln (x)=\frac{1}{x},f^{\prime }(1)=\frac{1}{1}=1$$
• Second derivative:
  $$f^{″}(x)=\frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{x^2},f^{″}(1)=-\frac{1}{1^2}=-1$$$$f^{″}(x)=\frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{x^2},f^{″}(1)=-\frac{1}{1^2}=-1$$f^(″)(x)=(d)/(dx)((1)/(x))=-(1)/(x^(2)),quadf^(″)(1)=-(1)/(1^(2))=-1f”(x) = \frac{d}{dx} \left( \frac{1}{x} \right) = -\frac{1}{x^2}, \quad f”(1) = -\frac{1}{1^2} = -1$$f^{″}(x)=\frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{x^2},f^{″}(1)=-\frac{1}{1^2}=-1$$
• Third derivative:
  $$f^{‴}(x)=\frac{d}{dx}(-\frac{1}{x^2})=\frac{2}{x^3},f^{‴}(1)=\frac{2}{1^3}=2$$$$f^{‴}(x)=\frac{d}{dx}\left(-\frac{1}{x^2}\right)=\frac{2}{x^3},f^{‴}(1)=\frac{2}{1^3}=2$$f^(‴)(x)=(d)/(dx)(-(1)/(x^(2)))=(2)/(x^(3)),quadf^(‴)(1)=(2)/(1^(3))=2f”'(x) = \frac{d}{dx} \left( -\frac{1}{x^2} \right) = \frac{2}{x^3}, \quad f”'(1) = \frac{2}{1^3} = 2$$f^{‴}(x)=\frac{d}{dx}(-\frac{1}{x^2})=\frac{2}{x^3},f^{‴}(1)=\frac{2}{1^3}=2$$
• Fourth derivative:
  $$f^{(4)}(x)=\frac{d}{dx}\left(\frac{2}{x^3}\right)=-\frac{6}{x^4},f^{(4)}(1)=-\frac{6}{1^4}=-6$$$$f^{(4)}(x)=\frac{d}{dx}\left(\frac{2}{x^3}\right)=-\frac{6}{x^4},f^{(4)}(1)=-\frac{6}{1^4}=-6$$f^((4))(x)=(d)/(dx)((2)/(x^(3)))=-(6)/(x^(4)),quadf^((4))(1)=-(6)/(1^(4))=-6f^{(4)}(x) = \frac{d}{dx} \left( \frac{2}{x^3} \right) = -\frac{6}{x^4}, \quad f^{(4)}(1) = -\frac{6}{1^4} = -6$$f^{(4)}(x)=\frac{d}{dx}\left(\frac{2}{x^3}\right)=-\frac{6}{x^4},f^{(4)}(1)=-\frac{6}{1^4}=-6$$
• Fifth derivative:
  $$f^{(5)}(x)=\frac{d}{dx}(-\frac{6}{x^4})=\frac{24}{x^5},f^{(5)}(1)=\frac{24}{1^5}=24$$$$f^{(5)}(x)=\frac{d}{dx}\left(-\frac{6}{x^4}\right)=\frac{24}{x^5},f^{(5)}(1)=\frac{24}{1^5}=24$$f^((5))(x)=(d)/(dx)(-(6)/(x^(4)))=(24)/(x^(5)),quadf^((5))(1)=(24)/(1^(5))=24f^{(5)}(x) = \frac{d}{dx} \left( -\frac{6}{x^4} \right) = \frac{24}{x^5}, \quad f^{(5)}(1) = \frac{24}{1^5} = 24$$f^{(5)}(x)=\frac{d}{dx}(-\frac{6}{x^4})=\frac{24}{x^5},f^{(5)}(1)=\frac{24}{1^5}=24$$

• First derivative: $f^{\prime }(x)=x^{-1}$$f^{\prime }(x)=x^{-1}$f^(‘)(x)=x^(-1)f'(x) = x^{-1}$f^{\prime }(x)=x^{-1}$
• Second derivative: $f^{″}(x)=-x^{-2}$$f^{″}(x)=-x^{-2}$f^(″)(x)=-x^(-2)f”(x) = -x^{-2}$f^{″}(x)=-x^{-2}$
• Third derivative: $f^{‴}(x)=2x^{-3}=2\cdot 1\cdot x^{-3}$$f^{‴}(x)=2x^{-3}=2\cdot 1\cdot x^{-3}$f^(‴)(x)=2x^(-3)=2*1*x^(-3)f”'(x) = 2x^{-3} = 2 \cdot 1 \cdot x^{-3}$f^{‴}(x)=2x^{-3}=2\cdot 1\cdot x^{-3}$
• Fourth derivative: $f^{(4)}(x)=-6x^{-4}=-3\cdot 2\cdot 1\cdot x^{-4}$$f^{(4)}(x)=-6x^{-4}=-3\cdot 2\cdot 1\cdot x^{-4}$f^((4))(x)=-6x^(-4)=-3*2*1*x^(-4)f^{(4)}(x) = -6x^{-4} = -3 \cdot 2 \cdot 1 \cdot x^{-4}$f^{(4)}(x)=-6x^{-4}=-3\cdot 2\cdot 1\cdot x^{-4}$
• Fifth derivative: $f^{(5)}(x)=24x^{-5}=4\cdot 3\cdot 2\cdot 1\cdot x^{-5}$$f^{(5)}(x)=24x^{-5}=4\cdot 3\cdot 2\cdot 1\cdot x^{-5}$f^((5))(x)=24x^(-5)=4*3*2*1*x^(-5)f^{(5)}(x) = 24x^{-5} = 4 \cdot 3 \cdot 2 \cdot 1 \cdot x^{-5}$f^{(5)}(x)=24x^{-5}=4\cdot 3\cdot 2\cdot 1\cdot x^{-5}$

Step 3: Construct the Taylor Series

• $f(1)=0$$f(1)=0$f(1)=0f(1) = 0$f(1)=0$
• $f^{\prime }(1)=1$$f^{\prime }(1)=1$f^(‘)(1)=1f'(1) = 1$f^{\prime }(1)=1$
• $f^{″}(1)=-1$$f^{″}(1)=-1$f^(″)(1)=-1f”(1) = -1$f^{″}(1)=-1$, so $\frac{f^{″}(1)}{2!}=\frac{-1}{2}=-\frac{1}{2}$$\frac{f^{″}(1)}{2!}=\frac{-1}{2}=-\frac{1}{2}$(f^(″)(1))/(2!)=(-1)/(2)=-(1)/(2)\frac{f”(1)}{2!} = \frac{-1}{2} = -\frac{1}{2}$\frac{f^{″}(1)}{2!}=\frac{-1}{2}=-\frac{1}{2}$
• $f^{‴}(1)=2$$f^{‴}(1)=2$f^(‴)(1)=2f”'(1) = 2$f^{‴}(1)=2$, so $\frac{f^{‴}(1)}{3!}=\frac{2}{6}=\frac{1}{3}$$\frac{f^{‴}(1)}{3!}=\frac{2}{6}=\frac{1}{3}$(f^(‴)(1))/(3!)=(2)/(6)=(1)/(3)\frac{f”'(1)}{3!} = \frac{2}{6} = \frac{1}{3}$\frac{f^{‴}(1)}{3!}=\frac{2}{6}=\frac{1}{3}$
• $f^{(4)}(1)=-6$$f^{(4)}(1)=-6$f^((4))(1)=-6f^{(4)}(1) = -6$f^{(4)}(1)=-6$, so $\frac{f^{(4)}(1)}{4!}=\frac{-6}{24}=-\frac{1}{4}$$\frac{f^{(4)}(1)}{4!}=\frac{-6}{24}=-\frac{1}{4}$(f^((4))(1))/(4!)=(-6)/(24)=-(1)/(4)\frac{f^{(4)}(1)}{4!} = \frac{-6}{24} = -\frac{1}{4}$\frac{f^{(4)}(1)}{4!}=\frac{-6}{24}=-\frac{1}{4}$
• $f^{(5)}(1)=24$$f^{(5)}(1)=24$f^((5))(1)=24f^{(5)}(1) = 24$f^{(5)}(1)=24$, so $\frac{f^{(5)}(1)}{5!}=\frac{24}{120}=\frac{1}{5}$$\frac{f^{(5)}(1)}{5!}=\frac{24}{120}=\frac{1}{5}$(f^((5))(1))/(5!)=(24)/(120)=(1)/(5)\frac{f^{(5)}(1)}{5!} = \frac{24}{120} = \frac{1}{5}$\frac{f^{(5)}(1)}{5!}=\frac{24}{120}=\frac{1}{5}$

Step 4: Evaluate $\ln (1.1)$$\ln (1.1)$ln(1.1)\ln(1.1)$\ln (1.1)$

• $n=1$$n=1$n=1n=1$n=1$: $\frac{(0.1{)}^1}{1}=0.1$$\frac{(0.1{)}^1}{1}=0.1$((0.1)^(1))/(1)=0.1\frac{(0.1)^1}{1} = 0.1$\frac{(0.1{)}^1}{1}=0.1$
• $n=2$$n=2$n=2n=2$n=2$: $-\frac{(0.1{)}^2}{2}=-\frac{0.01}{2}=-0.005$$-\frac{(0.1{)}^2}{2}=-\frac{0.01}{2}=-0.005$-((0.1)^(2))/(2)=-(0.01)/(2)=-0.005-\frac{(0.1)^2}{2} = -\frac{0.01}{2} = -0.005$-\frac{(0.1{)}^2}{2}=-\frac{0.01}{2}=-0.005$
• $n=3$$n=3$n=3n=3$n=3$: $\frac{(0.1{)}^3}{3}=\frac{0.001}{3}\approx 0.000333333$$\frac{(0.1{)}^3}{3}=\frac{0.001}{3}\approx 0.000333333$((0.1)^(3))/(3)=(0.001)/(3)~~0.000333333\frac{(0.1)^3}{3} = \frac{0.001}{3} \approx 0.000333333$\frac{(0.1{)}^3}{3}=\frac{0.001}{3}\approx 0.000333333$
• $n=4$$n=4$n=4n=4$n=4$: $-\frac{(0.1{)}^4}{4}=-\frac{0.0001}{4}=-0.000025$$-\frac{(0.1{)}^4}{4}=-\frac{0.0001}{4}=-0.000025$-((0.1)^(4))/(4)=-(0.0001)/(4)=-0.000025-\frac{(0.1)^4}{4} = -\frac{0.0001}{4} = -0.000025$-\frac{(0.1{)}^4}{4}=-\frac{0.0001}{4}=-0.000025$
• $n=5$$n=5$n=5n=5$n=5$: $\frac{(0.1{)}^5}{5}=\frac{0.00001}{5}=0.000002$$\frac{(0.1{)}^5}{5}=\frac{0.00001}{5}=0.000002$((0.1)^(5))/(5)=(0.00001)/(5)=0.000002\frac{(0.1)^5}{5} = \frac{0.00001}{5} = 0.000002$\frac{(0.1{)}^5}{5}=\frac{0.00001}{5}=0.000002$
• $n=6$$n=6$n=6n=6$n=6$: $-\frac{(0.1{)}^6}{6}=-\frac{0.000001}{6}\approx -0.00000016667$$-\frac{(0.1{)}^6}{6}=-\frac{0.000001}{6}\approx -0.00000016667$-((0.1)^(6))/(6)=-(0.000001)/(6)~~-0.00000016667-\frac{(0.1)^6}{6} = -\frac{0.000001}{6} \approx -0.00000016667$-\frac{(0.1{)}^6}{6}=-\frac{0.000001}{6}\approx -0.00000016667$

• First term: $0.1$$0.1$0.10.1$0.1$
• First two terms: $0.1-0.005=0.095$$0.1-0.005=0.095$0.1-0.005=0.0950.1 – 0.005 = 0.095$0.1-0.005=0.095$
• First three terms: $0.095+0.000333333\approx 0.095333333$$0.095+0.000333333\approx 0.095333333$0.095+0.000333333~~0.0953333330.095 + 0.000333333 \approx 0.095333333$0.095+0.000333333\approx 0.095333333$
• First four terms: $0.095333333-0.000025=0.095308333$$0.095333333-0.000025=0.095308333$0.095333333-0.000025=0.0953083330.095333333 – 0.000025 = 0.095308333$0.095333333-0.000025=0.095308333$
• First five terms: $0.095308333+0.000002=0.095310333$$0.095308333+0.000002=0.095310333$0.095308333+0.000002=0.0953103330.095308333 + 0.000002 = 0.095310333$0.095308333+0.000002=0.095310333$
• First six terms: $0.095310333-0.00000016667\approx 0.09531016667$$0.095310333-0.00000016667\approx 0.09531016667$0.095310333-0.00000016667~~0.095310166670.095310333 – 0.00000016667 \approx 0.09531016667$0.095310333-0.00000016667\approx 0.09531016667$

Step 5: Determine Terms Needed for Four Decimal Places

Step 6: Verify the Result

Final Answer

Find the equation of the right circular cylinder which passes through the circle $x^2+y^2+z^2=9,x-y+z=3$$x^2+y^2+z^2=9,x-y+z=3$x^(2)+y^(2)+z^(2)=9,x-y+z=3x^{2}+y^{2}+z^{2}=9,\; x-y+z=3$x^2+y^2+z^2=9,x-y+z=3$.

Answer:

Given:

Step 1: Find the Center and Radius of the Given Circle

1. Center of the Sphere:
   The sphere $x^2+y^2+z^2=9$$x^2+y^2+z^2=9$x^(2)+y^(2)+z^(2)=9x^{2} + y^{2} + z^{2} = 9$x^2+y^2+z^2=9$ has its center at the origin $(0,0,0)$$(0,0,0)$(0,0,0)(0, 0, 0)$(0,0,0)$ and radius $3$$3$33$3$.
2. Distance from the Center to the Plane:
   The plane is $x-y+z=3$$x-y+z=3$x-y+z=3x – y + z = 3$x-y+z=3$. The distance $d$$d$dd$d$ from the center $(0,0,0)$$(0,0,0)$(0,0,0)(0, 0, 0)$(0,0,0)$ to the plane is:
   $$d=\frac{|0-0+0-3|}{\sqrt{1^2+(-1{)}^2+1^2}}=\frac{3}{\sqrt{3}}=\sqrt{3}.$$$$d=\frac{|0-0+0-3|}{\sqrt{1^2+(-1{)}^2+1^2}}=\frac{3}{\sqrt{3}}=\sqrt{3}.$$d=(|0-0+0-3|)/(sqrt(1^(2)+(-1)^(2)+1^(2)))=(3)/(sqrt3)=sqrt3.d = \frac{|0 – 0 + 0 – 3|}{\sqrt{1^2 + (-1)^2 + 1^2}} = \frac{3}{\sqrt{3}} = \sqrt{3}.$$d=\frac{|0-0+0-3|}{\sqrt{1^2+(-1{)}^2+1^2}}=\frac{3}{\sqrt{3}}=\sqrt{3}.$$
3. Radius of the Circle:
   Using the Pythagorean theorem for the sphere and plane intersection:
   $$\text{Radius of the circle}=\sqrt{R^2-d^2}=\sqrt{9-3}=\sqrt{6}.$$$$\text{Radius of the circle}=\sqrt{R^2-d^2}=\sqrt{9-3}=\sqrt{6}.$$“Radius of the circle”=sqrt(R^(2)-d^(2))=sqrt(9-3)=sqrt6.\text{Radius of the circle} = \sqrt{R^2 – d^2} = \sqrt{9 – 3} = \sqrt{6}.$$\text{Radius of the circle}=\sqrt{R^2-d^2}=\sqrt{9-3}=\sqrt{6}.$$
4. Center of the Circle:
   The center of the circle lies along the normal from the sphere’s center to the plane. The normal vector to the plane $x-y+z=3$$x-y+z=3$x-y+z=3x – y + z = 3$x-y+z=3$ is $\mathbf{n}=(1,-1,1)$$\mathbf{n}=(1,-1,1)$n=(1,-1,1)\mathbf{n} = (1, -1, 1)$\mathbf{n}=(1,-1,1)$.
   The parametric equations for the line from the origin in the direction of $\mathbf{n}$$\mathbf{n}$n\mathbf{n}$\mathbf{n}$ are:
   $$x=t,y=-t,z=t.$$$$x=t,y=-t,z=t.$$x=t,quad y=-t,quad z=t.x = t, \quad y = -t, \quad z = t.$$x=t,y=-t,z=t.$$
   Substituting into the plane equation to find the foot of the perpendicular:
   $$t-(-t)+t=3\Rightarrow 3t=3\Rightarrow t=1.$$$$t-(-t)+t=3\Rightarrow 3t=3\Rightarrow t=1.$$t-(-t)+t=3=>3t=3=>t=1.t – (-t) + t = 3 \Rightarrow 3t = 3 \Rightarrow t = 1.$$t-(-t)+t=3\Rightarrow 3t=3\Rightarrow t=1.$$
   So, the center of the circle is $(1,-1,1)$$(1,-1,1)$(1,-1,1)(1, -1, 1)$(1,-1,1)$.

