SECTION-A

Let $G$$G$GG$G$ be a group of order 10 and $G^{\prime }$$G^{\prime }$G^(‘)G’$G^{\prime }$ be a group of order 6. Examine whether there exists a homomorphism of $G$$G$GG$G$ onto $G^{\prime }$$G^{\prime }$G^(‘)G’$G^{\prime }$.

Answer:

1. First Isomorphism Theorem:
   If $\varphi$$\varphi$phi\phi$\varphi$ is a surjective homomorphism, then:
   $$G/\ker (\varphi )\cong G^{\prime }.$$$$G/\ker (\varphi )\cong G^{\prime }.$$G//ker(phi)~=G^(‘).G / \ker(\phi) \cong G’.$$G/\ker (\varphi )\cong G^{\prime }.$$
   This implies:
   $$|G|/|\ker (\varphi )|=|G^{\prime }|⟹10/|\ker (\varphi )|=6.$$$$|G|/|\ker (\varphi )|=|G^{\prime }|⟹10/|\ker (\varphi )|=6.$$|G|//|ker(phi)|=|G^(‘)|Longrightarrow10//|ker(phi)|=6.|G| / |\ker(\phi)| = |G’| \implies 10 / |\ker(\phi)| = 6.$$|G|/|\ker (\varphi )|=|G^{\prime }|⟹10/|\ker (\varphi )|=6.$$
   Solving for $|\ker (\varphi )|$$|\ker (\varphi )|$|ker(phi)||\ker(\phi)|$|\ker (\varphi )|$, we get:
   $$|\ker (\varphi )|=\frac{10}{6}=\frac{5}{3}.$$$$|\ker (\varphi )|=\frac{10}{6}=\frac{5}{3}.$$|ker(phi)|=(10)/(6)=(5)/(3).|\ker(\phi)| = \frac{10}{6} = \frac{5}{3}.$$|\ker (\varphi )|=\frac{10}{6}=\frac{5}{3}.$$
   However, $|\ker (\varphi )|$$|\ker (\varphi )|$|ker(phi)||\ker(\phi)|$|\ker (\varphi )|$ must be an integer because it represents the order of a subgroup of $G$$G$GG$G$. Since $\frac{5}{3}$$\frac{5}{3}$(5)/(3)\frac{5}{3}$\frac{5}{3}$ is not an integer, this leads to a contradiction.
2. Lagrange’s Theorem Verification:
   The possible orders of $\ker (\varphi )$$\ker (\varphi )$ker(phi)\ker(\phi)$\ker (\varphi )$ (subgroups of $G$$G$GG$G$) are the divisors of 10: 1, 2, 5, 10.
   • If $|\ker (\varphi )|=1$$|\ker (\varphi )|=1$|ker(phi)|=1|\ker(\phi)| = 1$|\ker (\varphi )|=1$, then $|G^{\prime }|=10\ne 6$$|G^{\prime }|=10\ne 6$|G^(‘)|=10!=6|G’| = 10 \neq 6$|G^{\prime }|=10\ne 6$.
   • If $|\ker (\varphi )|=2$$|\ker (\varphi )|=2$|ker(phi)|=2|\ker(\phi)| = 2$|\ker (\varphi )|=2$, then $|G^{\prime }|=5\ne 6$$|G^{\prime }|=5\ne 6$|G^(‘)|=5!=6|G’| = 5 \neq 6$|G^{\prime }|=5\ne 6$.
   • If $|\ker (\varphi )|=5$$|\ker (\varphi )|=5$|ker(phi)|=5|\ker(\phi)| = 5$|\ker (\varphi )|=5$, then $|G^{\prime }|=2\ne 6$$|G^{\prime }|=2\ne 6$|G^(‘)|=2!=6|G’| = 2 \neq 6$|G^{\prime }|=2\ne 6$.
   • If $|\ker (\varphi )|=10$$|\ker (\varphi )|=10$|ker(phi)|=10|\ker(\phi)| = 10$|\ker (\varphi )|=10$, then $|G^{\prime }|=1\ne 6$$|G^{\prime }|=1\ne 6$|G^(‘)|=1!=6|G’| = 1 \neq 6$|G^{\prime }|=1\ne 6$.
   None of these cases satisfy $|G^{\prime }|=6$$|G^{\prime }|=6$|G^(‘)|=6|G’| = 6$|G^{\prime }|=6$.
3. Conclusion:
   Since there is no subgroup $\ker (\varphi )$$\ker (\varphi )$ker(phi)\ker(\phi)$\ker (\varphi )$ of $G$$G$GG$G$ such that $G/\ker (\varphi )$$G/\ker (\varphi )$G//ker(phi)G / \ker(\phi)$G/\ker (\varphi )$ has order 6, no surjective homomorphism exists from $G$$G$GG$G$ to $G^{\prime }$$G^{\prime }$G^(‘)G’$G^{\prime }$.

Express the ideal $4Z+6Z$$4Z+6Z$4Z+6Z4Z + 6Z$4Z+6Z$ as a principal ideal in the integral domain $Z$$Z$ZZ$Z$.

Answer:

1. Understand the Sum of Ideals:
   • The sum of two ideals $4\mathbb{Z}$$4\mathbb{Z}$4Z4\mathbb{Z}$4\mathbb{Z}$ and $6\mathbb{Z}$$6\mathbb{Z}$6Z6\mathbb{Z}$6\mathbb{Z}$ in $\mathbb{Z}$$\mathbb{Z}$Z\mathbb{Z}$\mathbb{Z}$ is the ideal generated by the union of their generators. In other words:$$4\mathbb{Z}+6\mathbb{Z}=\{4a+6b\mid a,b\in \mathbb{Z}\}$$$$4\mathbb{Z}+6\mathbb{Z}=\{4a+6b\mid a,b\in \mathbb{Z}\}$$4Z+6Z={4a+6b∣a,b inZ}4\mathbb{Z} + 6\mathbb{Z} = \{4a + 6b \mid a, b \in \mathbb{Z}\}$$4\mathbb{Z}+6\mathbb{Z}=\{4a+6b\mid a,b\in \mathbb{Z}\}$$
2. Find the Greatest Common Divisor (GCD):
   • The ideal $4\mathbb{Z}+6\mathbb{Z}$$4\mathbb{Z}+6\mathbb{Z}$4Z+6Z4\mathbb{Z} + 6\mathbb{Z}$4\mathbb{Z}+6\mathbb{Z}$ is generated by the greatest common divisor (GCD) of 4 and 6. This is a fundamental property of ideals in $\mathbb{Z}$$\mathbb{Z}$Z\mathbb{Z}$\mathbb{Z}$.
   • Compute the GCD of 4 and 6:$$gcd(4,6)=2$$$$gcd(4,6)=2$$gcd(4,6)=2\gcd(4, 6) = 2$$gcd(4,6)=2$$
3. Express as a Principal Ideal:
   • Since the GCD is 2, the ideal $4\mathbb{Z}+6\mathbb{Z}$$4\mathbb{Z}+6\mathbb{Z}$4Z+6Z4\mathbb{Z} + 6\mathbb{Z}$4\mathbb{Z}+6\mathbb{Z}$ can be expressed as the principal ideal generated by 2:$$4\mathbb{Z}+6\mathbb{Z}=2\mathbb{Z}$$$$4\mathbb{Z}+6\mathbb{Z}=2\mathbb{Z}$$4Z+6Z=2Z4\mathbb{Z} + 6\mathbb{Z} = 2\mathbb{Z}$$4\mathbb{Z}+6\mathbb{Z}=2\mathbb{Z}$$

Test the convergence of the series

Answer:

Step 1: Simplify the General Term

Step 2: Apply the Ratio Test

Step 3: Determine Convergence

• If $\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|<1$$\underset{n\to \mathrm{\infty }}{lim} \left|\frac{a_{n+1}}{a_n}\right|<1$lim_(n rarr oo)|(a_(n+1))/(a_(n))| < 1\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1$\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|<1$, the series converges.
• If $\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|>1$$\underset{n\to \mathrm{\infty }}{lim} \left|\frac{a_{n+1}}{a_n}\right|>1$lim_(n rarr oo)|(a_(n+1))/(a_(n))| > 1\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| > 1$\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|>1$, the series diverges.
• If $\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|=1$$\underset{n\to \mathrm{\infty }}{lim} \left|\frac{a_{n+1}}{a_n}\right|=1$lim_(n rarr oo)|(a_(n+1))/(a_(n))|=1\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 1$\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|=1$, the test is inconclusive.

• The series converges if $x^2<1$$x^2<1$x^(2) < 1x^2 < 1$x^2<1$ (i.e., $0<x<1$$0<x<1$0 < x < 10 < x < 1$0<x<1$).
• The series diverges if $x^2>1$$x^2>1$x^(2) > 1x^2 > 1$x^2>1$ (i.e., $x>1$$x>1$x > 1x > 1$x>1$).
• The test is inconclusive if $x=1$$x=1$x=1x = 1$x=1$.

Step 4: Check the Boundary Case $x=1$$x=1$x=1x = 1$x=1$

Final Conclusion

• The series converges for $0<x<1$$0<x<1$0 < x < 10 < x < 1$0<x<1$.
• The series diverges for $x\ge 1$$x\ge 1$x >= 1x \geq 1$x\ge 1$.

State the sufficient conditions for a function $f(z)=f(x+iy)=u(x,y)+iv(x,y)$$f(z)=f(x+iy)=u(x,y)+iv(x,y)$f(z)=f(x+iy)=u(x,y)+iv(x,y)f(z) = f(x + iy) = u(x, y) + i v(x, y)$f(z)=f(x+iy)=u(x,y)+iv(x,y)$ to be analytic in its domain. Hence, show that $f(z)=\log z$$f(z)=\log z$f(z)=log zf(z) = \log z$f(z)=\log z$ is analytic in its domain and find $\frac{df}{dz}$$\frac{df}{dz}$(df)/(dz)\frac{df}{dz}$\frac{df}{dz}$.

Answer:

Sufficient Conditions for Analyticity

1. Existence of Partial Derivatives:
   The partial derivatives $\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial v}{\partial x},\frac{\partial v}{\partial y}$$\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial v}{\partial x},\frac{\partial v}{\partial y}$(del u)/(del x),(del u)/(del y),(del v)/(del x),(del v)/(del y)\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}$\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial v}{\partial x},\frac{\partial v}{\partial y}$ exist and are continuous in $D$$D$DD$D$.
2. Cauchy-Riemann Equations:
   The functions $u(x,y)$$u(x,y)$u(x,y)u(x,y)$u(x,y)$ and $v(x,y)$$v(x,y)$v(x,y)v(x,y)$v(x,y)$ satisfy:
   $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.$$$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.$$(del u)/(del x)=(del v)/(del y),quad(del u)/(del y)=-(del v)/(del x).\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.$$

Analyticity of $f(z)=\log z$$f(z)=\log z$f(z)=log zf(z) = \log z$f(z)=\log z$

• $\ln |z|$$\ln |z|$ln |z|\ln |z|$\ln |z|$ is the natural logarithm of the modulus,
• $\arg z$$\arg z$arg z\arg z$\arg z$ is the argument (angle) of $z$$z$zz$z$.