Step 2: Determine the Axis of the Cylinder

Step 3: Find the Equation of the Cylinder

1. Distance from a Point $(x,y,z)$$(x,y,z)$(x,y,z)(x, y, z)$(x,y,z)$ to the Axis:
   The distance $D$$D$DD$D$ from a point $(x,y,z)$$(x,y,z)$(x,y,z)(x, y, z)$(x,y,z)$ to the line (axis) passing through $(1,-1,1)$$(1,-1,1)$(1,-1,1)(1, -1, 1)$(1,-1,1)$ in the direction $\mathbf{n}=(1,-1,1)$$\mathbf{n}=(1,-1,1)$n=(1,-1,1)\mathbf{n} = (1, -1, 1)$\mathbf{n}=(1,-1,1)$ is given by:
   $$D=\frac{\Vert \mathbf{n}\times \mathbf{v}\Vert }{\Vert \mathbf{n}\Vert },$$$$D=\frac{\Vert \mathbf{n}\times \mathbf{v}\Vert }{\Vert \mathbf{n}\Vert },$$D=(||nxxv||)/(||n||),D = \frac{\| \mathbf{n} \times \mathbf{v} \|}{\| \mathbf{n} \|},$$D=\frac{\Vert \mathbf{n}\times \mathbf{v}\Vert }{\Vert \mathbf{n}\Vert },$$
   where $\mathbf{v}=(x-1,y+1,z-1)$$\mathbf{v}=(x-1,y+1,z-1)$v=(x-1,y+1,z-1)\mathbf{v} = (x – 1, y + 1, z – 1)$\mathbf{v}=(x-1,y+1,z-1)$.
   Compute the cross product:
   $$\mathbf{n}\times \mathbf{v}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& -1& 1\\ x-1& y+1& z-1\end{array}\right|=\mathbf{i}((-1)(z-1)-1(y+1))-\mathbf{j}(1(z-1)-1(x-1))+\mathbf{k}(1(y+1)-(-1)(x-1)).$$$$\mathbf{n}\times \mathbf{v}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& -1& 1\\ x-1& y+1& z-1\end{array}\right|=\mathbf{i}\left((-1)(z-1)-1(y+1)\right)-\mathbf{j}\left(1(z-1)-1(x-1)\right)+\mathbf{k}\left(1(y+1)-(-1)(x-1)\right).$$nxxv=|[i,j,k],[1,-1,1],[x-1,y+1,z-1]|=i((-1)(z-1)-1(y+1))-j(1(z-1)-1(x-1))+k(1(y+1)-(-1)(x-1)).\mathbf{n} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ x – 1 & y + 1 & z – 1 \end{vmatrix} = \mathbf{i} \left( (-1)(z – 1) – 1(y + 1) \right) – \mathbf{j} \left( 1(z – 1) – 1(x – 1) \right) + \mathbf{k} \left( 1(y + 1) – (-1)(x – 1) \right).$$\mathbf{n}\times \mathbf{v}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& -1& 1\\ x-1& y+1& z-1\end{array}\right|=\mathbf{i}((-1)(z-1)-1(y+1))-\mathbf{j}(1(z-1)-1(x-1))+\mathbf{k}(1(y+1)-(-1)(x-1)).$$
   Simplifying:
   $$\mathbf{n}\times \mathbf{v}=(-(z-1)-(y+1))\mathbf{i}-((z-1)-(x-1))\mathbf{j}+((y+1)+(x-1))\mathbf{k}.$$$$\mathbf{n}\times \mathbf{v}=\left(-(z-1)-(y+1)\right)\mathbf{i}-\left((z-1)-(x-1)\right)\mathbf{j}+\left((y+1)+(x-1)\right)\mathbf{k}.$$nxxv=(-(z-1)-(y+1))i-((z-1)-(x-1))j+((y+1)+(x-1))k.\mathbf{n} \times \mathbf{v} = \left( – (z – 1) – (y + 1) \right) \mathbf{i} – \left( (z – 1) – (x – 1) \right) \mathbf{j} + \left( (y + 1) + (x – 1) \right) \mathbf{k}.$$\mathbf{n}\times \mathbf{v}=(-(z-1)-(y+1))\mathbf{i}-((z-1)-(x-1))\mathbf{j}+((y+1)+(x-1))\mathbf{k}.$$
   $$=(-y-z)\mathbf{i}-(z-x)\mathbf{j}+(x+y)\mathbf{k}.$$$$=(-y-z)\mathbf{i}-(z-x)\mathbf{j}+(x+y)\mathbf{k}.$$=(-y-z)i-(z-x)j+(x+y)k.= (-y – z) \mathbf{i} – (z – x) \mathbf{j} + (x + y) \mathbf{k}.$$=(-y-z)\mathbf{i}-(z-x)\mathbf{j}+(x+y)\mathbf{k}.$$
   The magnitude is:
   $$\Vert \mathbf{n}\times \mathbf{v}\Vert =\sqrt{(-y-z{)}^2+(-z+x{)}^2+(x+y{)}^2}.$$$$\Vert \mathbf{n}\times \mathbf{v}\Vert =\sqrt{(-y-z{)}^2+(-z+x{)}^2+(x+y{)}^2}.$$||nxxv||=sqrt((-y-z)^(2)+(-z+x)^(2)+(x+y)^(2)).\| \mathbf{n} \times \mathbf{v} \| = \sqrt{(-y – z)^2 + (-z + x)^2 + (x + y)^2}.$$\Vert \mathbf{n}\times \mathbf{v}\Vert =\sqrt{(-y-z{)}^2+(-z+x{)}^2+(x+y{)}^2}.$$
   $$=\sqrt{(y+z{)}^2+(z-x{)}^2+(x+y{)}^2}.$$$$=\sqrt{(y+z{)}^2+(z-x{)}^2+(x+y{)}^2}.$$=sqrt((y+z)^(2)+(z-x)^(2)+(x+y)^(2)).= \sqrt{(y + z)^2 + (z – x)^2 + (x + y)^2}.$$=\sqrt{(y+z{)}^2+(z-x{)}^2+(x+y{)}^2}.$$
   The magnitude of $\mathbf{n}$$\mathbf{n}$n\mathbf{n}$\mathbf{n}$ is:
   $$\Vert \mathbf{n}\Vert =\sqrt{1^2+(-1{)}^2+1^2}=\sqrt{3}.$$$$\Vert \mathbf{n}\Vert =\sqrt{1^2+(-1{)}^2+1^2}=\sqrt{3}.$$||n||=sqrt(1^(2)+(-1)^(2)+1^(2))=sqrt3.\| \mathbf{n} \| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}.$$\Vert \mathbf{n}\Vert =\sqrt{1^2+(-1{)}^2+1^2}=\sqrt{3}.$$
   Therefore:
   $$D=\frac{\sqrt{(y+z{)}^2+(z-x{)}^2+(x+y{)}^2}}{\sqrt{3}}.$$$$D=\frac{\sqrt{(y+z{)}^2+(z-x{)}^2+(x+y{)}^2}}{\sqrt{3}}.$$D=(sqrt((y+z)^(2)+(z-x)^(2)+(x+y)^(2)))/(sqrt3).D = \frac{\sqrt{(y + z)^2 + (z – x)^2 + (x + y)^2}}{\sqrt{3}}.$$D=\frac{\sqrt{(y+z{)}^2+(z-x{)}^2+(x+y{)}^2}}{\sqrt{3}}.$$
2. Equation of the Cylinder:
   The cylinder consists of all points $(x,y,z)$$(x,y,z)$(x,y,z)(x, y, z)$(x,y,z)$ such that the distance $D$$D$DD$D$ to the axis is equal to the radius $\sqrt{6}$$\sqrt{6}$sqrt6\sqrt{6}$\sqrt{6}$:
   $$\frac{\sqrt{(y+z{)}^2+(z-x{)}^2+(x+y{)}^2}}{\sqrt{3}}=\sqrt{6}.$$$$\frac{\sqrt{(y+z{)}^2+(z-x{)}^2+(x+y{)}^2}}{\sqrt{3}}=\sqrt{6}.$$(sqrt((y+z)^(2)+(z-x)^(2)+(x+y)^(2)))/(sqrt3)=sqrt6.\frac{\sqrt{(y + z)^2 + (z – x)^2 + (x + y)^2}}{\sqrt{3}} = \sqrt{6}.$$\frac{\sqrt{(y+z{)}^2+(z-x{)}^2+(x+y{)}^2}}{\sqrt{3}}=\sqrt{6}.$$
   Squaring both sides:
   $$\frac{(y+z{)}^2+(z-x{)}^2+(x+y{)}^2}{3}=6.$$$$\frac{(y+z{)}^2+(z-x{)}^2+(x+y{)}^2}{3}=6.$$((y+z)^(2)+(z-x)^(2)+(x+y)^(2))/(3)=6.\frac{(y + z)^2 + (z – x)^2 + (x + y)^2}{3} = 6.$$\frac{(y+z{)}^2+(z-x{)}^2+(x+y{)}^2}{3}=6.$$
   Multiply by 3:
   $$(y+z{)}^2+(z-x{)}^2+(x+y{)}^2=18.$$$$(y+z{)}^2+(z-x{)}^2+(x+y{)}^2=18.$$(y+z)^(2)+(z-x)^(2)+(x+y)^(2)=18.(y + z)^2 + (z – x)^2 + (x + y)^2 = 18.$$(y+z{)}^2+(z-x{)}^2+(x+y{)}^2=18.$$
   Expand each square:
   $$(y^2+2yz+z^2)+(z^2-2xz+x^2)+(x^2+2xy+y^2)=18.$$$$(y^2+2yz+z^2)+(z^2-2xz+x^2)+(x^2+2xy+y^2)=18.$$(y^(2)+2yz+z^(2))+(z^(2)-2xz+x^(2))+(x^(2)+2xy+y^(2))=18.(y^2 + 2yz + z^2) + (z^2 – 2xz + x^2) + (x^2 + 2xy + y^2) = 18.$$(y^2+2yz+z^2)+(z^2-2xz+x^2)+(x^2+2xy+y^2)=18.$$
   Combine like terms:
   $$2x^2+2y^2+2z^2+2xy+2yz-2xz=18.$$$$2x^2+2y^2+2z^2+2xy+2yz-2xz=18.$$2x^(2)+2y^(2)+2z^(2)+2xy+2yz-2xz=18.2x^2 + 2y^2 + 2z^2 + 2xy + 2yz – 2xz = 18.$$2x^2+2y^2+2z^2+2xy+2yz-2xz=18.$$
   Divide by 2:
   $$x^2+y^2+z^2+xy+yz-xz=9.$$$$x^2+y^2+z^2+xy+yz-xz=9.$$x^(2)+y^(2)+z^(2)+xy+yz-xz=9.x^2 + y^2 + z^2 + xy + yz – xz = 9.$$x^2+y^2+z^2+xy+yz-xz=9.$$

Final Answer:

Consider a linear operator $T$$T$TT$T$ on ${\mathbb{R}}^3$${\mathbb{R}}^3$R^(3)\mathbb{R}^{3}${\mathbb{R}}^3$ over $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$ defined by $T(x,y,z)=(2x,4x-y,2x+3y-z)$$T(x,y,z)=(2x,4x-y,2x+3y-z)$T(x,y,z)=(2x,4x-y,2x+3y-z)T(x, y, z)=(2x,\,4x-y,\,2x+3y-z)$T(x,y,z)=(2x,4x-y,2x+3y-z)$. Is $T$$T$TT$T$ invertible? If yes, justify your answer and find $T^{-1}$$T^{-1}$T^(-1)T^{-1}$T^{-1}$.

Answer:

1. Definition of $T$$T$TT$T$:

2. Matrix Representation of $T$$T$TT$T$:

3. Checking Invertibility:

4. Finding the Inverse $T^{-1}$$T^{-1}$T^(-1)T^{-1}$T^{-1}$:

5. Expression for $T^{-1}$$T^{-1}$T^(-1)T^{-1}$T^{-1}$:

Verification:

Final Answer:

If $u={\displaystyle \frac{x+y}{1-xy}}$$u={\displaystyle \frac{x+y}{1-xy}}$u=(x+y)/(1-xy)u=\dfrac{x+y}{1-xy}$u={\displaystyle \frac{x+y}{1-xy}}$ and $v={\tan }^{-1}x+{\tan }^{-1}y$$v={\tan }^{-1}x+{\tan }^{-1}y$v=tan^(-1)x+tan^(-1)yv=\tan^{-1}x+\tan^{-1}y$v={\tan }^{-1}x+{\tan }^{-1}y$, then find $\partial (u,v)/\partial (x,y)$$\partial (u,v)/\partial (x,y)$del(u,v)//del(x,y)\partial(u,v)/\partial(x,y)$\partial (u,v)/\partial (x,y)$. Are $u$$u$uu$u$ and $v$$v$vv$v$ functionally related? If yes, find the relationship.

Answer:

Given:

Step 1: Compute the Jacobian $\frac{\partial (u,v)}{\partial (x,y)}$$\frac{\partial (u,v)}{\partial (x,y)}$(del(u,v))/(del(x,y))\frac{\partial(u, v)}{\partial(x, y)}$\frac{\partial (u,v)}{\partial (x,y)}$

Partial Derivatives of $u$$u$uu$u$:

Partial Derivatives of $v$$v$vv$v$:

Compute the Jacobian Determinant:

Step 2: Determine Functional Relationship

Find the Relationship:

Final Answer:

Find the image of the line $x=3-6t,y=2t,z=3+2t$$x=3-6t,y=2t,z=3+2t$x=3-6t,y=2t,z=3+2tx=3-6t,\; y=2t,\; z=3+2t$x=3-6t,y=2t,z=3+2t$ in the plane $3x+4y-5z+26=0$$3x+4y-5z+26=0$3x+4y-5z+26=03x+4y-5z+26=0$3x+4y-5z+26=0$.

Answer:

Step 1: Understand the Line and Plane

Step 2: Reflection of a Point

• Line from $P(3,0,3)$$P(3,0,3)$P(3,0,3)P(3, 0, 3)$P(3,0,3)$ to the plane: The line through $P$$P$PP$P$ parallel to the normal $\overrightarrow{n}=(3,4,-5)$$\overrightarrow{n}=(3,4,-5)$vec(n)=(3,4,-5)\vec{n} = (3, 4, -5)$\overrightarrow{n}=(3,4,-5)$ is:
  $$x=3+3s,y=0+4s,z=3-5s$$$$x=3+3s,y=0+4s,z=3-5s$$x=3+3s,quad y=0+4s,quad z=3-5sx = 3 + 3s, \quad y = 0 + 4s, \quad z = 3 – 5s$$x=3+3s,y=0+4s,z=3-5s$$
• Intersection with the plane: Substitute into the plane equation:
  $$3(3+3s)+4(4s)-5(3-5s)+26=0$$$$3(3+3s)+4(4s)-5(3-5s)+26=0$$3(3+3s)+4(4s)-5(3-5s)+26=03(3 + 3s) + 4(4s) – 5(3 – 5s) + 26 = 0$$3(3+3s)+4(4s)-5(3-5s)+26=0$$
  $$9+9s+16s-15+25s+26=0$$$$9+9s+16s-15+25s+26=0$$9+9s+16 s-15+25 s+26=09 + 9s + 16s – 15 + 25s + 26 = 0$$9+9s+16s-15+25s+26=0$$
  $$50s+20=0⟹s=-\frac{20}{50}=-\frac{2}{5}$$$$50s+20=0⟹s=-\frac{20}{50}=-\frac{2}{5}$$50 s+20=0Longrightarrows=-(20)/(50)=-(2)/(5)50s + 20 = 0 \implies s = -\frac{20}{50} = -\frac{2}{5}$$50s+20=0⟹s=-\frac{20}{50}=-\frac{2}{5}$$
• Foot of the perpendicular (point $Q$$Q$QQ$Q$):
  $$x=3+3\cdot (-\frac{2}{5})=3-\frac{6}{5}=\frac{9}{5}$$$$x=3+3\cdot \left(-\frac{2}{5}\right)=3-\frac{6}{5}=\frac{9}{5}$$x=3+3*(-(2)/(5))=3-(6)/(5)=(9)/(5)x = 3 + 3 \cdot \left(-\frac{2}{5}\right) = 3 – \frac{6}{5} = \frac{9}{5}$$x=3+3\cdot (-\frac{2}{5})=3-\frac{6}{5}=\frac{9}{5}$$
  $$y=4\cdot (-\frac{2}{5})=-\frac{8}{5}$$$$y=4\cdot \left(-\frac{2}{5}\right)=-\frac{8}{5}$$y=4*(-(2)/(5))=-(8)/(5)y = 4 \cdot \left(-\frac{2}{5}\right) = -\frac{8}{5}$$y=4\cdot (-\frac{2}{5})=-\frac{8}{5}$$
  $$z=3-5\cdot (-\frac{2}{5})=3+2=5$$$$z=3-5\cdot \left(-\frac{2}{5}\right)=3+2=5$$z=3-5*(-(2)/(5))=3+2=5z = 3 – 5 \cdot \left(-\frac{2}{5}\right) = 3 + 2 = 5$$z=3-5\cdot (-\frac{2}{5})=3+2=5$$
  So, $Q=(\frac{9}{5},-\frac{8}{5},5)$$Q=\left(\frac{9}{5},-\frac{8}{5},5\right)$Q=((9)/(5),-(8)/(5),5)Q = \left( \frac{9}{5}, -\frac{8}{5}, 5 \right)$Q=(\frac{9}{5},-\frac{8}{5},5)$.
• Find the reflected point $P^{\prime }$$P^{\prime }$P^(‘)P’$P^{\prime }$: The foot $Q$$Q$QQ$Q$ is the midpoint of $P(3,0,3)$$P(3,0,3)$P(3,0,3)P(3, 0, 3)$P(3,0,3)$ and $P^{\prime }(x^{\prime },y^{\prime },z^{\prime })$$P^{\prime }(x^{\prime },y^{\prime },z^{\prime })$P^(‘)(x^(‘),y^(‘),z^(‘))P'(x’, y’, z’)$P^{\prime }(x^{\prime },y^{\prime },z^{\prime })$. Using the midpoint formula:
  $$\frac{3+x^{\prime }}{2}=\frac{9}{5}⟹3+x^{\prime }=\frac{18}{5}⟹x^{\prime }=\frac{18}{5}-3=\frac{18}{5}-\frac{15}{5}=\frac{3}{5}$$$$\frac{3+x^{\prime }}{2}=\frac{9}{5}⟹3+x^{\prime }=\frac{18}{5}⟹x^{\prime }=\frac{18}{5}-3=\frac{18}{5}-\frac{15}{5}=\frac{3}{5}$$(3+x^(‘))/(2)=(9)/(5)Longrightarrow3+x^(‘)=(18)/(5)Longrightarrowx^(‘)=(18)/(5)-3=(18)/(5)-(15)/(5)=(3)/(5)\frac{3 + x’}{2} = \frac{9}{5} \implies 3 + x’ = \frac{18}{5} \implies x’ = \frac{18}{5} – 3 = \frac{18}{5} – \frac{15}{5} = \frac{3}{5}$$\frac{3+x^{\prime }}{2}=\frac{9}{5}⟹3+x^{\prime }=\frac{18}{5}⟹x^{\prime }=\frac{18}{5}-3=\frac{18}{5}-\frac{15}{5}=\frac{3}{5}$$
  $$\frac{0+y^{\prime }}{2}=-\frac{8}{5}⟹y^{\prime }=-\frac{16}{5}$$$$\frac{0+y^{\prime }}{2}=-\frac{8}{5}⟹y^{\prime }=-\frac{16}{5}$$(0+y^(‘))/(2)=-(8)/(5)Longrightarrowy^(‘)=-(16)/(5)\frac{0 + y’}{2} = -\frac{8}{5} \implies y’ = -\frac{16}{5}$$\frac{0+y^{\prime }}{2}=-\frac{8}{5}⟹y^{\prime }=-\frac{16}{5}$$
  $$\frac{3+z^{\prime }}{2}=5⟹3+z^{\prime }=10⟹z^{\prime }=7$$$$\frac{3+z^{\prime }}{2}=5⟹3+z^{\prime }=10⟹z^{\prime }=7$$(3+z^(‘))/(2)=5Longrightarrow3+z^(‘)=10Longrightarrowz^(‘)=7\frac{3 + z’}{2} = 5 \implies 3 + z’ = 10 \implies z’ = 7$$\frac{3+z^{\prime }}{2}=5⟹3+z^{\prime }=10⟹z^{\prime }=7$$
  So, the reflected point is:
  $$P^{\prime }=(\frac{3}{5},-\frac{16}{5},7)$$$$P^{\prime }=\left(\frac{3}{5},-\frac{16}{5},7\right)$$P^(‘)=((3)/(5),-(16)/(5),7)P’ = \left( \frac{3}{5}, -\frac{16}{5}, 7 \right)$$P^{\prime }=(\frac{3}{5},-\frac{16}{5},7)$$

Step 3: Direction of the Reflected Line

• Compute $\overrightarrow{v}\cdot \overrightarrow{n}$$\overrightarrow{v}\cdot \overrightarrow{n}$vec(v)* vec(n)\vec{v} \cdot \vec{n}$\overrightarrow{v}\cdot \overrightarrow{n}$:$$\overrightarrow{d}\cdot \overrightarrow{n}=(-6)\cdot 3+2\cdot 4+2\cdot (-5)=-18+8-10=-20$$$$\overrightarrow{d}\cdot \overrightarrow{n}=(-6)\cdot 3+2\cdot 4+2\cdot (-5)=-18+8-10=-20$$vec(d)* vec(n)=(-6)*3+2*4+2*(-5)=-18+8-10=-20\vec{d} \cdot \vec{n} = (-6) \cdot 3 + 2 \cdot 4 + 2 \cdot (-5) = -18 + 8 – 10 = -20$$\overrightarrow{d}\cdot \overrightarrow{n}=(-6)\cdot 3+2\cdot 4+2\cdot (-5)=-18+8-10=-20$$
• Compute $\Vert \overrightarrow{n}{\Vert }^2$$\Vert \overrightarrow{n}{\Vert }^2$|| vec(n)||^(2)\|\vec{n}\|^2$\Vert \overrightarrow{n}{\Vert }^2$:$$\Vert \overrightarrow{n}{\Vert }^2=3^2+4^2+(-5{)}^2=9+16+25=50$$$$\Vert \overrightarrow{n}{\Vert }^2=3^2+4^2+(-5{)}^2=9+16+25=50$$|| vec(n)||^(2)=3^(2)+4^(2)+(-5)^(2)=9+16+25=50\|\vec{n}\|^2 = 3^2 + 4^2 + (-5)^2 = 9 + 16 + 25 = 50$$\Vert \overrightarrow{n}{\Vert }^2=3^2+4^2+(-5{)}^2=9+16+25=50$$
• Compute the projection term:$$2\frac{\overrightarrow{d}\cdot \overrightarrow{n}}{\Vert \overrightarrow{n}{\Vert }^2}\overrightarrow{n}=2\cdot \frac{-20}{50}\cdot (3,4,-5)=-\frac{40}{50}\cdot (3,4,-5)=-\frac{4}{5}\cdot (3,4,-5)=(-\frac{12}{5},-\frac{16}{5},4)$$$$2\frac{\overrightarrow{d}\cdot \overrightarrow{n}}{\Vert \overrightarrow{n}{\Vert }^2}\overrightarrow{n}=2\cdot \frac{-20}{50}\cdot (3,4,-5)=-\frac{40}{50}\cdot (3,4,-5)=-\frac{4}{5}\cdot (3,4,-5)=\left(-\frac{12}{5},-\frac{16}{5},4\right)$$2(( vec(d))*( vec(n)))/(||( vec(n))||^(2)) vec(n)=2*(-20)/(50)*(3,4,-5)=-(40)/(50)*(3,4,-5)=-(4)/(5)*(3,4,-5)=(-(12)/(5),-(16)/(5),4)2 \frac{\vec{d} \cdot \vec{n}}{\|\vec{n}\|^2} \vec{n} = 2 \cdot \frac{-20}{50} \cdot (3, 4, -5) = -\frac{40}{50} \cdot (3, 4, -5) = -\frac{4}{5} \cdot (3, 4, -5) = \left( -\frac{12}{5}, -\frac{16}{5}, 4 \right)$$2\frac{\overrightarrow{d}\cdot \overrightarrow{n}}{\Vert \overrightarrow{n}{\Vert }^2}\overrightarrow{n}=2\cdot \frac{-20}{50}\cdot (3,4,-5)=-\frac{40}{50}\cdot (3,4,-5)=-\frac{4}{5}\cdot (3,4,-5)=(-\frac{12}{5},-\frac{16}{5},4)$$
• Reflected direction vector:$${\overrightarrow{d}}^{\prime }=(-6,2,2)-(-\frac{12}{5},-\frac{16}{5},4)=(-6+\frac{12}{5},2+\frac{16}{5},2-4)$$$${\overrightarrow{d}}^{\prime }=(-6,2,2)-\left(-\frac{12}{5},-\frac{16}{5},4\right)=\left(-6+\frac{12}{5},2+\frac{16}{5},2-4\right)$$vec(d)^(‘)=(-6,2,2)-(-(12)/(5),-(16)/(5),4)=(-6+(12)/(5),2+(16)/(5),2-4)\vec{d}’ = (-6, 2, 2) – \left( -\frac{12}{5}, -\frac{16}{5}, 4 \right) = \left( -6 + \frac{12}{5}, 2 + \frac{16}{5}, 2 – 4 \right)$${\overrightarrow{d}}^{\prime }=(-6,2,2)-(-\frac{12}{5},-\frac{16}{5},4)=(-6+\frac{12}{5},2+\frac{16}{5},2-4)$$$$=(-\frac{30}{5}+\frac{12}{5},\frac{10}{5}+\frac{16}{5},-2)=(-\frac{18}{5},\frac{26}{5},-2)$$$$=\left(-\frac{30}{5}+\frac{12}{5},\frac{10}{5}+\frac{16}{5},-2\right)=\left(-\frac{18}{5},\frac{26}{5},-2\right)$$=(-(30)/(5)+(12)/(5),(10)/(5)+(16)/(5),-2)=(-(18)/(5),(26)/(5),-2)= \left( -\frac{30}{5} + \frac{12}{5}, \frac{10}{5} + \frac{16}{5}, -2 \right) = \left( -\frac{18}{5}, \frac{26}{5}, -2 \right)$$=(-\frac{30}{5}+\frac{12}{5},\frac{10}{5}+\frac{16}{5},-2)=(-\frac{18}{5},\frac{26}{5},-2)$$

Step 4: Parametric Equations of the Reflected Line

Step 5: Verify the Reflection

• Check if $P^{\prime }$$P^{\prime }$P^(‘)P’$P^{\prime }$ is the reflection: The midpoint $Q$$Q$QQ$Q$ is correct, and the vector from $Q$$Q$QQ$Q$ to $P^{\prime }$$P^{\prime }$P^(‘)P’$P^{\prime }$ should be opposite to $Q$$Q$QQ$Q$ to $P$$P$PP$P$, adjusted by the normal. This was computed correctly.
• Check if the reflected line is consistent: Reflect another point, e.g., at $t=1$$t=1$t=1t = 1$t=1$:$$P_1=(3-6,2,3+2)=(-3,2,5)$$$$P_1=(3-6,2,3+2)=(-3,2,5)$$P_(1)=(3-6,2,3+2)=(-3,2,5)P_1 = (3 – 6, 2, 3 + 2) = (-3, 2, 5)$$P_1=(3-6,2,3+2)=(-3,2,5)$$This computation is complex, so instead, verify the direction vector by ensuring the reflected line is consistent with the plane’s geometry. The reflected direction should satisfy symmetry across the plane, which we’ve computed.