1. Check the Cauchy-Riemann Equations

2. Continuity of Partial Derivatives

Derivative of $\log z$$\log z$log z\log z$\log z$

Final Answer

A person requires 24, 24, and 20 units of chemicals $A$$A$AA$A$, $B$$B$BB$B$, and $C$$C$CC$C$ respectively for his garden. Product $P$$P$PP$P$ contains 2, 4, and 1 units of chemicals $A$$A$AA$A$, $B$$B$BB$B$, and $C$$C$CC$C$ respectively per jar, and product $Q$$Q$QQ$Q$ contains 2, 1, and 5 units of chemicals $A$$A$AA$A$, $B$$B$BB$B$, and $C$$C$CC$C$ respectively per jar. If a jar of $P$$P$PP$P$ costs ₹30 and a jar of $Q$$Q$QQ$Q$ costs ₹50, then how many jars of each should be purchased in order to minimize the cost and meet the requirements?

Answer:

Problem Statement

• 24 units of chemical $A$$A$AA$A$,
• 24 units of chemical $B$$B$BB$B$,
• 20 units of chemical $C$$C$CC$C$.

• Product $P$$P$PP$P$ (per jar):
  • 2 units of $A$$A$AA$A$,
  • 4 units of $B$$B$BB$B$,
  • 1 unit of $C$$C$CC$C$.
  • Cost: ₹30 per jar.
• Product $Q$$Q$QQ$Q$ (per jar):
  • 2 units of $A$$A$AA$A$,
  • 1 unit of $B$$B$BB$B$,
  • 5 units of $C$$C$CC$C$.
  • Cost: ₹50 per jar.

---

Step 1: Define Variables

• $x$$x$xx$x$ = number of jars of $P$$P$PP$P$,
• $y$$y$yy$y$ = number of jars of $Q$$Q$QQ$Q$.

Step 2: Formulate Constraints

1. Chemical $A$$A$AA$A$:
   $$2x+2y\ge 24\Rightarrow x+y\ge 12.$$$$2x+2y\ge 24\Rightarrow x+y\ge 12.$$2x+2y >= 24quad=>quad x+y >= 12.2x + 2y \geq 24 \quad \Rightarrow \quad x + y \geq 12.$$2x+2y\ge 24\Rightarrow x+y\ge 12.$$
2. Chemical $B$$B$BB$B$:
   $$4x+y\ge 24.$$$$4x+y\ge 24.$$4x+y >= 24.4x + y \geq 24.$$4x+y\ge 24.$$
3. Chemical $C$$C$CC$C$:
   $$x+5y\ge 20.$$$$x+5y\ge 20.$$x+5y >= 20.x + 5y \geq 20.$$x+5y\ge 20.$$
4. Non-negativity:
   $$x\ge 0,y\ge 0.$$$$x\ge 0,y\ge 0.$$x >= 0,quad y >= 0.x \geq 0, \quad y \geq 0.$$x\ge 0,y\ge 0.$$

Step 3: Objective Function

Step 4: Solve the System of Inequalities

1. From $x+y\ge 12$$x+y\ge 12$x+y >= 12x + y \geq 12$x+y\ge 12$:
   • If $x=0$$x=0$x=0x = 0$x=0$, $y=12$$y=12$y=12y = 12$y=12$.
   • If $y=0$$y=0$y=0y = 0$y=0$, $x=12$$x=12$x=12x = 12$x=12$.
2. From $4x+y\ge 24$$4x+y\ge 24$4x+y >= 244x + y \geq 24$4x+y\ge 24$:
   • If $x=0$$x=0$x=0x = 0$x=0$, $y=24$$y=24$y=24y = 24$y=24$.
   • If $y=0$$y=0$y=0y = 0$y=0$, $x=6$$x=6$x=6x = 6$x=6$.
3. From $x+5y\ge 20$$x+5y\ge 20$x+5y >= 20x + 5y \geq 20$x+5y\ge 20$:
   • If $x=0$$x=0$x=0x = 0$x=0$, $y=4$$y=4$y=4y = 4$y=4$.
   • If $y=0$$y=0$y=0y = 0$y=0$, $x=20$$x=20$x=20x = 20$x=20$.