Final Answer

Let $V=M_{2\times 2}(\mathbb{R})$$V=M_{2\times 2}(\mathbb{R})$V=M_(2xx2)(R)V=M_{2 \times 2}(\mathbb{R})$V=M_{2\times 2}(\mathbb{R})$ denote a vector space over the field of real numbers. Find the matrix of the linear mapping $\varphi :V\to V$$\varphi :V\to V$phi:V rarr V\phi: V \rightarrow V$\varphi :V\to V$ given by

Answer:

Step 1: Understand the Vector Space and Basis

Step 2: Compute the Matrix of $\varphi$$\varphi$phi\phi$\varphi$

• For $E_{11}=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$$E_{11}=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$E_(11)=([1,0],[0,0])E_{11} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$E_{11}=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$:
  $$\varphi (E_{11})=\left(\begin{array}{cc}1& 2\\ 3& -1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}1\cdot 1+2\cdot 0& 1\cdot 0+2\cdot 0\\ 3\cdot 1+(-1)\cdot 0& 3\cdot 0+(-1)\cdot 0\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 3& 0\end{array}\right)$$$$\varphi (E_{11})=\left(\begin{array}{cc}1& 2\\ 3& -1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}1\cdot 1+2\cdot 0& 1\cdot 0+2\cdot 0\\ 3\cdot 1+(-1)\cdot 0& 3\cdot 0+(-1)\cdot 0\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 3& 0\end{array}\right)$$phi(E_(11))=([1,2],[3,-1])([1,0],[0,0])=([1*1+2*0,1*0+2*0],[3*1+(-1)*0,3*0+(-1)*0])=([1,0],[3,0])\phi(E_{11}) = \begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 \cdot 1 + 2 \cdot 0 & 1 \cdot 0 + 2 \cdot 0 \\ 3 \cdot 1 + (-1) \cdot 0 & 3 \cdot 0 + (-1) \cdot 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 3 & 0 \end{pmatrix}$$\varphi (E_{11})=\left(\begin{array}{cc}1& 2\\ 3& -1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}1\cdot 1+2\cdot 0& 1\cdot 0+2\cdot 0\\ 3\cdot 1+(-1)\cdot 0& 3\cdot 0+(-1)\cdot 0\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 3& 0\end{array}\right)$$
  Express in the basis:
  $$\left(\begin{array}{cc}1& 0\\ 3& 0\end{array}\right)=1\cdot E_{11}+0\cdot E_{12}+3\cdot E_{21}+0\cdot E_{22}$$$$\left(\begin{array}{cc}1& 0\\ 3& 0\end{array}\right)=1\cdot E_{11}+0\cdot E_{12}+3\cdot E_{21}+0\cdot E_{22}$$([1,0],[3,0])=1*E_(11)+0*E_(12)+3*E_(21)+0*E_(22)\begin{pmatrix} 1 & 0 \\ 3 & 0 \end{pmatrix} = 1 \cdot E_{11} + 0 \cdot E_{12} + 3 \cdot E_{21} + 0 \cdot E_{22}$$\left(\begin{array}{cc}1& 0\\ 3& 0\end{array}\right)=1\cdot E_{11}+0\cdot E_{12}+3\cdot E_{21}+0\cdot E_{22}$$
  Column 1: $(1,0,3,0)$$(1,0,3,0)$(1,0,3,0)(1, 0, 3, 0)$(1,0,3,0)$.
• For $E_{12}=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$$E_{12}=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$E_(12)=([0,1],[0,0])E_{12} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$E_{12}=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$:
  $$\varphi (E_{12})=\left(\begin{array}{cc}1& 2\\ 3& -1\end{array}\right)\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}1\cdot 0+2\cdot 0& 1\cdot 1+2\cdot 0\\ 3\cdot 0+(-1)\cdot 0& 3\cdot 1+(-1)\cdot 0\end{array}\right)=\left(\begin{array}{cc}0& 1\\ 0& 3\end{array}\right)$$$$\varphi (E_{12})=\left(\begin{array}{cc}1& 2\\ 3& -1\end{array}\right)\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}1\cdot 0+2\cdot 0& 1\cdot 1+2\cdot 0\\ 3\cdot 0+(-1)\cdot 0& 3\cdot 1+(-1)\cdot 0\end{array}\right)=\left(\begin{array}{cc}0& 1\\ 0& 3\end{array}\right)$$phi(E_(12))=([1,2],[3,-1])([0,1],[0,0])=([1*0+2*0,1*1+2*0],[3*0+(-1)*0,3*1+(-1)*0])=([0,1],[0,3])\phi(E_{12}) = \begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 \cdot 0 + 2 \cdot 0 & 1 \cdot 1 + 2 \cdot 0 \\ 3 \cdot 0 + (-1) \cdot 0 & 3 \cdot 1 + (-1) \cdot 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 3 \end{pmatrix}$$\varphi (E_{12})=\left(\begin{array}{cc}1& 2\\ 3& -1\end{array}\right)\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}1\cdot 0+2\cdot 0& 1\cdot 1+2\cdot 0\\ 3\cdot 0+(-1)\cdot 0& 3\cdot 1+(-1)\cdot 0\end{array}\right)=\left(\begin{array}{cc}0& 1\\ 0& 3\end{array}\right)$$
  $$\left(\begin{array}{cc}0& 1\\ 0& 3\end{array}\right)=0\cdot E_{11}+1\cdot E_{12}+0\cdot E_{21}+3\cdot E_{22}$$$$\left(\begin{array}{cc}0& 1\\ 0& 3\end{array}\right)=0\cdot E_{11}+1\cdot E_{12}+0\cdot E_{21}+3\cdot E_{22}$$([0,1],[0,3])=0*E_(11)+1*E_(12)+0*E_(21)+3*E_(22)\begin{pmatrix} 0 & 1 \\ 0 & 3 \end{pmatrix} = 0 \cdot E_{11} + 1 \cdot E_{12} + 0 \cdot E_{21} + 3 \cdot E_{22}$$\left(\begin{array}{cc}0& 1\\ 0& 3\end{array}\right)=0\cdot E_{11}+1\cdot E_{12}+0\cdot E_{21}+3\cdot E_{22}$$
  Column 2: $(0,1,0,3)$$(0,1,0,3)$(0,1,0,3)(0, 1, 0, 3)$(0,1,0,3)$.
• For $E_{21}=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$$E_{21}=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$E_(21)=([0,0],[1,0])E_{21} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$E_{21}=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$:
  $$\varphi (E_{21})=\left(\begin{array}{cc}1& 2\\ 3& -1\end{array}\right)\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}1\cdot 0+2\cdot 1& 1\cdot 0+2\cdot 0\\ 3\cdot 0+(-1)\cdot 1& 3\cdot 0+(-1)\cdot 0\end{array}\right)=\left(\begin{array}{cc}2& 0\\ -1& 0\end{array}\right)$$$$\varphi (E_{21})=\left(\begin{array}{cc}1& 2\\ 3& -1\end{array}\right)\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}1\cdot 0+2\cdot 1& 1\cdot 0+2\cdot 0\\ 3\cdot 0+(-1)\cdot 1& 3\cdot 0+(-1)\cdot 0\end{array}\right)=\left(\begin{array}{cc}2& 0\\ -1& 0\end{array}\right)$$phi(E_(21))=([1,2],[3,-1])([0,0],[1,0])=([1*0+2*1,1*0+2*0],[3*0+(-1)*1,3*0+(-1)*0])=([2,0],[-1,0])\phi(E_{21}) = \begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 \cdot 0 + 2 \cdot 1 & 1 \cdot 0 + 2 \cdot 0 \\ 3 \cdot 0 + (-1) \cdot 1 & 3 \cdot 0 + (-1) \cdot 0 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ -1 & 0 \end{pmatrix}$$\varphi (E_{21})=\left(\begin{array}{cc}1& 2\\ 3& -1\end{array}\right)\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}1\cdot 0+2\cdot 1& 1\cdot 0+2\cdot 0\\ 3\cdot 0+(-1)\cdot 1& 3\cdot 0+(-1)\cdot 0\end{array}\right)=\left(\begin{array}{cc}2& 0\\ -1& 0\end{array}\right)$$
  $$\left(\begin{array}{cc}2& 0\\ -1& 0\end{array}\right)=2\cdot E_{11}+0\cdot E_{12}+(-1)\cdot E_{21}+0\cdot E_{22}$$$$\left(\begin{array}{cc}2& 0\\ -1& 0\end{array}\right)=2\cdot E_{11}+0\cdot E_{12}+(-1)\cdot E_{21}+0\cdot E_{22}$$([2,0],[-1,0])=2*E_(11)+0*E_(12)+(-1)*E_(21)+0*E_(22)\begin{pmatrix} 2 & 0 \\ -1 & 0 \end{pmatrix} = 2 \cdot E_{11} + 0 \cdot E_{12} + (-1) \cdot E_{21} + 0 \cdot E_{22}$$\left(\begin{array}{cc}2& 0\\ -1& 0\end{array}\right)=2\cdot E_{11}+0\cdot E_{12}+(-1)\cdot E_{21}+0\cdot E_{22}$$
  Column 3: $(2,0,-1,0)$$(2,0,-1,0)$(2,0,-1,0)(2, 0, -1, 0)$(2,0,-1,0)$.
• For $E_{22}=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)$$E_{22}=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)$E_(22)=([0,0],[0,1])E_{22} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$E_{22}=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)$:
  $$\varphi (E_{22})=\left(\begin{array}{cc}1& 2\\ 3& -1\end{array}\right)\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}1\cdot 0+2\cdot 0& 1\cdot 0+2\cdot 1\\ 3\cdot 0+(-1)\cdot 0& 3\cdot 0+(-1)\cdot 1\end{array}\right)=\left(\begin{array}{cc}0& 2\\ 0& -1\end{array}\right)$$$$\varphi (E_{22})=\left(\begin{array}{cc}1& 2\\ 3& -1\end{array}\right)\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}1\cdot 0+2\cdot 0& 1\cdot 0+2\cdot 1\\ 3\cdot 0+(-1)\cdot 0& 3\cdot 0+(-1)\cdot 1\end{array}\right)=\left(\begin{array}{cc}0& 2\\ 0& -1\end{array}\right)$$phi(E_(22))=([1,2],[3,-1])([0,0],[0,1])=([1*0+2*0,1*0+2*1],[3*0+(-1)*0,3*0+(-1)*1])=([0,2],[0,-1])\phi(E_{22}) = \begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 \cdot 0 + 2 \cdot 0 & 1 \cdot 0 + 2 \cdot 1 \\ 3 \cdot 0 + (-1) \cdot 0 & 3 \cdot 0 + (-1) \cdot 1 \end{pmatrix} = \begin{pmatrix} 0 & 2 \\ 0 & -1 \end{pmatrix}$$\varphi (E_{22})=\left(\begin{array}{cc}1& 2\\ 3& -1\end{array}\right)\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}1\cdot 0+2\cdot 0& 1\cdot 0+2\cdot 1\\ 3\cdot 0+(-1)\cdot 0& 3\cdot 0+(-1)\cdot 1\end{array}\right)=\left(\begin{array}{cc}0& 2\\ 0& -1\end{array}\right)$$
  $$\left(\begin{array}{cc}0& 2\\ 0& -1\end{array}\right)=0\cdot E_{11}+2\cdot E_{12}+0\cdot E_{21}+(-1)\cdot E_{22}$$$$\left(\begin{array}{cc}0& 2\\ 0& -1\end{array}\right)=0\cdot E_{11}+2\cdot E_{12}+0\cdot E_{21}+(-1)\cdot E_{22}$$([0,2],[0,-1])=0*E_(11)+2*E_(12)+0*E_(21)+(-1)*E_(22)\begin{pmatrix} 0 & 2 \\ 0 & -1 \end{pmatrix} = 0 \cdot E_{11} + 2 \cdot E_{12} + 0 \cdot E_{21} + (-1) \cdot E_{22}$$\left(\begin{array}{cc}0& 2\\ 0& -1\end{array}\right)=0\cdot E_{11}+2\cdot E_{12}+0\cdot E_{21}+(-1)\cdot E_{22}$$
  Column 4: $(0,2,0,-1)$$(0,2,0,-1)$(0,2,0,-1)(0, 2, 0, -1)$(0,2,0,-1)$.

Step 3: Alternative Approach Using Vector Representation

Step 4: Find the Rank of $\varphi$$\varphi$phi\phi$\varphi$

• $R_3\leftarrow R_3-3R_1$$R_3\leftarrow R_3-3R_1$R_(3)larrR_(3)-3R_(1)R_3 \gets R_3 – 3R_1$R_3\leftarrow R_3-3R_1$:$$R_3=(3,0,-1,0)-3(1,0,2,0)=(0,0,-7,0)$$$$R_3=(3,0,-1,0)-3(1,0,2,0)=(0,0,-7,0)$$R_(3)=(3,0,-1,0)-3(1,0,2,0)=(0,0,-7,0)R_3 = (3, 0, -1, 0) – 3(1, 0, 2, 0) = (0, 0, -7, 0)$$R_3=(3,0,-1,0)-3(1,0,2,0)=(0,0,-7,0)$$
• $R_4\leftarrow R_4-3R_2$$R_4\leftarrow R_4-3R_2$R_(4)larrR_(4)-3R_(2)R_4 \gets R_4 – 3R_2$R_4\leftarrow R_4-3R_2$:$$R_4=(0,3,0,-1)-3(0,1,0,2)=(0,0,0,-7)$$$$R_4=(0,3,0,-1)-3(0,1,0,2)=(0,0,0,-7)$$R_(4)=(0,3,0,-1)-3(0,1,0,2)=(0,0,0,-7)R_4 = (0, 3, 0, -1) – 3(0, 1, 0, 2) = (0, 0, 0, -7)$$R_4=(0,3,0,-1)-3(0,1,0,2)=(0,0,0,-7)$$

Step 5: Determine Invertibility

Step 6: Optional Verification

Final Answer

Find the volume of the greatest cylinder which can be inscribed in a cone of height $h$$h$hh$h$ and semi-vertical angle $\alpha$$\alpha$alpha\alpha$\alpha$.

Answer:

Step 1: Set Up the Geometry of the Cone

Step 2: Define the Inscribed Cylinder

Step 3: Volume of the Cylinder

Step 4: Optimize the Volume

• Partial derivative with respect to $L$$L$LL$L$:
  $$\frac{\partial f}{\partial L}=\frac{\partial }{\partial L}[(H-L{)}^{2}L]=(H-L{)}^2\cdot 1+L\cdot 2(H-L)\cdot (-1)$$$$\frac{\partial f}{\partial L}=\frac{\partial }{\partial L}[(H-L{)}^{2}L]=(H-L{)}^2\cdot 1+L\cdot 2(H-L)\cdot (-1)$$(del f)/(del L)=(del)/(del L)[(H-L)^(2)L]=(H-L)^(2)*1+L*2(H-L)*(-1)\frac{\partial f}{\partial L} = \frac{\partial}{\partial L} [ (H – L)^2 L ] = (H – L)^2 \cdot 1 + L \cdot 2 (H – L) \cdot (-1)$$\frac{\partial f}{\partial L}=\frac{\partial }{\partial L}[(H-L{)}^{2}L]=(H-L{)}^2\cdot 1+L\cdot 2(H-L)\cdot (-1)$$
  $$=(H-L{)}^2-2L(H-L)=(H-L)[(H-L)-2L]=(H-L)(H-3L)$$$$=(H-L{)}^2-2L(H-L)=(H-L)[(H-L)-2L]=(H-L)(H-3L)$$=(H-L)^(2)-2L(H-L)=(H-L)[(H-L)-2L]=(H-L)(H-3L)= (H – L)^2 – 2L (H – L) = (H – L) [ (H – L) – 2L ] = (H – L) (H – 3L)$$=(H-L{)}^2-2L(H-L)=(H-L)[(H-L)-2L]=(H-L)(H-3L)$$
  Set to zero:
  $$(H-L)(H-3L)=0$$$$(H-L)(H-3L)=0$$(H-L)(H-3L)=0(H – L)(H – 3L) = 0$$(H-L)(H-3L)=0$$
  $$H-L=0\text{or}H-3L=0$$$$H-L=0\text{or}H-3L=0$$H-L=0quad”or”quad H-3L=0H – L = 0 \quad \text{or} \quad H – 3L = 0$$H-L=0\text{or}H-3L=0$$
  $$H=L\text{or}H=3L$$$$H=L\text{or}H=3L$$H=L quad”or”quad H=3LH = L \quad \text{or} \quad H = 3L$$H=L\text{or}H=3L$$
  Since $L<H$$L<H$L < HL < H$L<H$, discard $H=L$$H=L$H=LH = L$H=L$. Thus:
  $$H=3L$$$$H=3L$$H=3LH = 3L$$H=3L$$
• Substitute $H=3L$$H=3L$H=3LH = 3L$H=3L$ into the volume:
  $$f(L,H)=(3L-L{)}^{2}L=(2L{)}^{2}L=4L^2\cdot L=4L^3$$$$f(L,H)=(3L-L{)}^{2}L=(2L{)}^{2}L=4L^2\cdot L=4L^3$$f(L,H)=(3L-L)^(2)L=(2L)^(2)L=4L^(2)*L=4L^(3)f(L, H) = (3L – L)^2 L = (2L)^2 L = 4L^2 \cdot L = 4L^3$$f(L,H)=(3L-L{)}^{2}L=(2L{)}^{2}L=4L^2\cdot L=4L^3$$
  Constraints: $0<L<H=3L\le h$$0<L<H=3L\le h$0 < L < H=3L <= h0 < L < H = 3L \leq h$0<L<H=3L\le h$, so:
  $$3L\le h⟹L\le \frac{h}{3}$$$$3L\le h⟹L\le \frac{h}{3}$$3L <= hLongrightarrowL <= (h)/(3)3L \leq h \implies L \leq \frac{h}{3}$$3L\le h⟹L\le \frac{h}{3}$$
  Maximize:
  $$g(L)=4L^3,0<L\le \frac{h}{3}$$$$g(L)=4L^3,0<L\le \frac{h}{3}$$g(L)=4L^(3),quad0 < L <= (h)/(3)g(L) = 4L^3, \quad 0 < L \leq \frac{h}{3}$$g(L)=4L^3,0<L\le \frac{h}{3}$$
  Since $g(L)=4L^3$$g(L)=4L^3$g(L)=4L^(3)g(L) = 4L^3$g(L)=4L^3$ is increasing ($g^{\prime }(L)=12L^2>0$$g^{\prime }(L)=12L^2>0$g^(‘)(L)=12L^(2) > 0g'(L) = 12L^2 > 0$g^{\prime }(L)=12L^2>0$), the maximum occurs at the boundary $L=\frac{h}{3}$$L=\frac{h}{3}$L=(h)/(3)L = \frac{h}{3}$L=\frac{h}{3}$.
• Compute $H$$H$HH$H$ and $r$$r$rr$r$:
  $$H=3L=3\cdot \frac{h}{3}=h$$$$H=3L=3\cdot \frac{h}{3}=h$$H=3L=3*(h)/(3)=hH = 3L = 3 \cdot \frac{h}{3} = h$$H=3L=3\cdot \frac{h}{3}=h$$
  Base at:
  $$H-L=h-\frac{h}{3}=\frac{2h}{3}$$$$H-L=h-\frac{h}{3}=\frac{2h}{3}$$H-L=h-(h)/(3)=(2h)/(3)H – L = h – \frac{h}{3} = \frac{2h}{3}$$H-L=h-\frac{h}{3}=\frac{2h}{3}$$
  Radius:
  $$r=(H-L)\tan \alpha =\frac{2h}{3}\tan \alpha$$$$r=(H-L)\tan \alpha =\frac{2h}{3}\tan \alpha$$r=(H-L)tan alpha=(2h)/(3)tan alphar = (H – L) \tan \alpha = \frac{2h}{3} \tan \alpha$$r=(H-L)\tan \alpha =\frac{2h}{3}\tan \alpha$$
• Volume:
  $$V=\pi {(\frac{2h}{3}\tan \alpha )}^2\cdot \frac{h}{3}=\pi \cdot \frac{4h^2{\tan }^2\alpha }{9}\cdot \frac{h}{3}=\pi \cdot \frac{4h^3{\tan }^2\alpha }{27}=\frac{4\pi h^3{\tan }^2\alpha }{27}$$$$V=\pi {\left(\frac{2h}{3}\tan \alpha \right)}^2\cdot \frac{h}{3}=\pi \cdot \frac{4h^2{\tan }^2\alpha }{9}\cdot \frac{h}{3}=\pi \cdot \frac{4h^3{\tan }^2\alpha }{27}=\frac{4\pi h^3{\tan }^2\alpha }{27}$$V=pi((2h)/(3)tan alpha)^(2)*(h)/(3)=pi*(4h^(2)tan^(2)alpha)/(9)*(h)/(3)=pi*(4h^(3)tan^(2)alpha)/(27)=(4pih^(3)tan^(2)alpha)/(27)V = \pi \left( \frac{2h}{3} \tan \alpha \right)^2 \cdot \frac{h}{3} = \pi \cdot \frac{4h^2 \tan^2 \alpha}{9} \cdot \frac{h}{3} = \pi \cdot \frac{4h^3 \tan^2 \alpha}{27} = \frac{4\pi h^3 \tan^2 \alpha}{27}$$V=\pi {(\frac{2h}{3}\tan \alpha )}^2\cdot \frac{h}{3}=\pi \cdot \frac{4h^2{\tan }^2\alpha }{9}\cdot \frac{h}{3}=\pi \cdot \frac{4h^3{\tan }^2\alpha }{27}=\frac{4\pi h^3{\tan }^2\alpha }{27}$$

Step 5: Verify the Critical Point

Step 6: Geometric Verification

Final Answer

Find the vertex of the cone $4x^2-y^2+2z^2+2xy-3yz+12x-11y+6z+4=0$$4x^2-y^2+2z^2+2xy-3yz+12x-11y+6z+4=0$4x^(2)-y^(2)+2z^(2)+2xy-3yz+12 x-11 y+6z+4=04x^{2}-y^{2}+2z^{2}+2xy-3yz+12x-11y+6z+4=0$4x^2-y^2+2z^2+2xy-3yz+12x-11y+6z+4=0$.