• Intersection of $x+y=12$$x+y=12$x+y=12x + y = 12$x+y=12$ and $4x+y=24$$4x+y=24$4x+y=244x + y = 24$4x+y=24$:
  $$\{\begin{array}{l}x+y=12,\\ 4x+y=24.\end{array}$$$$\left\{\begin{array}{l}x+y=12,\\ 4x+y=24.\end{array}\right.$${[x+y=12″,”],[4x+y=24.]:}\begin{cases} x + y = 12, \\ 4x + y = 24. \end{cases}$$\{\begin{array}{l}x+y=12,\\ 4x+y=24.\end{array}$$
  Subtract the first equation from the second:
  $$3x=12\Rightarrow x=4,y=8.$$$$3x=12\Rightarrow x=4,y=8.$$3x=12quad=>quad x=4,quad y=8.3x = 12 \quad \Rightarrow \quad x = 4, \quad y = 8.$$3x=12\Rightarrow x=4,y=8.$$
  Point: $(4,8)$$(4,8)$(4,8)(4, 8)$(4,8)$.
• Intersection of $x+y=12$$x+y=12$x+y=12x + y = 12$x+y=12$ and $x+5y=20$$x+5y=20$x+5y=20x + 5y = 20$x+5y=20$:
  $$\{\begin{array}{l}x+y=12,\\ x+5y=20.\end{array}$$$$\left\{\begin{array}{l}x+y=12,\\ x+5y=20.\end{array}\right.$${[x+y=12″,”],[x+5y=20.]:}\begin{cases} x + y = 12, \\ x + 5y = 20. \end{cases}$$\{\begin{array}{l}x+y=12,\\ x+5y=20.\end{array}$$
  Subtract the first equation from the second:
  $$4y=8\Rightarrow y=2,x=10.$$$$4y=8\Rightarrow y=2,x=10.$$4y=8quad=>quad y=2,quad x=10.4y = 8 \quad \Rightarrow \quad y = 2, \quad x = 10.$$4y=8\Rightarrow y=2,x=10.$$
  Point: $(10,2)$$(10,2)$(10,2)(10, 2)$(10,2)$.
• Intersection of $4x+y=24$$4x+y=24$4x+y=244x + y = 24$4x+y=24$ and $x+5y=20$$x+5y=20$x+5y=20x + 5y = 20$x+5y=20$:
  $$\{\begin{array}{l}4x+y=24,\\ x+5y=20.\end{array}$$$$\left\{\begin{array}{l}4x+y=24,\\ x+5y=20.\end{array}\right.$${[4x+y=24″,”],[x+5y=20.]:}\begin{cases} 4x + y = 24, \\ x + 5y = 20. \end{cases}$$\{\begin{array}{l}4x+y=24,\\ x+5y=20.\end{array}$$
  Solve the first equation for $y$$y$yy$y$: $y=24-4x$$y=24-4x$y=24-4xy = 24 – 4x$y=24-4x$.
  Substitute into the second equation:
  $$x+5(24-4x)=20\Rightarrow x+120-20x=20\Rightarrow -19x=-100\Rightarrow x=\frac{100}{19}\approx 5.26.$$$$x+5(24-4x)=20\Rightarrow x+120-20x=20\Rightarrow -19x=-100\Rightarrow x=\frac{100}{19}\approx 5.26.$$x+5(24-4x)=20quad=>quad x+120-20 x=20quad=>quad-19 x=-100quad=>quad x=(100)/(19)~~5.26.x + 5(24 – 4x) = 20 \quad \Rightarrow \quad x + 120 – 20x = 20 \quad \Rightarrow \quad -19x = -100 \quad \Rightarrow \quad x = \frac{100}{19} \approx 5.26.$$x+5(24-4x)=20\Rightarrow x+120-20x=20\Rightarrow -19x=-100\Rightarrow x=\frac{100}{19}\approx 5.26.$$
  Then $y=24-4\left(\frac{100}{19}\right)=\frac{456-400}{19}=\frac{56}{19}\approx 2.95$$y=24-4\left(\frac{100}{19}\right)=\frac{456-400}{19}=\frac{56}{19}\approx 2.95$y=24-4((100)/(19))=(456-400)/(19)=(56)/(19)~~2.95y = 24 – 4 \left( \frac{100}{19} \right) = \frac{456 – 400}{19} = \frac{56}{19} \approx 2.95$y=24-4\left(\frac{100}{19}\right)=\frac{456-400}{19}=\frac{56}{19}\approx 2.95$.
  Point: $(\frac{100}{19},\frac{56}{19})$$\left(\frac{100}{19},\frac{56}{19}\right)$((100)/(19),(56)/(19))\left( \frac{100}{19}, \frac{56}{19} \right)$(\frac{100}{19},\frac{56}{19})$.

Step 5: Evaluate Cost at Critical Points

1. At $(4,8)$$(4,8)$(4,8)(4, 8)$(4,8)$:
   $$30(4)+50(8)=120+400=₹520.$$$$30(4)+50(8)=120+400=₹520.$$30(4)+50(8)=120+400=₹520.30(4) + 50(8) = 120 + 400 = ₹520.$$30(4)+50(8)=120+400=₹520.$$
2. At $(10,2)$$(10,2)$(10,2)(10, 2)$(10,2)$:
   $$30(10)+50(2)=300+100=₹400.$$$$30(10)+50(2)=300+100=₹400.$$30(10)+50(2)=300+100=₹400.30(10) + 50(2) = 300 + 100 = ₹400.$$30(10)+50(2)=300+100=₹400.$$
3. At $(\frac{100}{19},\frac{56}{19})$$\left(\frac{100}{19},\frac{56}{19}\right)$((100)/(19),(56)/(19))\left( \frac{100}{19}, \frac{56}{19} \right)$(\frac{100}{19},\frac{56}{19})$:
   $$30\left(\frac{100}{19}\right)+50\left(\frac{56}{19}\right)=\frac{3000}{19}+\frac{2800}{19}=\frac{5800}{19}\approx ₹305.26.$$$$30\left(\frac{100}{19}\right)+50\left(\frac{56}{19}\right)=\frac{3000}{19}+\frac{2800}{19}=\frac{5800}{19}\approx ₹305.26.$$30((100)/(19))+50((56)/(19))=(3000)/(19)+(2800)/(19)=(5800)/(19)≈₹305.26.30 \left( \frac{100}{19} \right) + 50 \left( \frac{56}{19} \right) = \frac{3000}{19} + \frac{2800}{19} = \frac{5800}{19} \approx ₹305.26.$$30\left(\frac{100}{19}\right)+50\left(\frac{56}{19}\right)=\frac{3000}{19}+\frac{2800}{19}=\frac{5800}{19}\approx ₹305.26.$$