Answer:

Given Equation:

Step 1: Rewrite the Equation in Matrix Form

• $\mathbf{X}=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$$\mathbf{X}=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$X=[[x],[y],[z]]\mathbf{X} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$\mathbf{X}=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$
• $A$$A$AA$A$ is the symmetric matrix of the quadratic terms.
• $B$$B$BB$B$ is the row matrix of the linear terms.
• $C$$C$CC$C$ is the constant term.

Step 2: Find the Vertex

Substituting $A$$A$AA$A$ and $B$$B$BB$B$:
[
$$\left[\begin{array}{ccc}4& 1& 0\\ 1& -1& -\frac{3}{2}\\ 0& -\frac{3}{2}& 2\end{array}\right]$$$$\left[\begin{array}{ccc}4& 1& 0\\ 1& -1& -\frac{3}{2}\\ 0& -\frac{3}{2}& 2\end{array}\right]$$[[4,1,0],[1,-1,-(3)/(2)],[0,-(3)/(2),2]]\begin{bmatrix} 4 & 1 & 0 \\ 1 & -1 & -\frac{3}{2} \\ 0 & -\frac{3}{2} & 2 \end{bmatrix}$$\left[\begin{array}{ccc}4& 1& 0\\ 1& -1& -\frac{3}{2}\\ 0& -\frac{3}{2}& 2\end{array}\right]$$
$$\left[\begin{array}{c}{x}_0\\ y_0\\ z_0\end{array}\right]$$$$\left[\begin{array}{c}{x}_0\\ y_0\\ z_0\end{array}\right]$$[[x_(0)],[y_(0)],[z_(0)]]\begin{bmatrix} x_0 \\ y_0 \\ z_0 \end{bmatrix}$$\left[\begin{array}{c}{x}_0\\ y_0\\ z_0\end{array}\right]$$
= -\frac{1}{2}
$$\left[\begin{array}{c}12\\ -11\\ 6\end{array}\right]$$$$\left[\begin{array}{c}12\\ -11\\ 6\end{array}\right]$$[[12],[-11],[6]]\begin{bmatrix} 12 \\ -11 \\ 6 \end{bmatrix}$$\left[\begin{array}{c}12\\ -11\\ 6\end{array}\right]$$

1. $4x_0+y_0=-6$$4x_0+y_0=-6$4x_(0)+y_(0)=-64x_0 + y_0 = -6$4x_0+y_0=-6$            (1)
2. $x_0-y_0-\frac{3}{2}{z}_0=\frac{11}{2}$$x_0-y_0-\frac{3}{2}{z}_0=\frac{11}{2}$x_(0)-y_(0)-(3)/(2)z_(0)=(11)/(2)x_0 – y_0 – \frac{3}{2}z_0 = \frac{11}{2}$x_0-y_0-\frac{3}{2}{z}_0=\frac{11}{2}$   (2)
3. $-\frac{3}{2}{y}_0+2z_0=-3$$-\frac{3}{2}{y}_0+2z_0=-3$-(3)/(2)y_(0)+2z_(0)=-3-\frac{3}{2}y_0 + 2z_0 = -3$-\frac{3}{2}{y}_0+2z_0=-3$        (3)

Step 3: Solve the System of Equations

Final Answer:

Let

Answer:

Finding the Eigenvalues of Matrix $A$$A$AA$A$

• $\lambda =-1$$\lambda =-1$lambda=-1\lambda = -1$\lambda =-1$:$$(-1{)}^3-6(-1{)}^2-15(-1)-8=-1-6+15-8=0$$$$(-1{)}^3-6(-1{)}^2-15(-1)-8=-1-6+15-8=0$$(-1)^(3)-6(-1)^(2)-15(-1)-8=-1-6+15-8=0(-1)^3 – 6(-1)^2 – 15(-1) – 8 = -1 – 6 + 15 – 8 = 0$$(-1{)}^3-6(-1{)}^2-15(-1)-8=-1-6+15-8=0$$So, $\lambda =-1$$\lambda =-1$lambda=-1\lambda = -1$\lambda =-1$ is a root.

Finding the Eigenvectors

1. Let $x=1$$x=1$x=1x = 1$x=1$, $z=0$$z=0$z=0z = 0$z=0$: $y=-2$$y=-2$y=-2y = -2$y=-2$$${\mathbf{v}}_2=\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)$$$${\mathbf{v}}_2=\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)$$v_(2)=([1],[-2],[0])\mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}$${\mathbf{v}}_2=\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)$$
2. Let $x=0$$x=0$x=0x = 0$x=0$, $z=1$$z=1$z=1z = 1$z=1$: $y=-2$$y=-2$y=-2y = -2$y=-2$$${\mathbf{v}}_3=\left(\begin{array}{c}0\\ -2\\ 1\end{array}\right)$$$${\mathbf{v}}_3=\left(\begin{array}{c}0\\ -2\\ 1\end{array}\right)$$v_(3)=([0],[-2],[1])\mathbf{v}_3 = \begin{pmatrix} 0 \\ -2 \\ 1 \end{pmatrix}$${\mathbf{v}}_3=\left(\begin{array}{c}0\\ -2\\ 1\end{array}\right)$$

Eigenvalues and Eigenvectors of $A^{-15}$$A^{-15}$A^(-15)A^{-15}$A^{-15}$

• For eigenvalue $8^{-15}$$8^{-15}$8^(-15)8^{-15}$8^{-15}$: ${\mathbf{v}}_1=\left(\begin{array}{c}2\\ 1\\ 2\end{array}\right)$${\mathbf{v}}_1=\left(\begin{array}{c}2\\ 1\\ 2\end{array}\right)$v_(1)=([2],[1],[2])\mathbf{v}_1 = \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}${\mathbf{v}}_1=\left(\begin{array}{c}2\\ 1\\ 2\end{array}\right)$
• For eigenvalue $-1$$-1$-1-1$-1$: ${\mathbf{v}}_2=\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)$${\mathbf{v}}_2=\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)$v_(2)=([1],[-2],[0])\mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}${\mathbf{v}}_2=\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)$ and ${\mathbf{v}}_3=\left(\begin{array}{c}0\\ -2\\ 1\end{array}\right)$${\mathbf{v}}_3=\left(\begin{array}{c}0\\ -2\\ 1\end{array}\right)$v_(3)=([0],[-2],[1])\mathbf{v}_3 = \begin{pmatrix} 0 \\ -2 \\ 1 \end{pmatrix}${\mathbf{v}}_3=\left(\begin{array}{c}0\\ -2\\ 1\end{array}\right)$

Final Answer

• Eigenvalue $8$$8$88$8$: Eigenvector $\left(\begin{array}{c}2\\ 1\\ 2\end{array}\right)$$\left(\begin{array}{c}2\\ 1\\ 2\end{array}\right)$([2],[1],[2])\begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}$\left(\begin{array}{c}2\\ 1\\ 2\end{array}\right)$
• Eigenvalue $-1$$-1$-1-1$-1$: Eigenvectors $\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)$$\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)$([1],[-2],[0])\begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}$\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)$ and $\left(\begin{array}{c}0\\ -2\\ 1\end{array}\right)$$\left(\begin{array}{c}0\\ -2\\ 1\end{array}\right)$([0],[-2],[1])\begin{pmatrix} 0 \\ -2 \\ 1 \end{pmatrix}$\left(\begin{array}{c}0\\ -2\\ 1\end{array}\right)$

• Eigenvalue $8^{-15}$$8^{-15}$8^(-15)8^{-15}$8^{-15}$: Eigenvector $\left(\begin{array}{c}2\\ 1\\ 2\end{array}\right)$$\left(\begin{array}{c}2\\ 1\\ 2\end{array}\right)$([2],[1],[2])\begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}$\left(\begin{array}{c}2\\ 1\\ 2\end{array}\right)$
• Eigenvalue $-1$$-1$-1-1$-1$: Eigenvectors $\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)$$\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)$([1],[-2],[0])\begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}$\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)$ and $\left(\begin{array}{c}0\\ -2\\ 1\end{array}\right)$$\left(\begin{array}{c}0\\ -2\\ 1\end{array}\right)$([0],[-2],[1])\begin{pmatrix} 0 \\ -2 \\ 1 \end{pmatrix}$\left(\begin{array}{c}0\\ -2\\ 1\end{array}\right)$

Using double integration, find the area lying inside the cardioid $r=a(1+\cos \theta )$$r=a(1+\cos \theta )$r=a(1+cos theta)r=a(1+\cos\theta)$r=a(1+\cos \theta )$ and outside the circle $r=a$$r=a$r=ar=a$r=a$.

Answer:

Solution: Finding the Area Inside the Cardioid and Outside the Circle

• Cardioid: $r=a(1+\cos \theta )$$r=a(1+\cos \theta )$r=a(1+cos theta)r = a(1 + \cos\theta)$r=a(1+\cos \theta )$
• Circle: $r=a$$r=a$r=ar = a$r=a$

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Step 1: Identify the Points of Intersection

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Step 2: Set Up the Double Integral

• Inner boundary (circle): $r=a$$r=a$r=ar = a$r=a$
• Outer boundary (cardioid): $r=a(1+\cos \theta )$$r=a(1+\cos \theta )$r=a(1+cos theta)r = a(1 + \cos\theta)$r=a(1+\cos \theta )$

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Step 3: Integrate with Respect to $r$$r$rr$r$

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Step 4: Integrate with Respect to $\theta$$\theta$theta\theta$\theta$

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Final Answer

Find the equation of the sphere which touches the plane $3x+2y-z+2=0$$3x+2y-z+2=0$3x+2y-z+2=03x+2y-z+2=0$3x+2y-z+2=0$ at the point $(1,-2,1)$$(1,-2,1)$(1,-2,1)(1,-2,1)$(1,-2,1)$ and cuts orthogonally the sphere $x^2+y^2+z^2-4x+6y+4=0$$x^2+y^2+z^2-4x+6y+4=0$x^(2)+y^(2)+z^(2)-4x+6y+4=0x^{2}+y^{2}+z^{2}-4x+6y+4=0$x^2+y^2+z^2-4x+6y+4=0$.

Answer:

Step 1: Equation of the Sphere Touching the Plane

• Normal Vector to the Plane: The coefficients of $x$$x$xx$x$, $y$$y$yy$y$, and $z$$z$zz$z$ in the plane equation give the normal vector $\mathbf{n}=(3,2,-1)$$\mathbf{n}=(3,2,-1)$n=(3,2,-1)\mathbf{n} = (3, 2, -1)$\mathbf{n}=(3,2,-1)$.
• Parametric Equations of the Line: The line through $(1,-2,1)$$(1,-2,1)$(1,-2,1)(1, -2, 1)$(1,-2,1)$ in the direction of $\mathbf{n}$$\mathbf{n}$n\mathbf{n}$\mathbf{n}$ is:$$x=1+3t,y=-2+2t,z=1-t$$$$x=1+3t,y=-2+2t,z=1-t$$x=1+3t,quad y=-2+2t,quad z=1-tx = 1 + 3t, \quad y = -2 + 2t, \quad z = 1 – t$$x=1+3t,y=-2+2t,z=1-t$$
• Center of the Sphere: Let the center be at $(1+3t,-2+2t,1-t)$$(1+3t,-2+2t,1-t)$(1+3t,-2+2t,1-t)(1 + 3t, -2 + 2t, 1 – t)$(1+3t,-2+2t,1-t)$.
• Radius: The distance from the center to the plane equals the radius $r$$r$rr$r$:$$r=\frac{|3(1+3t)+2(-2+2t)-(1-t)+2|}{\sqrt{3^2+2^2+(-1{)}^2}}=\frac{|3+9t-4+4t-1+t+2|}{\sqrt{14}}=\frac{|14t|}{\sqrt{14}}=\sqrt{14}|t|$$$$r=\frac{|3(1+3t)+2(-2+2t)-(1-t)+2|}{\sqrt{3^2+2^2+(-1{)}^2}}=\frac{|3+9t-4+4t-1+t+2|}{\sqrt{14}}=\frac{|14t|}{\sqrt{14}}=\sqrt{14}|t|$$r=(|3(1+3t)+2(-2+2t)-(1-t)+2|)/(sqrt(3^(2)+2^(2)+(-1)^(2)))=(|3+9t-4+4t-1+t+2|)/(sqrt14)=(|14 t|)/(sqrt14)=sqrt14|t|r = \frac{|3(1 + 3t) + 2(-2 + 2t) – (1 – t) + 2|}{\sqrt{3^2 + 2^2 + (-1)^2}} = \frac{|3 + 9t – 4 + 4t – 1 + t + 2|}{\sqrt{14}} = \frac{|14t|}{\sqrt{14}} = \sqrt{14}|t|$$r=\frac{|3(1+3t)+2(-2+2t)-(1-t)+2|}{\sqrt{3^2+2^2+(-1{)}^2}}=\frac{|3+9t-4+4t-1+t+2|}{\sqrt{14}}=\frac{|14t|}{\sqrt{14}}=\sqrt{14}|t|$$Since the sphere touches the plane, the distance equals the radius:$$\sqrt{(3t{)}^2+(2t{)}^2+(-t{)}^2}=\sqrt{14}|t|=r$$$$\sqrt{(3t{)}^2+(2t{)}^2+(-t{)}^2}=\sqrt{14}|t|=r$$sqrt((3t)^(2)+(2t)^(2)+(-t)^(2))=sqrt14|t|=r\sqrt{(3t)^2 + (2t)^2 + (-t)^2} = \sqrt{14}|t| = r$$\sqrt{(3t{)}^2+(2t{)}^2+(-t{)}^2}=\sqrt{14}|t|=r$$Thus, the radius $r=\sqrt{14}|t|$$r=\sqrt{14}|t|$r=sqrt14|t|r = \sqrt{14}|t|$r=\sqrt{14}|t|$.

Step 2: Orthogonal Condition with the Given Sphere

• Rewrite the Given Sphere in Standard Form:
  $$x^2-4x+y^2+6y+z^2=-4$$$$x^2-4x+y^2+6y+z^2=-4$$x^(2)-4x+y^(2)+6y+z^(2)=-4x^{2} – 4x + y^{2} + 6y + z^{2} = -4$$x^2-4x+y^2+6y+z^2=-4$$
  Completing the squares:
  $$(x-2{)}^2+(y+3{)}^2+z^2=-4+4+9=9$$$$(x-2{)}^2+(y+3{)}^2+z^2=-4+4+9=9$$(x-2)^(2)+(y+3)^(2)+z^(2)=-4+4+9=9(x – 2)^2 + (y + 3)^2 + z^2 = -4 + 4 + 9 = 9$$(x-2{)}^2+(y+3{)}^2+z^2=-4+4+9=9$$
  So, the given sphere has center $C_2=(2,-3,0)$$C_2=(2,-3,0)$C_(2)=(2,-3,0)C_2 = (2, -3, 0)$C_2=(2,-3,0)$ and radius $R=3$$R=3$R=3R = 3$R=3$.
• Condition for Orthogonal Intersection:
  Two spheres with centers $C_1$$C_1$C_(1)C_1$C_1$ and $C_2$$C_2$C_(2)C_2$C_2$, and radii $r$$r$rr$r$ and $R$$R$RR$R$, intersect orthogonally if:
  $$d^2=r^2+R^2$$$$d^2=r^2+R^2$$d^(2)=r^(2)+R^(2)d^2 = r^2 + R^2$$d^2=r^2+R^2$$
  where $d$$d$dd$d$ is the distance between $C_1$$C_1$C_(1)C_1$C_1$ and $C_2$$C_2$C_(2)C_2$C_2$.
• Distance Between Centers:
  $$C_1=(1+3t,-2+2t,1-t),C_2=(2,-3,0)$$$$C_1=(1+3t,-2+2t,1-t),C_2=(2,-3,0)$$C_(1)=(1+3t,-2+2t,1-t),quadC_(2)=(2,-3,0)C_1 = (1 + 3t, -2 + 2t, 1 – t), \quad C_2 = (2, -3, 0)$$C_1=(1+3t,-2+2t,1-t),C_2=(2,-3,0)$$
  $$d^2=(1+3t-2{)}^2+(-2+2t+3{)}^2+(1-t-0{)}^2=(3t-1{)}^2+(2t+1{)}^2+(1-t{)}^2$$$$d^2=(1+3t-2{)}^2+(-2+2t+3{)}^2+(1-t-0{)}^2=(3t-1{)}^2+(2t+1{)}^2+(1-t{)}^2$$d^(2)=(1+3t-2)^(2)+(-2+2t+3)^(2)+(1-t-0)^(2)=(3t-1)^(2)+(2t+1)^(2)+(1-t)^(2)d^2 = (1 + 3t – 2)^2 + (-2 + 2t + 3)^2 + (1 – t – 0)^2 = (3t – 1)^2 + (2t + 1)^2 + (1 – t)^2$$d^2=(1+3t-2{)}^2+(-2+2t+3{)}^2+(1-t-0{)}^2=(3t-1{)}^2+(2t+1{)}^2+(1-t{)}^2$$
  $$=9t^2-6t+1+4t^2+4t+1+t^2-2t+1=14t^2-4t+3$$$$=9t^2-6t+1+4t^2+4t+1+t^2-2t+1=14t^2-4t+3$$=9t^(2)-6t+1+4t^(2)+4t+1+t^(2)-2t+1=14t^(2)-4t+3= 9t^2 – 6t + 1 + 4t^2 + 4t + 1 + t^2 – 2t + 1 = 14t^2 – 4t + 3$$=9t^2-6t+1+4t^2+4t+1+t^2-2t+1=14t^2-4t+3$$
• Apply Orthogonal Condition:
  $$14t^2-4t+3=(\sqrt{14}|t|{)}^2+3^2=14t^2+9$$$$14t^2-4t+3=(\sqrt{14}|t|{)}^2+3^2=14t^2+9$$14t^(2)-4t+3=(sqrt14|t|)^(2)+3^(2)=14t^(2)+914t^2 – 4t + 3 = (\sqrt{14}|t|)^2 + 3^2 = 14t^2 + 9$$14t^2-4t+3=(\sqrt{14}|t|{)}^2+3^2=14t^2+9$$
  $$14t^2-4t+3=14t^2+9⟹-4t+3=9⟹-4t=6⟹t=-\frac{3}{2}$$$$14t^2-4t+3=14t^2+9⟹-4t+3=9⟹-4t=6⟹t=-\frac{3}{2}$$14t^(2)-4t+3=14t^(2)+9Longrightarrow-4t+3=9Longrightarrow-4t=6Longrightarrowt=-(3)/(2)14t^2 – 4t + 3 = 14t^2 + 9 \implies -4t + 3 = 9 \implies -4t = 6 \implies t = -\frac{3}{2}$$14t^2-4t+3=14t^2+9⟹-4t+3=9⟹-4t=6⟹t=-\frac{3}{2}$$

Step 3: Determine the Sphere’s Equation

• Center $C_1$$C_1$C_(1)C_1$C_1$:
  $$x=1+3(-\frac{3}{2})=1-\frac{9}{2}=-\frac{7}{2}$$$$x=1+3\left(-\frac{3}{2}\right)=1-\frac{9}{2}=-\frac{7}{2}$$x=1+3(-(3)/(2))=1-(9)/(2)=-(7)/(2)x = 1 + 3\left(-\frac{3}{2}\right) = 1 – \frac{9}{2} = -\frac{7}{2}$$x=1+3(-\frac{3}{2})=1-\frac{9}{2}=-\frac{7}{2}$$
  $$y=-2+2(-\frac{3}{2})=-2-3=-5$$$$y=-2+2\left(-\frac{3}{2}\right)=-2-3=-5$$y=-2+2(-(3)/(2))=-2-3=-5y = -2 + 2\left(-\frac{3}{2}\right) = -2 – 3 = -5$$y=-2+2(-\frac{3}{2})=-2-3=-5$$
  $$z=1-(-\frac{3}{2})=1+\frac{3}{2}=\frac{5}{2}$$$$z=1-\left(-\frac{3}{2}\right)=1+\frac{3}{2}=\frac{5}{2}$$z=1-(-(3)/(2))=1+(3)/(2)=(5)/(2)z = 1 – \left(-\frac{3}{2}\right) = 1 + \frac{3}{2} = \frac{5}{2}$$z=1-(-\frac{3}{2})=1+\frac{3}{2}=\frac{5}{2}$$
  So, $C_1=(-\frac{7}{2},-5,\frac{5}{2})$$C_1=\left(-\frac{7}{2},-5,\frac{5}{2}\right)$C_(1)=(-(7)/(2),-5,(5)/(2))C_1 = \left(-\frac{7}{2}, -5, \frac{5}{2}\right)$C_1=(-\frac{7}{2},-5,\frac{5}{2})$.
• Radius $r$$r$rr$r$:
  $$r=\sqrt{14}\times |-\frac{3}{2}|=\frac{3\sqrt{14}}{2}$$$$r=\sqrt{14}\times \left|-\frac{3}{2}\right|=\frac{3\sqrt{14}}{2}$$r=sqrt14xx|-(3)/(2)|=(3sqrt14)/(2)r = \sqrt{14} \times \left| -\frac{3}{2} \right| = \frac{3\sqrt{14}}{2}$$r=\sqrt{14}\times |-\frac{3}{2}|=\frac{3\sqrt{14}}{2}$$
• Equation of the Sphere:
  $${(x+\frac{7}{2})}^2+(y+5{)}^2+{(z-\frac{5}{2})}^2={\left(\frac{3\sqrt{14}}{2}\right)}^2$$$${\left(x+\frac{7}{2}\right)}^2+(y+5{)}^2+{\left(z-\frac{5}{2}\right)}^2={\left(\frac{3\sqrt{14}}{2}\right)}^2$$(x+(7)/(2))^(2)+(y+5)^(2)+(z-(5)/(2))^(2)=((3sqrt14)/(2))^(2)\left(x + \frac{7}{2}\right)^2 + (y + 5)^2 + \left(z – \frac{5}{2}\right)^2 = \left(\frac{3\sqrt{14}}{2}\right)^2$${(x+\frac{7}{2})}^2+(y+5{)}^2+{(z-\frac{5}{2})}^2={\left(\frac{3\sqrt{14}}{2}\right)}^2$$
  $${(x+\frac{7}{2})}^2+(y+5{)}^2+{(z-\frac{5}{2})}^2=\frac{126}{4}=\frac{63}{2}$$$${\left(x+\frac{7}{2}\right)}^2+(y+5{)}^2+{\left(z-\frac{5}{2}\right)}^2=\frac{126}{4}=\frac{63}{2}$$(x+(7)/(2))^(2)+(y+5)^(2)+(z-(5)/(2))^(2)=(126)/(4)=(63)/(2)\left(x + \frac{7}{2}\right)^2 + (y + 5)^2 + \left(z – \frac{5}{2}\right)^2 = \frac{126}{4} = \frac{63}{2}$${(x+\frac{7}{2})}^2+(y+5{)}^2+{(z-\frac{5}{2})}^2=\frac{126}{4}=\frac{63}{2}$$
  Multiplying through by 2 to eliminate fractions:
  $$2{(x+\frac{7}{2})}^2+2(y+5{)}^2+2{(z-\frac{5}{2})}^2=63$$$$2{\left(x+\frac{7}{2}\right)}^2+2(y+5{)}^2+2{\left(z-\frac{5}{2}\right)}^2=63$$2(x+(7)/(2))^(2)+2(y+5)^(2)+2(z-(5)/(2))^(2)=632\left(x + \frac{7}{2}\right)^2 + 2(y + 5)^2 + 2\left(z – \frac{5}{2}\right)^2 = 63$$2{(x+\frac{7}{2})}^2+2(y+5{)}^2+2{(z-\frac{5}{2})}^2=63$$
  Expanding:
  $$2(x^2+7x+\frac{49}{4})+2(y^2+10y+25)+2(z^2-5z+\frac{25}{4})=63$$$$2\left(x^2+7x+\frac{49}{4}\right)+2(y^2+10y+25)+2\left(z^2-5z+\frac{25}{4}\right)=63$$2(x^(2)+7x+(49)/(4))+2(y^(2)+10 y+25)+2(z^(2)-5z+(25)/(4))=632\left(x^2 + 7x + \frac{49}{4}\right) + 2(y^2 + 10y + 25) + 2\left(z^2 – 5z + \frac{25}{4}\right) = 63$$2(x^2+7x+\frac{49}{4})+2(y^2+10y+25)+2(z^2-5z+\frac{25}{4})=63$$
  $$2x^2+14x+\frac{49}{2}+2y^2+20y+50+2z^2-10z+\frac{25}{2}=63$$$$2x^2+14x+\frac{49}{2}+2y^2+20y+50+2z^2-10z+\frac{25}{2}=63$$2x^(2)+14 x+(49)/(2)+2y^(2)+20 y+50+2z^(2)-10 z+(25)/(2)=632x^2 + 14x + \frac{49}{2} + 2y^2 + 20y + 50 + 2z^2 – 10z + \frac{25}{2} = 63$$2x^2+14x+\frac{49}{2}+2y^2+20y+50+2z^2-10z+\frac{25}{2}=63$$
  Combine like terms:
  $$2x^2+2y^2+2z^2+14x+20y-10z+(\frac{49}{2}+50+\frac{25}{2})=63$$$$2x^2+2y^2+2z^2+14x+20y-10z+\left(\frac{49}{2}+50+\frac{25}{2}\right)=63$$2x^(2)+2y^(2)+2z^(2)+14 x+20 y-10 z+((49)/(2)+50+(25)/(2))=632x^2 + 2y^2 + 2z^2 + 14x + 20y – 10z + \left(\frac{49}{2} + 50 + \frac{25}{2}\right) = 63$$2x^2+2y^2+2z^2+14x+20y-10z+(\frac{49}{2}+50+\frac{25}{2})=63$$
  $$2x^2+2y^2+2z^2+14x+20y-10z+87=63$$$$2x^2+2y^2+2z^2+14x+20y-10z+87=63$$2x^(2)+2y^(2)+2z^(2)+14 x+20 y-10 z+87=632x^2 + 2y^2 + 2z^2 + 14x + 20y – 10z + 87 = 63$$2x^2+2y^2+2z^2+14x+20y-10z+87=63$$
  $$2x^2+2y^2+2z^2+14x+20y-10z+24=0$$$$2x^2+2y^2+2z^2+14x+20y-10z+24=0$$2x^(2)+2y^(2)+2z^(2)+14 x+20 y-10 z+24=02x^2 + 2y^2 + 2z^2 + 14x + 20y – 10z + 24 = 0$$2x^2+2y^2+2z^2+14x+20y-10z+24=0$$
  Divide the entire equation by 2:
  $$x^2+y^2+z^2+7x+10y-5z+12=0$$$$x^2+y^2+z^2+7x+10y-5z+12=0$$x^(2)+y^(2)+z^(2)+7x+10 y-5z+12=0x^2 + y^2 + z^2 + 7x + 10y – 5z + 12 = 0$$x^2+y^2+z^2+7x+10y-5z+12=0$$

Final Answer

Find the orthogonal trajectories of the family of curves $r=c(\sec \theta +\tan \theta )$$r=c(\sec \theta +\tan \theta )$r=c(sec theta+tan theta)r=c(\sec\theta+\tan\theta)$r=c(\sec \theta +\tan \theta )$, where $c$$c$cc$c$ is a parameter.

Answer:

Step 1: Rewrite the Given Family of Curves

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Step 2: Differentiate to Find the Slope

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Step 3: Find the Slope of the Orthogonal Trajectories

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Step 4: Solve the Differential Equation

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Final Answer

Solve the integral equation $y(t)=\cos t+{\int }_0^{t}y(x)\cos (t-x)dx$$y(t)=\cos t+{\int }_0^t y(x)\cos (t-x)dx$y(t)=cos t+int_(0)^(t)y(x)cos(t-x)dxy(t)=\cos t+\int_{0}^{t} y(x)\cos(t-x)\,dx$y(t)=\cos t+{\int }_0^{t}y(x)\cos (t-x)dx$ using Laplace transform.

Answer:

Solution:

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Step 1: Take the Laplace Transform of Both Sides

• $\mathcal{L}\{\cos t\}=\frac{s}{s^2+1}$$\mathcal{L}\{\cos t\}=\frac{s}{s^2+1}$L{cos t}=(s)/(s^(2)+1)\mathcal{L}\{\cos t\} = \frac{s}{s^2 + 1}$\mathcal{L}\{\cos t\}=\frac{s}{s^2+1}$.
• The integral ${\int }_0^{t}y(x)\cos (t-x)dx$${\int }_0^t y(x)\cos (t-x)dx$int_(0)^(t)y(x)cos(t-x)dx\int_{0}^{t} y(x) \cos(t – x) \, dx${\int }_0^{t}y(x)\cos (t-x)dx$ is a convolution of $y(t)$$y(t)$y(t)y(t)$y(t)$ and $\cos t$$\cos t$cos t\cos t$\cos t$. The Laplace transform of a convolution is the product of the Laplace transforms:$$\mathcal{L}\{{\int }_0^{t}y(x)\cos (t-x)dx\}=\mathcal{L}\{y(t)\}\cdot \mathcal{L}\{\cos t\}=Y(s)\cdot \frac{s}{s^2+1}.$$$$\mathcal{L}\left\{{\int }_0^t y(x)\cos (t-x)dx\right\}=\mathcal{L}\{y(t)\}\cdot \mathcal{L}\{\cos t\}=Y(s)\cdot \frac{s}{s^2+1}.$$L{int_(0)^(t)y(x)cos(t-x)dx}=L{y(t)}*L{cos t}=Y(s)*(s)/(s^(2)+1).\mathcal{L}\left\{ \int_{0}^{t} y(x) \cos(t – x) \, dx \right\} = \mathcal{L}\{y(t)\} \cdot \mathcal{L}\{\cos t\} = Y(s) \cdot \frac{s}{s^2 + 1}.$$\mathcal{L}\{{\int }_0^{t}y(x)\cos (t-x)dx\}=\mathcal{L}\{y(t)\}\cdot \mathcal{L}\{\cos t\}=Y(s)\cdot \frac{s}{s^2+1}.$$

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Step 2: Solve for $Y(s)$$Y(s)$Y(s)Y(s)$Y(s)$

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Step 3: Inverse Laplace Transform to Find $y(t)$$y(t)$y(t)y(t)$y(t)$

1. ${\mathcal{L}}^{-1}\left\{\frac{s-\frac{1}{2}}{{(s-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\right\}=e^{t/2}\cos \left(\frac{\sqrt{3}}{2}t\right)$${\mathcal{L}}^{-1}\left\{\frac{s-\frac{1}{2}}{{\left(s-\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\right\}=e^{t/2}\cos \left(\frac{\sqrt{3}}{2}t\right)$L^(-1){(s-(1)/(2))/((s-(1)/(2))^(2)+((sqrt3)/(2))^(2))}=e^(t//2)cos((sqrt3)/(2)t)\mathcal{L}^{-1}\left\{ \frac{s – \frac{1}{2}}{\left(s – \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \right\} = e^{t/2} \cos\left(\frac{\sqrt{3}}{2} t\right)${\mathcal{L}}^{-1}\left\{\frac{s-\frac{1}{2}}{{(s-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\right\}=e^{t/2}\cos \left(\frac{\sqrt{3}}{2}t\right)$.
2. ${\mathcal{L}}^{-1}\left\{\frac{\frac{1}{2}}{{(s-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\right\}=\frac{1}{2}\cdot \frac{2}{\sqrt{3}}{e}^{t/2}\sin \left(\frac{\sqrt{3}}{2}t\right)=\frac{1}{\sqrt{3}}{e}^{t/2}\sin \left(\frac{\sqrt{3}}{2}t\right)$${\mathcal{L}}^{-1}\left\{\frac{\frac{1}{2}}{{\left(s-\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\right\}=\frac{1}{2}\cdot \frac{2}{\sqrt{3}}{e}^{t/2}\sin \left(\frac{\sqrt{3}}{2}t\right)=\frac{1}{\sqrt{3}}{e}^{t/2}\sin \left(\frac{\sqrt{3}}{2}t\right)$L^(-1){((1)/(2))/((s-(1)/(2))^(2)+((sqrt3)/(2))^(2))}=(1)/(2)*(2)/(sqrt3)e^(t//2)sin((sqrt3)/(2)t)=(1)/(sqrt3)e^(t//2)sin((sqrt3)/(2)t)\mathcal{L}^{-1}\left\{ \frac{\frac{1}{2}}{\left(s – \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \right\} = \frac{1}{2} \cdot \frac{2}{\sqrt{3}} e^{t/2} \sin\left(\frac{\sqrt{3}}{2} t\right) = \frac{1}{\sqrt{3}} e^{t/2} \sin\left(\frac{\sqrt{3}}{2} t\right)${\mathcal{L}}^{-1}\left\{\frac{\frac{1}{2}}{{(s-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\right\}=\frac{1}{2}\cdot \frac{2}{\sqrt{3}}{e}^{t/2}\sin \left(\frac{\sqrt{3}}{2}t\right)=\frac{1}{\sqrt{3}}{e}^{t/2}\sin \left(\frac{\sqrt{3}}{2}t\right)$.

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Final Answer:

At any time $t$$t$tt$t$ (in seconds), the coterminous edges of a variable parallelepiped are represented by the vectors

Answer:

Problem Statement

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Part 1: Rate of Change of the Vectorial Area

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Part 2: Rate of Change of the Volume at $t=1$$t=1$t=1t = 1$t=1$

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Final Answers

1. Rate of change of the vectorial area:

2. Rate of change of the volume at $t=1$$t=1$t=1t = 1$t=1$:

A solid hemisphere rests in equilibrium on a solid sphere of equal radius. Determine the stability of the equilibrium in the two situations—(i) when the curved surface and (ii) when the flat surface of the hemisphere rests on the sphere.

Answer:

Problem Statement

1. Case (i): The curved surface of the hemisphere rests on the sphere.
2. Case (ii): The flat surface of the hemisphere rests on the sphere.

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Key Concepts

• Stable if the potential energy is at a minimum (small displacements increase the potential energy).
• Unstable if the potential energy is at a maximum (small displacements decrease the potential energy).

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Case (i): Curved Surface of the Hemisphere Rests on the Sphere

1. Geometry and Center of Mass (CM)

• Let the radius of both the hemisphere and the sphere be $R$$R$RR$R$.
• The hemisphere is placed such that its curved surface is in contact with the sphere.
• The CM of the hemisphere is at a height $h=\frac{3R}{8}$$h=\frac{3R}{8}$h=(3R)/(8)h = \frac{3R}{8}$h=\frac{3R}{8}$ from its flat face (standard result for a solid hemisphere).

2. Equilibrium Position

• In equilibrium, the hemisphere sits symmetrically on the sphere, with the CM directly above the contact point.
• The height of the CM above the contact point is:$$H=R+\frac{3R}{8}=\frac{11R}{8}.$$$$H=R+\frac{3R}{8}=\frac{11R}{8}.$$H=R+(3R)/(8)=(11 R)/(8).H = R + \frac{3R}{8} = \frac{11R}{8}.$$H=R+\frac{3R}{8}=\frac{11R}{8}.$$

3. Stability Analysis

• If the hemisphere is slightly tilted, the CM moves horizontally and vertically.
• The new height of the CM is $H^{\prime }=R\cos \theta +\frac{3R}{8}\cos \theta$$H^{\prime }=R\cos \theta +\frac{3R}{8}\cos \theta$H^(‘)=R cos theta+(3R)/(8)cos thetaH’ = R \cos \theta + \frac{3R}{8} \cos \theta$H^{\prime }=R\cos \theta +\frac{3R}{8}\cos \theta$, where $\theta$$\theta$theta\theta$\theta$ is the tilt angle.
• The potential energy $U$$U$UU$U$ is proportional to $H^{\prime }$$H^{\prime }$H^(‘)H’$H^{\prime }$, so:$$U(\theta )\propto (R+\frac{3R}{8})\cos \theta =\frac{11R}{8}\cos \theta .$$$$U(\theta )\propto \left(R+\frac{3R}{8}\right)\cos \theta =\frac{11R}{8}\cos \theta .$$U(theta)prop(R+(3R)/(8))cos theta=(11 R)/(8)cos theta.U(\theta) \propto \left( R + \frac{3R}{8} \right) \cos \theta = \frac{11R}{8} \cos \theta.$$U(\theta )\propto (R+\frac{3R}{8})\cos \theta =\frac{11R}{8}\cos \theta .$$
• For small $\theta$$\theta$theta\theta$\theta$, $\cos \theta \approx 1-\frac{{\theta }^2}{2}$$\cos \theta \approx 1-\frac{{\theta }^2}{2}$cos theta~~1-(theta^(2))/(2)\cos \theta \approx 1 – \frac{\theta^2}{2}$\cos \theta \approx 1-\frac{{\theta }^2}{2}$, so:$$U(\theta )\approx \text{constant}-\frac{11R}{16}{\theta }^2.$$$$U(\theta )\approx \text{constant}-\frac{11R}{16}{\theta }^2.$$U(theta)~~”constant”-(11 R)/(16)theta^(2).U(\theta) \approx \text{constant} – \frac{11R}{16} \theta^2.$$U(\theta )\approx \text{constant}-\frac{11R}{16}{\theta }^2.$$
• The potential energy decreases for small displacements ($\theta \ne 0$$\theta \ne 0$theta!=0\theta \neq 0$\theta \ne 0$), indicating unstable equilibrium.

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Case (ii): Flat Surface of the Hemisphere Rests on the Sphere

1. Geometry and Center of Mass (CM)

• The hemisphere is now placed with its flat surface on the sphere.
• The CM of the hemisphere is at a height $h=\frac{3R}{8}$$h=\frac{3R}{8}$h=(3R)/(8)h = \frac{3R}{8}$h=\frac{3R}{8}$ from the flat face.

2. Equilibrium Position

• The CM of the hemisphere is at a height $H=R-\frac{3R}{8}=\frac{5R}{8}$$H=R-\frac{3R}{8}=\frac{5R}{8}$H=R-(3R)/(8)=(5R)/(8)H = R – \frac{3R}{8} = \frac{5R}{8}$H=R-\frac{3R}{8}=\frac{5R}{8}$ above the contact point.

3. Stability Analysis

• If the hemisphere is slightly tilted, the CM moves horizontally and vertically.
• The new height of the CM is $H^{\prime }=R-\frac{3R}{8}\cos \theta$$H^{\prime }=R-\frac{3R}{8}\cos \theta$H^(‘)=R-(3R)/(8)cos thetaH’ = R – \frac{3R}{8} \cos \theta$H^{\prime }=R-\frac{3R}{8}\cos \theta$.
• The potential energy $U$$U$UU$U$ is proportional to $H^{\prime }$$H^{\prime }$H^(‘)H’$H^{\prime }$, so:$$U(\theta )\propto R-\frac{3R}{8}\cos \theta .$$$$U(\theta )\propto R-\frac{3R}{8}\cos \theta .$$U(theta)prop R-(3R)/(8)cos theta.U(\theta) \propto R – \frac{3R}{8} \cos \theta.$$U(\theta )\propto R-\frac{3R}{8}\cos \theta .$$
• For small $\theta$$\theta$theta\theta$\theta$, $\cos \theta \approx 1-\frac{{\theta }^2}{2}$$\cos \theta \approx 1-\frac{{\theta }^2}{2}$cos theta~~1-(theta^(2))/(2)\cos \theta \approx 1 – \frac{\theta^2}{2}$\cos \theta \approx 1-\frac{{\theta }^2}{2}$, so:$$U(\theta )\approx \text{constant}+\frac{3R}{16}{\theta }^2.$$$$U(\theta )\approx \text{constant}+\frac{3R}{16}{\theta }^2.$$U(theta)~~”constant”+(3R)/(16)theta^(2).U(\theta) \approx \text{constant} + \frac{3R}{16} \theta^2.$$U(\theta )\approx \text{constant}+\frac{3R}{16}{\theta }^2.$$
• The potential energy increases for small displacements ($\theta \ne 0$$\theta \ne 0$theta!=0\theta \neq 0$\theta \ne 0$), indicating stable equilibrium.