• Try $(5,3)$$(5,3)$(5,3)(5, 3)$(5,3)$:
  • Check constraints:$$2(5)+2(3)=16\ge 24?\text{No (16 < 24)}.\text{Not feasible.}$$$$2(5)+2(3)=16\ge 24?\text{No (16 < 24)}.\text{Not feasible.}$$2(5)+2(3)=16 >= 24?quad”No (16 < 24)”.quad”Not feasible.”2(5) + 2(3) = 16 \geq 24? \quad \text{No (16 < 24)}. \quad \text{Not feasible.}$$2(5)+2(3)=16\ge 24?\text{No (16 < 24)}.\text{Not feasible.}$$
• Try $(6,2)$$(6,2)$(6,2)(6, 2)$(6,2)$:
  • Check constraints:$$2(6)+2(2)=16\ge 24?\text{No (16 < 24)}.\text{Not feasible.}$$$$2(6)+2(2)=16\ge 24?\text{No (16 < 24)}.\text{Not feasible.}$$2(6)+2(2)=16 >= 24?quad”No (16 < 24)”.quad”Not feasible.”2(6) + 2(2) = 16 \geq 24? \quad \text{No (16 < 24)}. \quad \text{Not feasible.}$$2(6)+2(2)=16\ge 24?\text{No (16 < 24)}.\text{Not feasible.}$$
• Try $(6,6)$$(6,6)$(6,6)(6, 6)$(6,6)$:
  • Check constraints:$$\begin{gathered} 2(6)+2(6)=24\ge 24\text{(OK)}, \\ 4(6)+6=30\ge 24\text{(OK)}, \\ 6+5(6)=36\ge 20\text{(OK)}. \end{gathered}$$$$\begin{gathered} 2(6)+2(6)=24\ge 24\text{(OK)}, \\ 4(6)+6=30\ge 24\text{(OK)}, \\ 6+5(6)=36\ge 20\text{(OK)}. \end{gathered}$$2(6)+2(6)=24 >= 24quad(OK),4(6)+6=30 >= 24quad(OK),6+5(6)=36 >= 20quad(OK).2(6) + 2(6) = 24 \geq 24 \quad \text{(OK)}, \\ 4(6) + 6 = 30 \geq 24 \quad \text{(OK)}, \\ 6 + 5(6) = 36 \geq 20 \quad \text{(OK)}.$$\begin{gathered} 2(6)+2(6)=24\ge 24\text{(OK)}, \\ 4(6)+6=30\ge 24\text{(OK)}, \\ 6+5(6)=36\ge 20\text{(OK)}. \end{gathered}$$
  • Cost:$$30(6)+50(6)=180+300=₹480.$$$$30(6)+50(6)=180+300=₹480.$$30(6)+50(6)=180+300=₹480.30(6) + 50(6) = 180 + 300 = ₹480.$$30(6)+50(6)=180+300=₹480.$$
• Try $(4,8)$$(4,8)$(4,8)(4, 8)$(4,8)$: (Already computed: ₹520)
• Try $(10,2)$$(10,2)$(10,2)(10, 2)$(10,2)$: (Already computed: ₹400)

• Try $(8,4)$$(8,4)$(8,4)(8, 4)$(8,4)$:
  • Check constraints:$$\begin{gathered} 2(8)+2(4)=24\ge 24\text{(OK)}, \\ 4(8)+4=36\ge 24\text{(OK)}, \\ 8+5(4)=28\ge 20\text{(OK)}. \end{gathered}$$$$\begin{gathered} 2(8)+2(4)=24\ge 24\text{(OK)}, \\ 4(8)+4=36\ge 24\text{(OK)}, \\ 8+5(4)=28\ge 20\text{(OK)}. \end{gathered}$$2(8)+2(4)=24 >= 24quad(OK),4(8)+4=36 >= 24quad(OK),8+5(4)=28 >= 20quad(OK).2(8) + 2(4) = 24 \geq 24 \quad \text{(OK)}, \\ 4(8) + 4 = 36 \geq 24 \quad \text{(OK)}, \\ 8 + 5(4) = 28 \geq 20 \quad \text{(OK)}.$$\begin{gathered} 2(8)+2(4)=24\ge 24\text{(OK)}, \\ 4(8)+4=36\ge 24\text{(OK)}, \\ 8+5(4)=28\ge 20\text{(OK)}. \end{gathered}$$
  • Cost:$$30(8)+50(4)=240+200=₹440.$$$$30(8)+50(4)=240+200=₹440.$$30(8)+50(4)=240+200=₹440.30(8) + 50(4) = 240 + 200 = ₹440.$$30(8)+50(4)=240+200=₹440.$$

Verification of $(10,2)$$(10,2)$(10,2)(10, 2)$(10,2)$:

• Chemical $A$$A$AA$A$: $2(10)+2(2)=24$$2(10)+2(2)=24$2(10)+2(2)=242(10) + 2(2) = 24$2(10)+2(2)=24$ (OK),
• Chemical $B$$B$BB$B$: $4(10)+2=42\ge 24$$4(10)+2=42\ge 24$4(10)+2=42 >= 244(10) + 2 = 42 \geq 24$4(10)+2=42\ge 24$ (OK),
• Chemical $C$$C$CC$C$: $10+5(2)=20\ge 20$$10+5(2)=20\ge 20$10+5(2)=20 >= 2010 + 5(2) = 20 \geq 20$10+5(2)=20\ge 20$ (OK).

Conclusion

• 10 jars of $P$$P$PP$P$,
• 2 jars of $Q$$Q$QQ$Q$,
  with a total cost of ₹400.

(a) Prove that a non-commutative group of order $2p$$2p$2p2p$2p$, where $p$$p$pp$p$ is an odd prime, must have a subgroup of order $p$$p$pp$p$.

Answer:

Step 1: Use Sylow’s Theorems

• Sylow’s Third Theorem states that:
  $$n_p\equiv 1(\mathrm{mod}p)\text{and}{n}_p\mid 2p.$$$$n_p\equiv 1(\mathrm{mod}p)\text{and}{n}_p\mid 2p.$$n_(p)-=1quad(modp)quad”and”quadn_(p)∣2p.n_p \equiv 1 \pmod{p} \quad \text{and} \quad n_p \mid 2p.$$n_p\equiv 1(\mathrm{mod}p)\text{and}{n}_p\mid 2p.$$
  Since $p$$p$pp$p$ is prime and $p>2$$p>2$p > 2p > 2$p>2$, the divisors of $2p$$2p$2p2p$2p$ are $1,2,p,2p$$1,2,p,2p$1,2,p,2p1, 2, p, 2p$1,2,p,2p$. The condition $n_p\equiv 1(\mathrm{mod}p)$$n_p\equiv 1(\mathrm{mod}p)$n_(p)-=1(modp)n_p \equiv 1 \pmod{p}$n_p\equiv 1(\mathrm{mod}p)$ implies:
  $$n_p=1\text{or}{n}_p=p+1.$$$$n_p=1\text{or}{n}_p=p+1.$$n_(p)=1quad”or”quadn_(p)=p+1.n_p = 1 \quad \text{or} \quad n_p = p + 1.$$n_p=1\text{or}{n}_p=p+1.$$
  But $p+1$$p+1$p+1p + 1$p+1$ does not divide $2p$$2p$2p2p$2p$ unless $p=2$$p=2$p=2p = 2$p=2$, which is excluded since $p$$p$pp$p$ is odd. Thus:
  $$n_p=1.$$$$n_p=1.$$n_(p)=1.n_p = 1.$$n_p=1.$$
• Conclusion: There is exactly one Sylow $p$$p$pp$p$-subgroup, say $H$$H$HH$H$, of order $p$$p$pp$p$.

Step 2: Show $H$$H$HH$H$ is Normal

Step 3: Non-commutativity Implies $G$$G$GG$G$ is Not Cyclic

• $⟨r⟩$$⟨r⟩$(:r:)\langle r \rangle$⟨r⟩$ is the cyclic subgroup of order $p$$p$pp$p$,
• $⟨s⟩$$⟨s⟩$(:s:)\langle s \rangle$⟨s⟩$ is a subgroup of order $2$$2$22$2$.

Final Conclusion

• A normal subgroup $H$$H$HH$H$ of order $p$$p$pp$p$ (the Sylow $p$$p$pp$p$-subgroup),
• And another subgroup of order $2$$2$22$2$, but the question focuses on the subgroup of order $p$$p$pp$p$.

Using the method of Lagrange’s multipliers, find the minimum and maximum distances of the point $P(2,6,3)$$P(2,6,3)$P(2,6,3)P(2, 6, 3)$P(2,6,3)$ from the sphere $x^2+y^2+z^2=4$$x^2+y^2+z^2=4$x^(2)+y^(2)+z^(2)=4x^2 + y^2 + z^2 = 4$x^2+y^2+z^2=4$.

Answer:

Step 1: Define the Distance Function

Step 2: Set Up the Constraint

Step 3: Apply the Method of Lagrange Multipliers

Step 4: Solve for $y$$y$yy$y$ and $z$$z$zz$z$ in Terms of $x$$x$xx$x$

Step 5: Substitute into the Sphere Equation

Step 6: Find Corresponding $y$$y$yy$y$ and $z$$z$zz$z$

Step 7: Compute the Distances

Step 8: Determine Minimum and Maximum Distances

• Minimum distance: $D=\sqrt{25}=5$$D=\sqrt{25}=5$D=sqrt25=5D = \sqrt{25} = 5$D=\sqrt{25}=5$.
• Maximum distance: $D=\sqrt{81}=9$$D=\sqrt{81}=9$D=sqrt81=9D = \sqrt{81} = 9$D=\sqrt{81}=9$.