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Final Answers

1. Curved surface on sphere: The equilibrium is unstable.
   $\boxed{{\displaystyle \text{Unstable}}}$$\boxed{{\displaystyle \text{Unstable}}}$“Unstable”\boxed{\text{Unstable}}$\boxed{{\displaystyle \text{Unstable}}}$
2. Flat surface on sphere: The equilibrium is stable.
   $\boxed{{\displaystyle \text{Stable}}}$$\boxed{{\displaystyle \text{Stable}}}$“Stable”\boxed{\text{Stable}}$\boxed{{\displaystyle \text{Stable}}}$

(i) Let $C$$C$CC$C$ be a plane curve $\overline{r}(t)=f(t)\hat{i}+g(t)\hat{j}$$\overline{r}(t)=f(t)\widehat{i}+g(t)\widehat{j}$bar(r)(t)=f(t) hat(i)+g(t) hat(j)\bar{r}(t)=f(t)\hat{i}+g(t)\hat{j}$\overline{r}(t)=f(t)\hat{i}+g(t)\hat{j}$, where $f$$f$ff$f$ and $g$$g$gg$g$ have second-order derivatives. Show that the curvature at a point is given by

Answer:

Part (i): Curvature of a Plane Curve

1. First Derivative (Velocity Vector):
   $${\overline{r}}^{\prime }(t)=f^{\prime }(t)\hat{i}+g^{\prime }(t)\hat{j}.$$$${\overline{r}}^{\prime }(t)=f^{\prime }(t)\widehat{i}+g^{\prime }(t)\widehat{j}.$$bar(r)^(‘)(t)=f^(‘)(t) hat(i)+g^(‘)(t) hat(j).\bar{r}'(t) = f'(t)\hat{i} + g'(t)\hat{j}.$${\overline{r}}^{\prime }(t)=f^{\prime }(t)\hat{i}+g^{\prime }(t)\hat{j}.$$
   The speed $v(t)$$v(t)$v(t)v(t)$v(t)$ is:
   $$v(t)=\Vert {\overline{r}}^{\prime }(t)\Vert =\sqrt{[f^{\prime }(t){]}^2+[g^{\prime }(t){]}^2}.$$$$v(t)=\Vert {\overline{r}}^{\prime }(t)\Vert =\sqrt{[f^{\prime }(t){]}^2+[g^{\prime }(t){]}^2}.$$v(t)=|| bar(r)^(‘)(t)||=sqrt([f^(‘)(t)]^(2)+[g^(‘)(t)]^(2)).v(t) = \|\bar{r}'(t)\| = \sqrt{[f'(t)]^{2} + [g'(t)]^{2}}.$$v(t)=\Vert {\overline{r}}^{\prime }(t)\Vert =\sqrt{[f^{\prime }(t){]}^2+[g^{\prime }(t){]}^2}.$$
2. Second Derivative (Acceleration Vector):
   $${\overline{r}}^{″}(t)=f^{″}(t)\hat{i}+g^{″}(t)\hat{j}.$$$${\overline{r}}^{″}(t)=f^{″}(t)\widehat{i}+g^{″}(t)\widehat{j}.$$bar(r)^(″)(t)=f^(″)(t) hat(i)+g^(″)(t) hat(j).\bar{r}”(t) = f”(t)\hat{i} + g”(t)\hat{j}.$${\overline{r}}^{″}(t)=f^{″}(t)\hat{i}+g^{″}(t)\hat{j}.$$
3. Curvature Formula:
   The curvature $\kappa$$\kappa$kappa\kappa$\kappa$ is given by:
   $$\kappa =\frac{\Vert {\overline{r}}^{\prime }(t)\times {\overline{r}}^{″}(t)\Vert }{\Vert {\overline{r}}^{\prime }(t){\Vert }^3}.$$$$\kappa =\frac{\Vert {\overline{r}}^{\prime }(t)\times {\overline{r}}^{″}(t)\Vert }{\Vert {\overline{r}}^{\prime }(t){\Vert }^3}.$$kappa=(|| bar(r)^(‘)(t)xx bar(r)^(″)(t)||)/(|| bar(r)^(‘)(t)||^(3)).\kappa = \frac{\|\bar{r}'(t) \times \bar{r}”(t)\|}{\|\bar{r}'(t)\|^3}.$$\kappa =\frac{\Vert {\overline{r}}^{\prime }(t)\times {\overline{r}}^{″}(t)\Vert }{\Vert {\overline{r}}^{\prime }(t){\Vert }^3}.$$
   For plane curves, the cross product simplifies to:
   $${\overline{r}}^{\prime }(t)\times {\overline{r}}^{″}(t)=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ f^{\prime }(t)& g^{\prime }(t)& 0\\ f^{″}(t)& g^{″}(t)& 0\end{array}\right|=(f^{\prime }(t)g^{″}(t)-g^{\prime }(t)f^{″}(t))\hat{k}.$$$${\overline{r}}^{\prime }(t)\times {\overline{r}}^{″}(t)=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ f^{\prime }(t)& g^{\prime }(t)& 0\\ f^{″}(t)& g^{″}(t)& 0\end{array}\right|=\left(f^{\prime }(t)g^{″}(t)-g^{\prime }(t)f^{″}(t)\right)\widehat{k}.$$bar(r)^(‘)(t)xx bar(r)^(″)(t)=|[ hat(i), hat(j), hat(k)],[f^(‘)(t),g^(‘)(t),0],[f^(″)(t),g^(″)(t),0]|=(f^(‘)(t)g^(″)(t)-g^(‘)(t)f^(″)(t)) hat(k).\bar{r}'(t) \times \bar{r}”(t) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ f'(t) & g'(t) & 0 \\ f”(t) & g”(t) & 0 \end{vmatrix} = \left(f'(t)g”(t) – g'(t)f”(t)\right)\hat{k}.$${\overline{r}}^{\prime }(t)\times {\overline{r}}^{″}(t)=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ f^{\prime }(t)& g^{\prime }(t)& 0\\ f^{″}(t)& g^{″}(t)& 0\end{array}\right|=(f^{\prime }(t)g^{″}(t)-g^{\prime }(t)f^{″}(t))\hat{k}.$$
   Thus:
   $$\Vert {\overline{r}}^{\prime }(t)\times {\overline{r}}^{″}(t)\Vert =|f^{\prime }(t)g^{″}(t)-g^{\prime }(t)f^{″}(t)|.$$$$\Vert {\overline{r}}^{\prime }(t)\times {\overline{r}}^{″}(t)\Vert =\left|f^{\prime }(t)g^{″}(t)-g^{\prime }(t)f^{″}(t)\right|.$$|| bar(r)^(‘)(t)xx bar(r)^(″)(t)||=|f^(‘)(t)g^(″)(t)-g^(‘)(t)f^(″)(t)|.\|\bar{r}'(t) \times \bar{r}”(t)\| = \left|f'(t)g”(t) – g'(t)f”(t)\right|.$$\Vert {\overline{r}}^{\prime }(t)\times {\overline{r}}^{″}(t)\Vert =|f^{\prime }(t)g^{″}(t)-g^{\prime }(t)f^{″}(t)|.$$
   Substituting back:
   $$\kappa =\frac{|f^{\prime }(t)g^{″}(t)-g^{\prime }(t)f^{″}(t)|}{{([f^{\prime }(t){]}^2+[g^{\prime }(t){]}^2)}^{3/2}}.$$$$\kappa =\frac{\left|f^{\prime }(t)g^{″}(t)-g^{\prime }(t)f^{″}(t)\right|}{{\left([f^{\prime }(t){]}^2+[g^{\prime }(t){]}^2\right)}^{3/2}}.$$kappa=(|f^(‘)(t)g^(″)(t)-g^(‘)(t)f^(″)(t)|)/(([f^(‘)(t)]^(2)+[g^(‘)(t)]^(2))^(3//2)).\kappa = \frac{\left|f'(t)g”(t) – g'(t)f”(t)\right|}{\left([f'(t)]^{2} + [g'(t)]^{2}\right)^{3/2}}.$$\kappa =\frac{|f^{\prime }(t)g^{″}(t)-g^{\prime }(t)f^{″}(t)|}{{([f^{\prime }(t){]}^2+[g^{\prime }(t){]}^2)}^{3/2}}.$$

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Part (ii): Principal Normals and Torsion

1. Principal Normal Definition:
   The principal normal $\hat{N}(s)$$\widehat{N}(s)$hat(N)(s)\hat{N}(s)$\hat{N}(s)$ is the unit vector in the direction of the derivative of the tangent vector $\hat{T}(s)$$\widehat{T}(s)$hat(T)(s)\hat{T}(s)$\hat{T}(s)$:
   $$\hat{N}(s)=\frac{{\hat{T}}^{\prime }(s)}{\Vert {\hat{T}}^{\prime }(s)\Vert }.$$$$\widehat{N}(s)=\frac{{\widehat{T}}^{\prime }(s)}{\Vert {\widehat{T}}^{\prime }(s)\Vert }.$$hat(N)(s)=( hat(T)^(‘)(s))/(|| hat(T)^(‘)(s)||).\hat{N}(s) = \frac{\hat{T}'(s)}{\|\hat{T}'(s)\|}.$$\hat{N}(s)=\frac{{\hat{T}}^{\prime }(s)}{\Vert {\hat{T}}^{\prime }(s)\Vert }.$$
   For a curve parametrized by arc length $s$$s$ss$s$, the Frenet-Serret formulas give:
   $${\hat{T}}^{\prime }(s)=\kappa \hat{N}(s).$$$${\widehat{T}}^{\prime }(s)=\kappa \widehat{N}(s).$$hat(T)^(‘)(s)=kappa hat(N)(s).\hat{T}'(s) = \kappa \hat{N}(s).$${\hat{T}}^{\prime }(s)=\kappa \hat{N}(s).$$
2. Intersection Condition:
   Consider two consecutive points $\overline{r}(s)$$\overline{r}(s)$bar(r)(s)\bar{r}(s)$\overline{r}(s)$ and $\overline{r}(s+\mathrm{\Delta }s)$$\overline{r}(s+\mathrm{\Delta }s)$bar(r)(s+Delta s)\bar{r}(s + \Delta s)$\overline{r}(s+\mathrm{\Delta }s)$. The principal normals at these points are $\hat{N}(s)$$\widehat{N}(s)$hat(N)(s)\hat{N}(s)$\hat{N}(s)$ and $\hat{N}(s+\mathrm{\Delta }s)$$\widehat{N}(s+\mathrm{\Delta }s)$hat(N)(s+Delta s)\hat{N}(s + \Delta s)$\hat{N}(s+\mathrm{\Delta }s)$, respectively. The lines along these normals are:
   $$\text{At}s:\overline{r}(s)+\lambda \hat{N}(s),\lambda \in \mathbb{R},$$$$\text{At }s:\overline{r}(s)+\lambda \widehat{N}(s),\lambda \in \mathbb{R},$$“At “s:quad bar(r)(s)+lambda hat(N)(s),quad lambda inR,\text{At } s: \quad \bar{r}(s) + \lambda \hat{N}(s), \quad \lambda \in \mathbb{R},$$\text{At }s:\overline{r}(s)+\lambda \hat{N}(s),\lambda \in \mathbb{R},$$
   $$\text{At}s+\mathrm{\Delta }s:\overline{r}(s+\mathrm{\Delta }s)+\mu \hat{N}(s+\mathrm{\Delta }s),\mu \in \mathbb{R}.$$$$\text{At }s+\mathrm{\Delta }s:\overline{r}(s+\mathrm{\Delta }s)+\mu \widehat{N}(s+\mathrm{\Delta }s),\mu \in \mathbb{R}.$$“At “s+Delta s:quad bar(r)(s+Delta s)+mu hat(N)(s+Delta s),quad mu inR.\text{At } s + \Delta s: \quad \bar{r}(s + \Delta s) + \mu \hat{N}(s + \Delta s), \quad \mu \in \mathbb{R}.$$\text{At }s+\mathrm{\Delta }s:\overline{r}(s+\mathrm{\Delta }s)+\mu \hat{N}(s+\mathrm{\Delta }s),\mu \in \mathbb{R}.$$
   For these lines to intersect, there must exist $\lambda$$\lambda$lambda\lambda$\lambda$ and $\mu$$\mu$mu\mu$\mu$ such that:
   $$\overline{r}(s)+\lambda \hat{N}(s)=\overline{r}(s+\mathrm{\Delta }s)+\mu \hat{N}(s+\mathrm{\Delta }s).$$$$\overline{r}(s)+\lambda \widehat{N}(s)=\overline{r}(s+\mathrm{\Delta }s)+\mu \widehat{N}(s+\mathrm{\Delta }s).$$bar(r)(s)+lambda hat(N)(s)= bar(r)(s+Delta s)+mu hat(N)(s+Delta s).\bar{r}(s) + \lambda \hat{N}(s) = \bar{r}(s + \Delta s) + \mu \hat{N}(s + \Delta s).$$\overline{r}(s)+\lambda \hat{N}(s)=\overline{r}(s+\mathrm{\Delta }s)+\mu \hat{N}(s+\mathrm{\Delta }s).$$
   Expanding $\overline{r}(s+\mathrm{\Delta }s)$$\overline{r}(s+\mathrm{\Delta }s)$bar(r)(s+Delta s)\bar{r}(s + \Delta s)$\overline{r}(s+\mathrm{\Delta }s)$ to first order:
   $$\overline{r}(s+\mathrm{\Delta }s)\approx \overline{r}(s)+\mathrm{\Delta }s\hat{T}(s).$$$$\overline{r}(s+\mathrm{\Delta }s)\approx \overline{r}(s)+\mathrm{\Delta }s\widehat{T}(s).$$bar(r)(s+Delta s)~~ bar(r)(s)+Delta s hat(T)(s).\bar{r}(s + \Delta s) \approx \bar{r}(s) + \Delta s \hat{T}(s).$$\overline{r}(s+\mathrm{\Delta }s)\approx \overline{r}(s)+\mathrm{\Delta }s\hat{T}(s).$$
   Thus, the intersection condition becomes:
   $$\lambda \hat{N}(s)-\mu \hat{N}(s+\mathrm{\Delta }s)\approx \mathrm{\Delta }s\hat{T}(s).$$$$\lambda \widehat{N}(s)-\mu \widehat{N}(s+\mathrm{\Delta }s)\approx \mathrm{\Delta }s\widehat{T}(s).$$lambda hat(N)(s)-mu hat(N)(s+Delta s)~~Delta s hat(T)(s).\lambda \hat{N}(s) – \mu \hat{N}(s + \Delta s) \approx \Delta s \hat{T}(s).$$\lambda \hat{N}(s)-\mu \hat{N}(s+\mathrm{\Delta }s)\approx \mathrm{\Delta }s\hat{T}(s).$$
   Expanding $\hat{N}(s+\mathrm{\Delta }s)$$\widehat{N}(s+\mathrm{\Delta }s)$hat(N)(s+Delta s)\hat{N}(s + \Delta s)$\hat{N}(s+\mathrm{\Delta }s)$ using the Frenet-Serret formula ${\hat{N}}^{\prime }(s)=-\kappa \hat{T}(s)+\tau \hat{B}(s)$${\widehat{N}}^{\prime }(s)=-\kappa \widehat{T}(s)+\tau \widehat{B}(s)$hat(N)^(‘)(s)=-kappa hat(T)(s)+tau hat(B)(s)\hat{N}'(s) = -\kappa \hat{T}(s) + \tau \hat{B}(s)${\hat{N}}^{\prime }(s)=-\kappa \hat{T}(s)+\tau \hat{B}(s)$:
   $$\hat{N}(s+\mathrm{\Delta }s)\approx \hat{N}(s)+\mathrm{\Delta }s(-\kappa \hat{T}(s)+\tau \hat{B}(s)).$$$$\widehat{N}(s+\mathrm{\Delta }s)\approx \widehat{N}(s)+\mathrm{\Delta }s(-\kappa \widehat{T}(s)+\tau \widehat{B}(s)).$$hat(N)(s+Delta s)~~ hat(N)(s)+Delta s(-kappa hat(T)(s)+tau hat(B)(s)).\hat{N}(s + \Delta s) \approx \hat{N}(s) + \Delta s (-\kappa \hat{T}(s) + \tau \hat{B}(s)).$$\hat{N}(s+\mathrm{\Delta }s)\approx \hat{N}(s)+\mathrm{\Delta }s(-\kappa \hat{T}(s)+\tau \hat{B}(s)).$$
   Substituting back:
   $$\lambda \hat{N}(s)-\mu (\hat{N}(s)-\mathrm{\Delta }s\kappa \hat{T}(s)+\mathrm{\Delta }s\tau \hat{B}(s))\approx \mathrm{\Delta }s\hat{T}(s).$$$$\lambda \widehat{N}(s)-\mu \left(\widehat{N}(s)-\mathrm{\Delta }s\kappa \widehat{T}(s)+\mathrm{\Delta }s\tau \widehat{B}(s)\right)\approx \mathrm{\Delta }s\widehat{T}(s).$$lambda hat(N)(s)-mu(( hat(N))(s)-Delta s kappa( hat(T))(s)+Delta s tau( hat(B))(s))~~Delta s hat(T)(s).\lambda \hat{N}(s) – \mu \left(\hat{N}(s) – \Delta s \kappa \hat{T}(s) + \Delta s \tau \hat{B}(s)\right) \approx \Delta s \hat{T}(s).$$\lambda \hat{N}(s)-\mu (\hat{N}(s)-\mathrm{\Delta }s\kappa \hat{T}(s)+\mathrm{\Delta }s\tau \hat{B}(s))\approx \mathrm{\Delta }s\hat{T}(s).$$
   Collecting terms:
   $$(\lambda -\mu )\hat{N}(s)+\mu \mathrm{\Delta }s\kappa \hat{T}(s)-\mu \mathrm{\Delta }s\tau \hat{B}(s)\approx \mathrm{\Delta }s\hat{T}(s).$$$$(\lambda -\mu )\widehat{N}(s)+\mu \mathrm{\Delta }s\kappa \widehat{T}(s)-\mu \mathrm{\Delta }s\tau \widehat{B}(s)\approx \mathrm{\Delta }s\widehat{T}(s).$$(lambda-mu) hat(N)(s)+mu Delta s kappa hat(T)(s)-mu Delta s tau hat(B)(s)~~Delta s hat(T)(s).(\lambda – \mu)\hat{N}(s) + \mu \Delta s \kappa \hat{T}(s) – \mu \Delta s \tau \hat{B}(s) \approx \Delta s \hat{T}(s).$$(\lambda -\mu )\hat{N}(s)+\mu \mathrm{\Delta }s\kappa \hat{T}(s)-\mu \mathrm{\Delta }s\tau \hat{B}(s)\approx \mathrm{\Delta }s\hat{T}(s).$$
   For this to hold, the coefficients of $\hat{N}(s)$$\widehat{N}(s)$hat(N)(s)\hat{N}(s)$\hat{N}(s)$, $\hat{T}(s)$$\widehat{T}(s)$hat(T)(s)\hat{T}(s)$\hat{T}(s)$, and $\hat{B}(s)$$\widehat{B}(s)$hat(B)(s)\hat{B}(s)$\hat{B}(s)$ must match:
   $$\lambda -\mu =0,\mu \mathrm{\Delta }s\kappa =\mathrm{\Delta }s,-\mu \mathrm{\Delta }s\tau =0.$$$$\lambda -\mu =0,\mu \mathrm{\Delta }s\kappa =\mathrm{\Delta }s,-\mu \mathrm{\Delta }s\tau =0.$$lambda-mu=0,quad mu Delta s kappa=Delta s,quad-mu Delta s tau=0.\lambda – \mu = 0, \quad \mu \Delta s \kappa = \Delta s, \quad -\mu \Delta s \tau = 0.$$\lambda -\mu =0,\mu \mathrm{\Delta }s\kappa =\mathrm{\Delta }s,-\mu \mathrm{\Delta }s\tau =0.$$
   From the third equation, $\tau =0$$\tau =0$tau=0\tau = 0$\tau =0$ (since $\mu \ne 0$$\mu \ne 0$mu!=0\mu \neq 0$\mu \ne 0$ and $\mathrm{\Delta }s\ne 0$$\mathrm{\Delta }s\ne 0$Delta s!=0\Delta s \neq 0$\mathrm{\Delta }s\ne 0$).

A regular tetrahedron, formed of six light rods, each of length $l$$l$ll$l$, rests on a smooth horizontal plane. A ring of weight $W$$W$WW$W$ and radius $r$$r$rr$r$ is supported by the slant sides. Using the principle of virtual work, find the stress in any of the horizontal sides.

Answer:

Problem Statement

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Step 1: Understand the Geometry

1. Regular Tetrahedron:
   • A regular tetrahedron has 4 triangular faces, 6 edges (all of length $l$$l$ll$l$), and 4 vertices.
   • Let the tetrahedron rest on a horizontal plane with one face (base) on the plane and the other three faces (slant sides) meeting at the apex.
2. Ring Placement:
   • The ring of weight $W$$W$WW$W$ and radius $r$$r$rr$r$ is placed symmetrically on the three slant sides.
   • The ring touches all three slant sides, and its center lies along the vertical line from the apex to the base.

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Step 2: Define Variables and Forces

1. Forces:
   • Let $T$$T$TT$T$ be the stress (tension) in any of the three horizontal rods (edges of the base).
   • The ring exerts a normal force $N$$N$NN$N$ on each of the three slant sides. By symmetry, these forces are equal.
   • The weight $W$$W$WW$W$ of the ring is balanced by the vertical components of the normal forces $N$$N$NN$N$.
2. Geometry of Forces:
   • The angle $\theta$$\theta$theta\theta$\theta$ between a slant side and the horizontal plane can be determined from the tetrahedron’s geometry.
   • For a regular tetrahedron with edge length $l$$l$ll$l$, the height $h$$h$hh$h$ of the apex above the base is:$$h=\sqrt{\frac{2}{3}}l.$$$$h=\sqrt{\frac{2}{3}}l.$$h=sqrt((2)/(3))l.h = \sqrt{\frac{2}{3}} l.$$h=\sqrt{\frac{2}{3}}l.$$
   • The angle $\theta$$\theta$theta\theta$\theta$ satisfies:$$\sin \theta =\frac{h}{l}=\sqrt{\frac{2}{3}},\cos \theta =\sqrt{\frac{1}{3}}.$$$$\sin \theta =\frac{h}{l}=\sqrt{\frac{2}{3}},\cos \theta =\sqrt{\frac{1}{3}}.$$sin theta=(h)/(l)=sqrt((2)/(3)),quad cos theta=sqrt((1)/(3)).\sin \theta = \frac{h}{l} = \sqrt{\frac{2}{3}}, \quad \cos \theta = \sqrt{\frac{1}{3}}.$$\sin \theta =\frac{h}{l}=\sqrt{\frac{2}{3}},\cos \theta =\sqrt{\frac{1}{3}}.$$
3. Equilibrium of the Ring:
   • The vertical forces balance:$$3N\cos \theta =W⟹N=\frac{W}{3\cos \theta }=\frac{W\sqrt{3}}{3}.$$$$3N\cos \theta =W⟹N=\frac{W}{3\cos \theta }=\frac{W\sqrt{3}}{3}.$$3N cos theta=WLongrightarrowN=(W)/(3cos theta)=(Wsqrt3)/(3).3N \cos \theta = W \implies N = \frac{W}{3 \cos \theta} = \frac{W \sqrt{3}}{3}.$$3N\cos \theta =W⟹N=\frac{W}{3\cos \theta }=\frac{W\sqrt{3}}{3}.$$
   • The horizontal components of $N$$N$NN$N$ are balanced by the stresses in the horizontal rods.