Final Answer

Evaluate ${\int }_0^{2\pi }\frac{\cos 2\theta }{5+4\cos \theta }d\theta$${\int }_0^{2\pi } \frac{\cos 2\theta }{5+4\cos \theta }d\theta$int_(0)^(2pi)(cos 2theta)/(5+4cos theta)d theta\int_0^{2\pi} \frac{\cos 2\theta}{5 + 4 \cos \theta} d\theta${\int }_0^{2\pi }\frac{\cos 2\theta }{5+4\cos \theta }d\theta$ using contour integration.

Answer:

Step 1: Parameterize the Integral Using $z=e^{i\theta }$$z=e^{i\theta }$z=e^(i theta)z = e^{i\theta}$z=e^{i\theta }$

Step 2: Factor the Denominator

Step 3: Identify Poles Inside the Unit Circle

1. $z=0$$z=0$z=0z = 0$z=0$ (order 2),
2. $z=-\frac{1}{2}$$z=-\frac{1}{2}$z=-(1)/(2)z = -\frac{1}{2}$z=-\frac{1}{2}$ (from $2z+1=0$$2z+1=0$2z+1=02z + 1 = 0$2z+1=0$),
3. $z=-2$$z=-2$z=-2z = -2$z=-2$ (from $z+2=0$$z+2=0$z+2=0z + 2 = 0$z+2=0$).

Step 4: Compute Residues

Residue at $z=0$$z=0$z=0z = 0$z=0$:

Residue at $z=-\frac{1}{2}$$z=-\frac{1}{2}$z=-(1)/(2)z = -\frac{1}{2}$z=-\frac{1}{2}$:

Step 5: Apply the Residue Theorem

Final Answer

(a) Prove that $x^2+1$$x^2+1$x^(2)+1x^2 + 1$x^2+1$ is an irreducible polynomial in $Z_3[x]$$Z_3[x]$Z_(3)[x]Z_3[x]$Z_3[x]$. Further show that the quotient ring $\frac{Z_3[x]}{⟨x^2+1⟩}$$\frac{Z_3[x]}{⟨x^2+1⟩}$(Z_(3)[x])/((:x^(2)+1:))\frac{Z_3[x]}{\langle x^2 + 1 \rangle}$\frac{Z_3[x]}{⟨x^2+1⟩}$ is a field of 9 elements.

Answer:

Proof that $x^2+1$$x^2+1$x^(2)+1x^2 + 1$x^2+1$ is Irreducible in ${\mathbb{Z}}_3[x]$${\mathbb{Z}}_3[x]$Z_(3)[x]\mathbb{Z}_3[x]${\mathbb{Z}}_3[x]$

• At $x=0$$x=0$x=0x = 0$x=0$: $0^2+1=1\ne 0$$0^2+1=1\ne 0$0^(2)+1=1!=00^2 + 1 = 1 \neq 0$0^2+1=1\ne 0$,
• At $x=1$$x=1$x=1x = 1$x=1$: $1^2+1=2\ne 0$$1^2+1=2\ne 0$1^(2)+1=2!=01^2 + 1 = 2 \neq 0$1^2+1=2\ne 0$,
• At $x=2$$x=2$x=2x = 2$x=2$: $2^2+1=4\equiv 1\ne 0$$2^2+1=4\equiv 1\ne 0$2^(2)+1=4-=1!=02^2 + 1 = 4 \equiv 1 \neq 0$2^2+1=4\equiv 1\ne 0$.

Proof that $\frac{{\mathbb{Z}}_3[x]}{⟨x^2+1⟩}$$\frac{{\mathbb{Z}}_3[x]}{⟨x^2+1⟩}$(Z_(3)[x])/((:x^(2)+1:))\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}$\frac{{\mathbb{Z}}_3[x]}{⟨x^2+1⟩}$ is a Field of 9 Elements

• Addition and Multiplication are performed modulo $x^2+1$$x^2+1$x^(2)+1x^2 + 1$x^2+1$, with $x^2\equiv -1\equiv 2$$x^2\equiv -1\equiv 2$x^(2)-=-1-=2x^2 \equiv -1 \equiv 2$x^2\equiv -1\equiv 2$.
• Inverses: Every non-zero element $a+bx$$a+bx$a+bxa + bx$a+bx$ has a multiplicative inverse because $x^2+1$$x^2+1$x^(2)+1x^2 + 1$x^2+1$ is irreducible. For example:
  • The inverse of $1+x$$1+x$1+x1 + x$1+x$ is found by solving $(1+x)(c+dx)\equiv 1\mathrm{mod}(x^2+1)$$(1+x)(c+dx)\equiv 1\mathrm{mod}(x^2+1)$(1+x)(c+dx)-=1mod(x^(2)+1)(1 + x)(c + dx) \equiv 1 \mod (x^2 + 1)$(1+x)(c+dx)\equiv 1\mathrm{mod}(x^2+1)$, leading to $c+d=1$$c+d=1$c+d=1c + d = 1$c+d=1$ and $c-d=0$$c-d=0$c-d=0c – d = 0$c-d=0$, so $c=d=2$$c=d=2$c=d=2c = d = 2$c=d=2$. Thus, $(1+x{)}^{-1}=2+2x$$(1+x{)}^{-1}=2+2x$(1+x)^(-1)=2+2x(1 + x)^{-1} = 2 + 2x$(1+x{)}^{-1}=2+2x$.

Conclusion

• $x^2+1$$x^2+1$x^(2)+1x^2 + 1$x^2+1$ is irreducible in ${\mathbb{Z}}_3[x]$${\mathbb{Z}}_3[x]$Z_(3)[x]\mathbb{Z}_3[x]${\mathbb{Z}}_3[x]$ because it has no roots in ${\mathbb{Z}}_3$${\mathbb{Z}}_3$Z_(3)\mathbb{Z}_3${\mathbb{Z}}_3$.
• The quotient ring $\frac{{\mathbb{Z}}_3[x]}{⟨x^2+1⟩}$$\frac{{\mathbb{Z}}_3[x]}{⟨x^2+1⟩}$(Z_(3)[x])/((:x^(2)+1:))\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}$\frac{{\mathbb{Z}}_3[x]}{⟨x^2+1⟩}$ is a field with 9 elements, as it satisfies all field axioms and has the correct cardinality.

Prove that $u(x,y)=e^x(x\cos y-y\sin y)$$u(x,y)=e^x(x\cos y-y\sin y)$u(x,y)=e^(x)(x cos y-y sin y)u(x, y) = e^x (x \cos y – y \sin y)$u(x,y)=e^x(x\cos y-y\sin y)$ is harmonic. Find its conjugate harmonic function $v(x,y)$$v(x,y)$v(x,y)v(x, y)$v(x,y)$ and express the corresponding analytic function $f(z)$$f(z)$f(z)f(z)$f(z)$ in terms of $z$$z$zz$z$.

Answer:

Step 1: Verify that $u(x,y)=e^x(x\cos y-y\sin y)$$u(x,y)=e^x(x\cos y-y\sin y)$u(x,y)=e^(x)(x cos y-y sin y)u(x, y) = e^x (x \cos y – y \sin y)$u(x,y)=e^x(x\cos y-y\sin y)$ is Harmonic

Compute the First Partial Derivatives:

Compute the Second Partial Derivatives:

Check Laplace’s Equation:

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Step 2: Find the Conjugate Harmonic Function $v(x,y)$$v(x,y)$v(x,y)v(x, y)$v(x,y)$

From $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$(del u)/(del x)=(del v)/(del y)\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$:

From $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$(del u)/(del y)=-(del v)/(del x)\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$:

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Step 3: Express the Analytic Function $f(z)$$f(z)$f(z)f(z)$f(z)$ in Terms of $z$$z$zz$z$

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Final Answer

1. $u(x,y)=e^x(x\cos y-y\sin y)$$u(x,y)=e^x(x\cos y-y\sin y)$u(x,y)=e^(x)(x cos y-y sin y)u(x, y) = e^x (x \cos y – y \sin y)$u(x,y)=e^x(x\cos y-y\sin y)$ is harmonic.
2. The conjugate harmonic function is:$$v(x,y)=e^x((x+1)\sin y+y\cos y).$$$$v(x,y)=e^x\left((x+1)\sin y+y\cos y\right).$$v(x,y)=e^(x)((x+1)sin y+y cos y).v(x, y) = e^x \left( (x + 1) \sin y + y \cos y \right).$$v(x,y)=e^x((x+1)\sin y+y\cos y).$$
3. The corresponding analytic function is:$$f(z)=e^z(z+i).$$$$f(z)=e^z(z+i).$$f(z)=e^(z)(z+i).f(z) = e^z (z + i).$$f(z)=e^z(z+i).$$

Solve the following linear programming problem by the Big M method:

Answer:

Step 1: Convert Inequalities to Standard Form

• Minimize $Z=2x_1+3x_2$$Z=2x_1+3x_2$Z=2x_(1)+3x_(2)Z = 2x_1 + 3x_2$Z=2x_1+3x_2$
• Subject to:$$\begin{array}{rl}{x}_1+x_2& \ge 9,\\ x_1+2x_2& \ge 15,\\ 2x_1-3x_2& \le 9,\\ x_1,x_2& \ge 0.\end{array}$$$$\begin{array}{r}{x}_1+x_2\ge 9,\\ x_1+2x_2\ge 15,\\ 2x_1-3x_2\le 9,\\ x_1,x_2\ge 0.\end{array}$${:[x_(1)+x_(2) >= 9″,”],[x_(1)+2x_(2) >= 15″,”],[2x_(1)-3x_(2) <= 9″,”],[x_(1)”,”x_(2) >= 0.]:}\begin{aligned} x_1 + x_2 &\geq 9, \\ x_1 + 2x_2 &\geq 15, \\ 2x_1 – 3x_2 &\leq 9, \\ x_1, x_2 &\geq 0. \end{aligned}$$\begin{array}{rl}{x}_1+x_2& \ge 9,\\ x_1+2x_2& \ge 15,\\ 2x_1-3x_2& \le 9,\\ x_1,x_2& \ge 0.\end{array}$$

1. For $x_1+x_2\ge 9$$x_1+x_2\ge 9$x_(1)+x_(2) >= 9x_1 + x_2 \geq 9$x_1+x_2\ge 9$:
   Subtract surplus variable $s_1$$s_1$s_(1)s_1$s_1$ and add artificial variable $A_1$$A_1$A_(1)A_1$A_1$:
   $$x_1+x_2-s_1+A_1=9.$$$$x_1+x_2-s_1+A_1=9.$$x_(1)+x_(2)-s_(1)+A_(1)=9.x_1 + x_2 – s_1 + A_1 = 9.$$x_1+x_2-s_1+A_1=9.$$
2. For $x_1+2x_2\ge 15$$x_1+2x_2\ge 15$x_(1)+2x_(2) >= 15x_1 + 2x_2 \geq 15$x_1