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Step 3: Apply the Principle of Virtual Work

1. Virtual Displacement:
   • Imagine a small virtual displacement where the apex is raised by $\delta y$$\delta y$delta y\delta y$\delta y$, causing the horizontal rods to stretch by $\delta x$$\delta x$delta x\delta x$\delta x$.
   • The relationship between $\delta y$$\delta y$delta y\delta y$\delta y$ and $\delta x$$\delta x$delta x\delta x$\delta x$ is derived from the geometry:$$\delta x=\delta y\cot \theta =\delta y\sqrt{\frac{1}{2}}.$$$$\delta x=\delta y\cot \theta =\delta y\sqrt{\frac{1}{2}}.$$delta x=delta y cot theta=delta ysqrt((1)/(2)).\delta x = \delta y \cot \theta = \delta y \sqrt{\frac{1}{2}}.$$\delta x=\delta y\cot \theta =\delta y\sqrt{\frac{1}{2}}.$$
2. Work Done:
   • The work done by the weight $W$$W$WW$W$ is:$$\delta W_{\text{weight}}=-W\delta y.$$$$\delta W_{\text{weight}}=-W\delta y.$$deltaW_(“weight”)=-W delta y.\delta W_{\text{weight}} = -W \delta y.$$\delta W_{\text{weight}}=-W\delta y.$$
   • The work done by the stresses $T$$T$TT$T$ in the three horizontal rods is:$$\delta W_{\text{stress}}=3T\delta x=3T\delta y\sqrt{\frac{1}{2}}.$$$$\delta W_{\text{stress}}=3T\delta x=3T\delta y\sqrt{\frac{1}{2}}.$$deltaW_(“stress”)=3T delta x=3T delta ysqrt((1)/(2)).\delta W_{\text{stress}} = 3T \delta x = 3T \delta y \sqrt{\frac{1}{2}}.$$\delta W_{\text{stress}}=3T\delta x=3T\delta y\sqrt{\frac{1}{2}}.$$
3. Principle of Virtual Work:
   • The total virtual work must be zero for equilibrium:$$\delta W_{\text{weight}}+\delta W_{\text{stress}}=0.$$$$\delta W_{\text{weight}}+\delta W_{\text{stress}}=0.$$deltaW_(“weight”)+deltaW_(“stress”)=0.\delta W_{\text{weight}} + \delta W_{\text{stress}} = 0.$$\delta W_{\text{weight}}+\delta W_{\text{stress}}=0.$$$$-W\delta y+3T\delta y\sqrt{\frac{1}{2}}=0.$$$$-W\delta y+3T\delta y\sqrt{\frac{1}{2}}=0.$$-W delta y+3T delta ysqrt((1)/(2))=0.-W \delta y + 3T \delta y \sqrt{\frac{1}{2}} = 0.$$-W\delta y+3T\delta y\sqrt{\frac{1}{2}}=0.$$$$3T\sqrt{\frac{1}{2}}=W⟹T=\frac{W}{3}\sqrt{2}.$$$$3T\sqrt{\frac{1}{2}}=W⟹T=\frac{W}{3}\sqrt{2}.$$3Tsqrt((1)/(2))=WLongrightarrowT=(W)/(3)sqrt2.3T \sqrt{\frac{1}{2}} = W \implies T = \frac{W}{3} \sqrt{2}.$$3T\sqrt{\frac{1}{2}}=W⟹T=\frac{W}{3}\sqrt{2}.$$

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Step 4: Final Answer

A particle executes simple harmonic motion such that in two of its positions, velocities are $u$$u$uu$u$ and $v$$v$vv$v$, and the two corresponding accelerations are $f_1$$f_1$f_(1)f_{1}$f_1$ and $f_2$$f_2$f_(2)f_{2}$f_2$. For what value(s) of $k$$k$kk$k$, the distance between the two positions is $k(v^2-u^2)$$k\left(v^2-u^2\right)$k(v^(2)-u^(2))k\left(v^{2}-u^{2}\right)$k(v^2-u^2)$? Show also that the amplitude of the motion is

Answer:

Step-by-Step Solution

Part 1: Finding the value(s) of $k$$k$kk$k$

• At position $x_1$$x_1$x_(1)x_1$x_1$:
  $$u^2={\omega }^2(A^2-x_1^2),f_1=-{\omega }^2{x}_1.$$$$u^2={\omega }^2(A^2-x_1^2),f_1=-{\omega }^2{x}_1.$$u^(2)=omega^(2)(A^(2)-x_(1)^(2)),quadf_(1)=-omega^(2)x_(1).u^2 = \omega^2 (A^2 – x_1^2), \quad f_1 = -\omega^2 x_1.$$u^2={\omega }^2(A^2-x_1^2),f_1=-{\omega }^2{x}_1.$$
• At position $x_2$$x_2$x_(2)x_2$x_2$:
  $$v^2={\omega }^2(A^2-x_2^2),f_2=-{\omega }^2{x}_2.$$$$v^2={\omega }^2(A^2-x_2^2),f_2=-{\omega }^2{x}_2.$$v^(2)=omega^(2)(A^(2)-x_(2)^(2)),quadf_(2)=-omega^(2)x_(2).v^2 = \omega^2 (A^2 – x_2^2), \quad f_2 = -\omega^2 x_2.$$v^2={\omega }^2(A^2-x_2^2),f_2=-{\omega }^2{x}_2.$$

Part 2: Deriving the Amplitude

Edge Cases

• If $f_1+f_2=0$$f_1+f_2=0$f_(1)+f_(2)=0f_1 + f_2 = 0$f_1+f_2=0$, then $k$$k$kk$k$ is undefined unless the distance $|x_2-x_1|=0$$|x_2-x_1|=0$|x_(2)-x_(1)|=0|x_2 – x_1| = 0$|x_2-x_1|=0$, which implies $x_1=x_2$$x_1=x_2$x_(1)=x_(2)x_1 = x_2$x_1=x_2$, but this contradicts distinct positions unless velocities and accelerations align specially.
• If $f_2^2-f_1^2=0$$f_2^2-f_1^2=0$f_(2)^(2)-f_(1)^(2)=0f_2^2 – f_1^2 = 0$f_2^2-f_1^2=0$, the amplitude expression is undefined, suggesting the points may not be distinct or parameters are equal, which we assume is not the case for a valid SHM.

Final Answer

Find the second solution of the differential equation $x{y}^{″}+(x-1)y^{\prime }-y=0$$x{y}^{″}+(x-1)y^{\prime }-y=0$xy^(″)+(x-1)y^(‘)-y=0x y”+(x-1) y’-y = 0$x{y}^{″}+(x-1)y^{\prime }-y=0$ using $u(x)=-e^{-x}$$u(x)=-e^{-x}$u(x)=-e^(-x)u(x)=-e^{-x}$u(x)=-e^{-x}$ as one of the solutions.

Answer:

Find the general solution of the differential equation $x^2{y}^{″}-2x{y}^{\prime }+2y=x^3\sin x$$x^2{y}^{″}-2x{y}^{\prime }+2y=x^3\sin x$x^(2)y^(″)-2xy^(‘)+2y=x^(3)sin xx^{2} y”-2x y’+2y = x^{3}\sin x$x^2{y}^{″}-2x{y}^{\prime }+2y=x^3\sin x$ by the method of variation of parameters.

Answer:

Given Differential Equation:

Step 1: Solve the Homogeneous Equation

Step 2: Apply Variation of Parameters

Step 3: Construct the Particular Solution

Step 4: General Solution

Final Answer:

State the uniqueness theorem for the existence of a unique solution of the initial value problem ${\displaystyle \frac{dy}{dx}}=f(x,y),y(x_0)=y_0$${\displaystyle \frac{dy}{dx}}=f(x,y),y(x_0)=y_0$(dy)/(dx)=f(x,y),y(x_(0))=y_(0)\dfrac{dy}{dx}=f(x,y),\; y(x_{0})=y_{0}${\displaystyle \frac{dy}{dx}}=f(x,y),y(x_0)=y_0$ in the rectangular region $R:|x-x_0|\le a,|y-y_0|\le b$$R:|x-x_0|\le a,|y-y_0|\le b$R:|x-x_(0)| <= a,|y-y_(0)| <= bR:\,|x-x_{0}|\le a,\;|y-y_{0}|\le b$R:|x-x_0|\le a,|y-y_0|\le b$. Test the existence and uniqueness of the solution of the initial value problem ${\displaystyle \frac{dy}{dx}}=2\sqrt{y},y(1)=0$${\displaystyle \frac{dy}{dx}}=2\sqrt{y},y(1)=0$(dy)/(dx)=2sqrty,y(1)=0\dfrac{dy}{dx}=2\sqrt{y},\; y(1)=0${\displaystyle \frac{dy}{dx}}=2\sqrt{y},y(1)=0$, in a suitable rectangle $R$$R$RR$R$. If more than one solution exists, then find all the solutions.

Answer:

Uniqueness Theorem for Initial Value Problems (IVP)

1. Existence of $f(x,y)$$f(x,y)$f(x,y)f(x, y)$f(x,y)$:
   The function $f(x,y)$$f(x,y)$f(x,y)f(x, y)$f(x,y)$ is continuous in $R$$R$RR$R$.
2. Lipschitz Condition (Uniqueness):
   There exists a constant $L>0$$L>0$L > 0L > 0$L>0$ such that for all $(x,y_1),(x,y_2)\in R$$(x,y_1),(x,y_2)\in R$(x,y_(1)),(x,y_(2))in R(x, y_1), (x, y_2) \in R$(x,y_1),(x,y_2)\in R$,
   $$|f(x,y_1)-f(x,y_2)|\le L|y_1-y_2|.$$$$|f(x,y_1)-f(x,y_2)|\le L|y_1-y_2|.$$|f(x,y_(1))-f(x,y_(2))| <= L|y_(1)-y_(2)|.|f(x, y_1) – f(x, y_2)| \leq L |y_1 – y_2|.$$|f(x,y_1)-f(x,y_2)|\le L|y_1-y_2|.$$
   (This is satisfied if $\frac{\partial f}{\partial y}$$\frac{\partial f}{\partial y}$(del f)/(del y)\frac{\partial f}{\partial y}$\frac{\partial f}{\partial y}$ is continuous and bounded in $R$$R$RR$R$.)

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Testing the IVP: $\frac{dy}{dx}=2\sqrt{y},y(1)=0$$\frac{dy}{dx}=2\sqrt{y},y(1)=0$(dy)/(dx)=2sqrty,quad y(1)=0\frac{dy}{dx} = 2\sqrt{y}, \quad y(1) = 0$\frac{dy}{dx}=2\sqrt{y},y(1)=0$

1. Check Existence and Uniqueness in a Rectangle $R$$R$RR$R$:

• Function: $f(x,y)=2\sqrt{y}$$f(x,y)=2\sqrt{y}$f(x,y)=2sqrtyf(x, y) = 2\sqrt{y}$f(x,y)=2\sqrt{y}$.
• Continuity:
  $f(x,y)$$f(x,y)$f(x,y)f(x, y)$f(x,y)$ is continuous for $y\ge 0$$y\ge 0$y >= 0y \geq 0$y\ge 0$. At $y=0$$y=0$y=0y = 0$y=0$, $f(x,0)=0$$f(x,0)=0$f(x,0)=0f(x, 0) = 0$f(x,0)=0$, so it’s continuous in any rectangle $R$$R$RR$R$ containing $(1,0)$$(1,0)$(1,0)(1, 0)$(1,0)$ with $y\ge 0$$y\ge 0$y >= 0y \geq 0$y\ge 0$.
• Lipschitz Condition:
  Compute the partial derivative:
  $$\frac{\partial f}{\partial y}=\frac{1}{\sqrt{y}}.$$$$\frac{\partial f}{\partial y}=\frac{1}{\sqrt{y}}.$$(del f)/(del y)=(1)/(sqrty).\frac{\partial f}{\partial y} = \frac{1}{\sqrt{y}}.$$\frac{\partial f}{\partial y}=\frac{1}{\sqrt{y}}.$$
  This is unbounded as $y\to 0^{+}$$y\to 0^{+}$y rarr0^(+)y \to 0^+$y\to 0^{+}$. Thus, $f(x,y)$$f(x,y)$f(x,y)f(x, y)$f(x,y)$ does not satisfy a Lipschitz condition in any rectangle containing $y=0$$y=0$y=0y = 0$y=0$.

2. Find All Solutions:

• For $x<1$$x<1$x < 1x < 1$x<1$, $y=0⟹y^{\prime }=0=2\sqrt{0}$$y=0⟹y^{\prime }=0=2\sqrt{0}$y=0Longrightarrowy^(‘)=0=2sqrt0y = 0 \implies y’ = 0 = 2\sqrt{0}$y=0⟹y^{\prime }=0=2\sqrt{0}$.
• For $x>1$$x>1$x > 1x > 1$x>1$, $y=(x-1{)}^2⟹y^{\prime }=2(x-1)=2\sqrt{(x-1{)}^2}=2\sqrt{y}$$y=(x-1{)}^2⟹y^{\prime }=2(x-1)=2\sqrt{(x-1{)}^2}=2\sqrt{y}$y=(x-1)^(2)Longrightarrowy^(‘)=2(x-1)=2sqrt((x-1)^(2))=2sqrtyy = (x – 1)^2 \implies y’ = 2(x – 1) = 2\sqrt{(x – 1)^2} = 2\sqrt{y}$y=(x-1{)}^2⟹y^{\prime }=2(x-1)=2\sqrt{(x-1{)}^2}=2\sqrt{y}$.
• At $x=1$$x=1$x=1x = 1$x=1$, both branches give $y(1)=0$$y(1)=0$y(1)=0y(1) = 0$y(1)=0$, and the derivative from the right is $y^{\prime }(1^{+})=0$$y^{\prime }(1^{+})=0$y^(‘)(1^(+))=0y'(1^+) = 0$y^{\prime }(1^{+})=0$, matching the left derivative $y^{\prime }(1^{-})=0$$y^{\prime }(1^{-})=0$y^(‘)(1^(-))=0y'(1^-) = 0$y^{\prime }(1^{-})=0$.

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Final Answer:

1. The trivial solution:
   $$y(x)=0\text{(for all}x).$$$$y(x)=0\text{(for all }x).$$y(x)=0quad(for all x).y(x) = 0 \quad \text{(for all } x).$$y(x)=0\text{(for all }x).$$
2. A family of piecewise solutions for any $c\ge 1$$c\ge 1$c >= 1c \geq 1$c\ge 1$:
   $$y(x)=\{\begin{array}{ll}0& \text{if}x\le c,\\ (x-c{)}^2& \text{if}x\ge c.\end{array}$$$$y(x)=\left\{\begin{array}{l}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if }x\le c,\\ (x-c{)}^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if }x\ge c.\end{array}\right.$$y(x)={[0,”if “x <= c”,”],[(x-c)^(2),”if “x >= c.]:}y(x) = \begin{cases} 0 & \text{if } x \leq c, \\ (x – c)^2 & \text{if } x \geq c. \end{cases}$$y(x)=\{\begin{array}{ll}0& \text{if }x\le c,\\ (x-c{)}^2& \text{if }x\ge c.\end{array}$$

A heavy particle hanging vertically from a fixed point by a light inextensible string of length $l$$l$ll$l$ starts to move with initial velocity $u$$u$uu$u$ in a circle so as to make a complete revolution in a vertical plane. Show that the sum of tensions at the ends of any diameter is constant.

Answer:

Problem Statement:

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Approach:

1. Understand the Motion:
   • The particle moves in a vertical circle of radius $l$$l$ll$l$.
   • The forces acting on the particle are:
     • Tension $T$$T$TT$T$ in the string (always radially inward).
     • Gravitational force $mg$$mg$mgmg$mg$ (vertically downward).
   • The motion is governed by the interplay of tension and gravity.
2. Energy Considerations:
   • Since the string is inextensible and no other non-conservative forces are acting, mechanical energy is conserved.
   • The total energy $E$$E$EE$E$ at any point is the sum of kinetic and potential energy.
3. Tension in the String:
   • The tension $T$$T$TT$T$ varies with the position of the particle in the circle.
   • At any point, the net radial force provides the centripetal acceleration:$$T-mg\cos \theta =\frac{m{v}^2}{l},$$$$T-mg\cos \theta =\frac{m{v}^2}{l},$$T-mg cos theta=(mv^(2))/(l),T – mg \cos \theta = \frac{mv^2}{l},$$T-mg\cos \theta =\frac{m{v}^2}{l},$$where $\theta$$\theta$theta\theta$\theta$ is the angle the string makes with the vertical, and $v$$v$vv$v$ is the speed of the particle at that point.
4. Sum of Tensions at Ends of a Diameter:
   • Consider two points on the circle that are diametrically opposite (e.g., top and bottom).
   • Calculate the tensions $T_1$$T_1$T_(1)T_1$T_1$ and $T_2$$T_2$T_(2)T_2$T_2$ at these points.
   • Show that $T_1+T_2$$T_1+T_2$T_(1)+T_(2)T_1 + T_2$T_1+T_2$ is constant.

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Detailed Solution:

1. Conservation of Energy:

• At the lowest point (A):
  • Height $h=0$$h=0$h=0h = 0$h=0$.
  • Speed $v=u$$v=u$v=uv = u$v=u$.
  • Total energy $E=\frac{1}{2}m{u}^2$$E=\frac{1}{2}m{u}^2$E=(1)/(2)mu^(2)E = \frac{1}{2} m u^2$E=\frac{1}{2}m{u}^2$.
• At any other point at an angle $\theta$$\theta$theta\theta$\theta$ from the vertical:
  • Height $h=l(1-\cos \theta )$$h=l(1-\cos \theta )$h=l(1-cos theta)h = l (1 – \cos \theta)$h=l(1-\cos \theta )$.
  • Speed $v$$v$vv$v$.
  • Total energy $E=\frac{1}{2}m{v}^2+mgl(1-\cos \theta )$$E=\frac{1}{2}m{v}^2+mgl(1-\cos \theta )$E=(1)/(2)mv^(2)+mgl(1-cos theta)E = \frac{1}{2} m v^2 + m g l (1 – \cos \theta)$E=\frac{1}{2}m{v}^2+mgl(1-\cos \theta )$.

2. Tension at Any Point:

3. Tensions at Ends of a Diameter:

• Point 1 (Bottom, $\theta =0$$\theta =0$theta=0\theta = 0$\theta =0$):$$T_1=\frac{m{u}^2}{l}-2mg+3mg\cos 0=\frac{m{u}^2}{l}-2mg+3mg=\frac{m{u}^2}{l}+mg.$$$$T_1=\frac{m{u}^2}{l}-2mg+3mg\cos 0=\frac{m{u}^2}{l}-2mg+3mg=\frac{m{u}^2}{l}+mg.$$T_(1)=(mu^(2))/(l)-2mg+3mg cos 0=(mu^(2))/(l)-2mg+3mg=(mu^(2))/(l)+mg.T_1 = \frac{m u^2}{l} – 2 m g + 3 m g \cos 0 = \frac{m u^2}{l} – 2 m g + 3 m g = \frac{m u^2}{l} + m g.$$T_1=\frac{m{u}^2}{l}-2mg+3mg\cos 0=\frac{m{u}^2}{l}-2mg+3mg=\frac{m{u}^2}{l}+mg.$$
• Point 2 (Top, $\theta =\pi$$\theta =\pi$theta=pi\theta = \pi$\theta =\pi$):$$T_2=\frac{m{u}^2}{l}-2mg+3mg\cos \pi =\frac{m{u}^2}{l}-2mg-3mg=\frac{m{u}^2}{l}-5mg.$$$$T_2=\frac{m{u}^2}{l}-2mg+3mg\cos \pi =\frac{m{u}^2}{l}-2mg-3mg=\frac{m{u}^2}{l}-5mg.$$T_(2)=(mu^(2))/(l)-2mg+3mg cos pi=(mu^(2))/(l)-2mg-3mg=(mu^(2))/(l)-5mg.T_2 = \frac{m u^2}{l} – 2 m g + 3 m g \cos \pi = \frac{m u^2}{l} – 2 m g – 3 m g = \frac{m u^2}{l} – 5 m g.$$T_2=\frac{m{u}^2}{l}-2mg+3mg\cos \pi =\frac{m{u}^2}{l}-2mg-3mg=\frac{m{u}^2}{l}-5mg.$$

• At angle $\theta$$\theta$theta\theta$\theta$:$$T(\theta )=\frac{m{u}^2}{l}-2mg+3mg\cos \theta .$$$$T(\theta )=\frac{m{u}^2}{l}-2mg+3mg\cos \theta .$$T(theta)=(mu^(2))/(l)-2mg+3mg cos theta.T(\theta) = \frac{m u^2}{l} – 2 m g + 3 m g \cos \theta.$$T(\theta )=\frac{m{u}^2}{l}-2mg+3mg\cos \theta .$$
• At angle $\theta +\pi$$\theta +\pi$theta+pi\theta + \pi$\theta +\pi$:$$T(\theta +\pi )=\frac{m{u}^2}{l}-2mg+3mg\cos (\theta +\pi )=\frac{m{u}^2}{l}-2mg-3mg\cos \theta .$$$$T(\theta +\pi )=\frac{m{u}^2}{l}-2mg+3mg\cos (\theta +\pi )=\frac{m{u}^2}{l}-2mg-3mg\cos \theta .$$T(theta+pi)=(mu^(2))/(l)-2mg+3mg cos(theta+pi)=(mu^(2))/(l)-2mg-3mg cos theta.T(\theta + \pi) = \frac{m u^2}{l} – 2 m g + 3 m g \cos (\theta + \pi) = \frac{m u^2}{l} – 2 m g – 3 m g \cos \theta.$$T(\theta +\pi )=\frac{m{u}^2}{l}-2mg+3mg\cos (\theta +\pi )=\frac{m{u}^2}{l}-2mg-3mg\cos \theta .$$

4. Condition for Complete Revolution:

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Final Answer:

State Stokes’ theorem and verify it for the vector field $\overrightarrow{F}=xy\hat{i}+yz\hat{j}+zx\hat{k}$$\overrightarrow{F}=xy\widehat{i}+yz\widehat{j}+zx\widehat{k}$vec(F)=xy hat(i)+yz hat(j)+zx hat(k)\vec{F}=x y \hat{i}+y z \hat{j}+z x \hat{k}$\overrightarrow{F}=xy\hat{i}+yz\hat{j}+zx\hat{k}$ over the surface $S$$S$SS$S$, which is the upwardly oriented part of the cylinder $z=1-x^2$$z=1-x^2$z=1-x^(2)z=1-x^{2}$z=1-x^2$, for $0\le x\le 1,-2\le y\le 2$$0\le x\le 1,-2\le y\le 2$0 <= x <= 1,-2 <= y <= 20 \le x \le 1,\; -2 \le y \le 2$0\le x\le 1,-2\le y\le 2$.

Answer:

Verification Details

Step 1: Compute the Curl of $\overrightarrow{F}$$\overrightarrow{F}$vec(F)\vec{F}$\overrightarrow{F}$

• $\frac{\partial }{\partial y}(zx)=0$$\frac{\partial }{\partial y}(zx)=0$(del)/(del y)(zx)=0\frac{\partial}{\partial y}(zx) = 0$\frac{\partial }{\partial y}(zx)=0$, $\frac{\partial }{\partial z}(yz)=y$$\frac{\partial }{\partial z}(yz)=y$(del)/(del z)(yz)=y\frac{\partial}{\partial z}(yz) = y$\frac{\partial }{\partial z}(yz)=y$, so the $\hat{i}$$\widehat{i}$hat(i)\hat{i}$\hat{i}$-component is $0-y=-y$$0-y=-y$0-y=-y0 – y = -y$0-y=-y$.
• $\frac{\partial }{\partial x}(zx)=z$$\frac{\partial }{\partial x}(zx)=z$(del)/(del x)(zx)=z\frac{\partial}{\partial x}(zx) = z$\frac{\partial }{\partial x}(zx)=z$, $\frac{\partial }{\partial z}(xy)=0$$\frac{\partial }{\partial z}(xy)=0$(del)/(del z)(xy)=0\frac{\partial}{\partial z}(xy) = 0$\frac{\partial }{\partial z}(xy)=0$, so the $\hat{j}$$\widehat{j}$hat(j)\hat{j}$\hat{j}$-component is $-(z-0)=-z$$-(z-0)=-z$-(z-0)=-z-(z – 0) = -z$-(z-0)=-z$.
• $\frac{\partial }{\partial x}(yz)=0$$\frac{\partial }{\partial x}(yz)=0$(del)/(del x)(yz)=0\frac{\partial}{\partial x}(yz) = 0$\frac{\partial }{\partial x}(yz)=0$, $\frac{\partial }{\partial y}(xy)=x$$\frac{\partial }{\partial y}(xy)=x$(del)/(del y)(xy)=x\frac{\partial}{\partial y}(xy) = x$\frac{\partial }{\partial y}(xy)=x$, so the $\hat{k}$$\widehat{k}$hat(k)\hat{k}$\hat{k}$-component is $0-x=-x$$0-x=-x$0-x=-x0 – x = -x$0-x=-x$.

Step 2: Surface Integral of $\mathrm{\nabla }\times \overrightarrow{F}$$\mathrm{\nabla }\times \overrightarrow{F}$grad xx vec(F)\nabla \times \vec{F}$\mathrm{\nabla }\times \overrightarrow{F}$ over $S$$S$SS$S$

Step 3: Line Integral of $\overrightarrow{F}$$\overrightarrow{F}$vec(F)\vec{F}$\overrightarrow{F}$ over $\partial S$$\partial S$del S\partial S$\partial S$

• $C_1$$C_1$C_(1)C_1$C_1$: From $(0,-2,1)$$(0,-2,1)$(0,-2,1)(0, -2, 1)$(0,-2,1)$ to $(1,-2,0)$$(1,-2,0)$(1,-2,0)(1, -2, 0)$(1,-2,0)$ along $y=-2$$y=-2$y=-2y = -2$y=-2$, $z=1-x^2$$z=1-x^2$z=1-x^(2)z = 1 – x^2$z=1-x^2$, $x$$x$xx$x$ from 0 to 1.
• $C_2$$C_2$C_(2)C_2$C_2$: From $(1,-2,0)$$(1,-2,0)$(1,-2,0)(1, -2, 0)$(1,-2,0)$ to $(1,2,0)$$(1,2,0)$(1,2,0)(1, 2, 0)$(1,2,0)$ along $x=1$$x=1$x=1x = 1$x=1$, $z=0$$z=0$z=0z = 0$z=0$, $y$$y$yy$y$ from $-2$$-2$-2-2$-2$ to 2.
• $C_3$$C_3$C_(3)C_3$C_3$: From $(1,2,0)$$(1,2,0)$(1,2,0)(1, 2, 0)$(1,2,0)$ to $(0,2,1)$$(0,2,1)$(0,2,1)(0, 2, 1)$(0,2,1)$ along $y=2$$y=2$y=2y = 2$y=2$, $z=1-x^2$$z=1-x^2$z=1-x^(2)z = 1 – x^2$z=1-x^2$, $x$$x$xx$x$ from 1 to 0.
• $C_4$$C_4$C_(4)C_4$C_4$: From $(0,2,1)$$(0,2,1)$(0,2,1)(0, 2, 1)$(0,2,1)$ to $(0,-2,1)$$(0,-2,1)$(0,-2,1)(0, -2, 1)$(0,-2,1)$ along $x=0$$x=0$x=0x = 0$x=0$, $z=1$$z=1$z=1z = 1$z=1$, $y$$y$yy$y$ from 2 to $-2$$-2$-2-2$-2$.

• Segment $C_1$$C_1$C_(1)C_1$C_1$: $\overrightarrow{F}=(-2x,-2(1-x^2),x(1-x^2))=(-2x,-2+2x^2,x-x^3)$$\overrightarrow{F}=(-2x,-2(1-x^2),x(1-x^2))=(-2x,-2+2x^2,x-x^3)$vec(F)=(-2x,-2(1-x^(2)),x(1-x^(2)))=(-2x,-2+2x^(2),x-x^(3))\vec{F} = (-2x, -2(1 – x^2), x(1 – x^2)) = (-2x, -2 + 2x^2, x – x^3)$\overrightarrow{F}=(-2x,-2(1-x^2),x(1-x^2))=(-2x,-2+2x^2,x-x^3)$, $d\overrightarrow{r}=(dx,0,-2xdx)$$d\overrightarrow{r}=(dx,0,-2xdx)$d vec(r)=(dx,0,-2xdx)d\vec{r} = (dx, 0, -2x dx)$d\overrightarrow{r}=(dx,0,-2xdx)$.$$\overrightarrow{F}\cdot d\overrightarrow{r}=(-2x)(dx)+(x-x^3)(-2xdx)=-2xdx-2x^2(1-x^2)dx=(-2x-2x^2+2x^4)dx.$$$$\overrightarrow{F}\cdot d\overrightarrow{r}=(-2x)(dx)+(x-x^3)(-2xdx)=-2xdx-2x^2(1-x^2)dx=(-2x-2x^2+2x^4)dx.$$vec(F)*d vec(r)=(-2x)(dx)+(x-x^(3))(-2xdx)=-2xdx-2x^(2)(1-x^(2))dx=(-2x-2x^(2)+2x^(4))dx.\vec{F} \cdot d\vec{r} = (-2x)(dx) + (x – x^3)(-2x dx) = -2x dx – 2x^2(1 – x^2) dx = (-2x – 2x^2 + 2x^4) dx.$$\overrightarrow{F}\cdot d\overrightarrow{r}=(-2x)(dx)+(x-x^3)(-2xdx)=-2xdx-2x^2(1-x^2)dx=(-2x-2x^2+2x^4)dx.$$Integrate from $x=0$$x=0$x=0x = 0$x=0$ to $x=1$$x=1$x=1x = 1$x=1$:$${\int }_0^1(-2x-2x^2+2x^4)dx=2{\int }_0^1(-x-x^2+x^4)dx=2{[-\frac{x^2}{2}-\frac{x^3}{3}+\frac{x^5}{5}]}_0^1=2(-\frac{1}{2}-\frac{1}{3}+\frac{1}{5})=2(-\frac{19}{30})=-\frac{19}{15}.$$$${\int }_0^1 (-2x-2x^2+2x^4)dx=2{\int }_0^1 (-x-x^2+x^4)dx=2{\left[-\frac{x^2}{2}-\frac{x^3}{3}+\frac{x^5}{5}\right]}_0^1=2\left(-\frac{1}{2}-\frac{1}{3}+\frac{1}{5}\right)=2\left(-\frac{19}{30}\right)=-\frac{19}{15}.$$int_(0)^(1)(-2x-2x^(2)+2x^(4))dx=2int_(0)^(1)(-x-x^(2)+x^(4))dx=2[-(x^(2))/(2)-(x^(3))/(3)+(x^(5))/(5)]_(0)^(1)=2(-(1)/(2)-(1)/(3)+(1)/(5))=2(-(19)/(30))=-(19)/(15).\int_{0}^{1} (-2x – 2x^2 + 2x^4) dx = 2 \int_{0}^{1} (-x – x^2 + x^4) dx = 2 \left[ -\frac{x^2}{2} – \frac{x^3}{3} + \frac{x^5}{5} \right]_{0}^{1} = 2 \left( -\frac{1}{2} – \frac{1}{3} + \frac{1}{5} \right) = 2 \left( -\frac{19}{30} \right) = -\frac{19}{15}.$${\int }_0^1(-2x-2x^2+2x^4)dx=2{\int }_0^1(-x-x^2+x^4)dx=2{[-\frac{x^2}{2}-\frac{x^3}{3}+\frac{x^5}{5}]}_0^1=2(-\frac{1}{2}-\frac{1}{3}+\frac{1}{5})=2(-\frac{19}{30})=-\frac{19}{15}.$$
• Segment $C_2$$C_2$C_(2)C_2$C_2$: $\overrightarrow{F}=(y,0,0)$$\overrightarrow{F}=(y,0,0)$vec(F)=(y,0,0)\vec{F} = (y, 0, 0)$\overrightarrow{F}=(y,0,0)$, $d\overrightarrow{r}=(0,dy,0)$$d\overrightarrow{r}=(0,dy,0)$d vec(r)=(0,dy,0)d\vec{r} = (0, dy, 0)$d\overrightarrow{r}=(0,dy,0)$. Since $F_y=0$$F_y=0$F_(y)=0F_y = 0$F_y=0$, $\overrightarrow{F}\cdot d\overrightarrow{r}=0$$\overrightarrow{F}\cdot d\overrightarrow{r}=0$vec(F)*d vec(r)=0\vec{F} \cdot d\vec{r} = 0$\overrightarrow{F}\cdot d\overrightarrow{r}=0$, so the integral is 0.
• Segment $C_3$$C_3$C_(3)C_3$C_3$: $\overrightarrow{F}=(2x,2(1-x^2),x(1-x^2))=(2x,2-2x^2,x-x^3)$$\overrightarrow{F}=(2x,2(1-x^2),x(1-x^2))=(2x,2-2x^2,x-x^3)$vec(F)=(2x,2(1-x^(2)),x(1-x^(2)))=(2x,2-2x^(2),x-x^(3))\vec{F} = (2x, 2(1 – x^2), x(1 – x^2)) = (2x, 2 – 2x^2, x – x^3)$\overrightarrow{F}=(2x,2(1-x^2),x(1-x^2))=(2x,2-2x^2,x-x^3)$, $d\overrightarrow{r}=(dx,0,-2xdx)$$d\overrightarrow{r}=(dx,0,-2xdx)$d vec(r)=(dx,0,-2xdx)d\vec{r} = (dx, 0, -2x dx)$d\overrightarrow{r}=(dx,0,-2xdx)$ (with $x$$x$xx$x$ decreasing from 1 to 0).$$\overrightarrow{F}\cdot d\overrightarrow{r}=(2x)(dx)+(x-x^3)(-2xdx)=2xdx-2x^2(1-x^2)dx=(2x-2x^2+2x^4)dx.$$$$\overrightarrow{F}\cdot d\overrightarrow{r}=(2x)(dx)+(x-x^3)(-2xdx)=2xdx-2x^2(1-x^2)dx=(2x-2x^2+2x^4)dx.$$vec(F)*d vec(r)=(2x)(dx)+(x-x^(3))(-2xdx)=2xdx-2x^(2)(1-x^(2))dx=(2x-2x^(2)+2x^(4))dx.\vec{F} \cdot d\vec{r} = (2x)(dx) + (x – x^3)(-2x dx) = 2x dx – 2x^2(1 – x^2) dx = (2x – 2x^2 + 2x^4) dx.$$\overrightarrow{F}\cdot d\overrightarrow{r}=(2x)(dx)+(x-x^3)(-2xdx)=2xdx-2x^2(1-x^2)dx=(2x-2x^2+2x^4)dx.$$Integrate from $x=1$$x=1$x=1x = 1$x=1$ to $x=0$$x=0$x=0x = 0$x=0$:$${\int }_1^0(2x-2x^2+2x^4)dx=-{\int }_0^1(2x-2x^2+2x^4)dx=-{[x^2-\frac{2x^3}{3}+\frac{2x^5}{5}]}_0^1=-(1-\frac{2}{3}+\frac{2}{5})=-\frac{11}{15}.$$$${\int }_1^0 (2x-2x^2+2x^4)dx=-{\int }_0^1 (2x-2x^2+2x^4)dx=-{\left[x^2-\frac{2x^3}{3}+\frac{2x^5}{5}\right]}_0^1=-\left(1-\frac{2}{3}+\frac{2}{5}\right)=-\frac{11}{15}.$$int_(1)^(0)(2x-2x^(2)+2x^(4))dx=-int_(0)^(1)(2x-2x^(2)+2x^(4))dx=-[x^(2)-(2x^(3))/(3)+(2x^(5))/(5)]_(0)^(1)=-(1-(2)/(3)+(2)/(5))=-(11)/(15).\int_{1}^{0} (2x – 2x^2 + 2x^4) dx = -\int_{0}^{1} (2x – 2x^2 + 2x^4) dx = -\left[ x^2 – \frac{2x^3}{3} + \frac{2x^5}{5} \right]_{0}^{1} = -\left( 1 – \frac{2}{3} + \frac{2}{5} \right) = -\frac{11}{15}.$${\int }_1^0(2x-2x^2+2x^4)dx=-{\int }_0^1(2x-2x^2+2x^4)dx=-{[x^2-\frac{2x^3}{3}+\frac{2x^5}{5}]}_0^1=-(1-\frac{2}{3}+\frac{2}{5})=-\frac{11}{15}.$$
• Segment $C_4$$C_4$C_(4)C_4$C_4$: $\overrightarrow{F}=(0,y,0)$$\overrightarrow{F}=(0,y,0)$vec(F)=(0,y,0)\vec{F} = (0, y, 0)$\overrightarrow{F}=(0,y,0)$, $d\overrightarrow{r}=(0,dy,0)$$d\overrightarrow{r}=(0,dy,0)$d vec(r)=(0,dy,0)d\vec{r} = (0, dy, 0)$d\overrightarrow{r}=(0,dy,0)$. Since $F_y=y$$F_y=y$F_(y)=yF_y = y$F_y=y$, but $y$$y$yy$y$ decreases from 2 to $-2$$-2$-2-2$-2$:$${\int }_2^{-2}ydy={\left[\frac{y^2}{2}\right]}_2^{-2}=\frac{(-2{)}^2}{2}-\frac{2^2}{2}=2-2=0.$$$${\int }_2^{-2} ydy={\left[\frac{y^2}{2}\right]}_2^{-2}=\frac{(-2{)}^2}{2}-\frac{2^2}{2}=2-2=0.$$int_(2)^(-2)ydy=[(y^(2))/(2)]_(2)^(-2)=((-2)^(2))/(2)-(2^(2))/(2)=2-2=0.\int_{2}^{-2} y dy = \left[ \frac{y^2}{2} \right]_{2}^{-2} = \frac{(-2)^2}{2} – \frac{2^2}{2} = 2 – 2 = 0.$${\int }_2^{-2}ydy={\left[\frac{y^2}{2}\right]}_2^{-2}=\frac{(-2{)}^2}{2}-\frac{2^2}{2}=2-2=0.$$

Using Laplace transform, solve the initial value problem

Answer:

Given:

Step 1: Take the Laplace Transform of Both Sides

• $\mathcal{L}\{y^{″}\}=s^{2}Y(s)-sy(0)-y^{\prime }(0)=s^{2}Y(s)$$\mathcal{L}\{y^{″}\}=s^{2}Y(s)-sy(0)-y^{\prime }(0)=s^{2}Y(s)$L{y^(″)}=s^(2)Y(s)-sy(0)-y^(‘)(0)=s^(2)Y(s)\mathcal{L}\{y”\} = s^2 Y(s) – s y(0) – y'(0) = s^2 Y(s)$\mathcal{L}\{y^{″}\}=s^{2}Y(s)-sy(0)-y^{\prime }(0)=s^{2}Y(s)$,
• $\mathcal{L}\{y^{\prime }\}=sY(s)-y(0)=sY(s)$$\mathcal{L}\{y^{\prime }\}=sY(s)-y(0)=sY(s)$L{y^(‘)}=sY(s)-y(0)=sY(s)\mathcal{L}\{y’\} = s Y(s) – y(0) = s Y(s)$\mathcal{L}\{y^{\prime }\}=sY(s)-y(0)=sY(s)$,
• $\mathcal{L}\{y\}=Y(s)$$\mathcal{L}\{y\}=Y(s)$L{y}=Y(s)\mathcal{L}\{y\} = Y(s)$\mathcal{L}\{y\}=Y(s)$,
• $\mathcal{L}\{\delta (t-2)\}=e^{-2s}$$\mathcal{L}\{\delta (t-2)\}=e^{-2s}$L{delta(t-2)}=e^(-2s)\mathcal{L}\{\delta(t-2)\} = e^{-2s}$\mathcal{L}\{\delta (t-2)\}=e^{-2s}$,

Step 2: Solve for $Y(s)$$Y(s)$Y(s)Y(s)$Y(s)$

Step 3: Complete the Square in the Denominator

Step 4: Find the Inverse Laplace Transform

Final Solution:

Boxed Answer:

Using Gauss divergence theorem, evaluate the integral

Answer:

Step 1: Compute the Divergence of $\overrightarrow{F}$$\overrightarrow{F}$vec(F)\vec{F}$\overrightarrow{F}$

Step 2: Apply the Gauss Divergence Theorem

Step 3: Set Up the Volume Integral

• $r$$r$rr$r$ from $0$$0$00$0$ to $4$$4$44$4$,
• $\theta$$\theta$theta\theta$\theta$ from $0$$0$00$0$ to $2\pi$$2\pi$2pi2\pi$2\pi$,
• $z$$z$zz$z$ from $1$$1$11$1$ to $5$$5$55$5$.

Step 4: Evaluate the Integral

Final Answer:

A particle moves with a central acceleration $\mu (\frac{3}{r^3}+\frac{d^2}{r^5})$$\mu (\frac{3}{r^3}+\frac{d^2}{r^5})$mu((3)/(r^(3))+(d^(2))/(r^(5)))\mu\bigl(\tfrac{3}{r^{3}}+\tfrac{d^{2}}{r^{5}}\bigr)$\mu (\frac{3}{r^3}+\frac{d^2}{r^5})$ being projected from a distance $d$$d$dd$d$ at an angle ${45}^{\circ }$${45}^{\circ }$45^(@)45^{\circ}${45}^{\circ }$ with a velocity equal to that in a circle at the same distance. Prove that the time it takes to reach the centre of force is ${\displaystyle \frac{d^2}{\sqrt{2\mu }}}(2-\frac{\pi }{2})$${\displaystyle \frac{d^2}{\sqrt{2\mu }}}(2-\frac{\pi }{2})$(d^(2))/(sqrt(2mu))(2-(pi)/(2))\dfrac{d^{2}}{\sqrt{2\mu}}\bigl(2-\tfrac{\pi}{2}\bigr)${\displaystyle \frac{d^2}{\sqrt{2\mu }}}(2-\frac{\pi }{2})$.

Answer:

Given:

• Central acceleration: $a=\mu (\frac{3}{r^3}+\frac{d^2}{r^5})$$a=\mu \left(\frac{3}{r^3}+\frac{d^2}{r^5}\right)$a=mu((3)/(r^(3))+(d^(2))/(r^(5)))a = \mu\left(\frac{3}{r^{3}} + \frac{d^{2}}{r^{5}}\right)$a=\mu (\frac{3}{r^3}+\frac{d^2}{r^5})$
• Initial conditions:
  • The particle is projected from a distance $r=d$$r=d$r=dr = d$r=d$.
  • The projection angle is ${45}^{\circ }$${45}^{\circ }$45^(@)45^\circ${45}^{\circ }$.
  • The initial velocity $v$$v$vv$v$ is equal to the velocity required for circular motion at distance $d$$d$dd$d$.

Step 1: Determine the Initial Velocity

Step 2: Resolve the Velocity Components

Step 3: Use Conservation of Angular Momentum

Step 4: Apply the Radial Equation of Motion

Step 5: Energy Consideration

Step 6: Time Calculation

Final Answer: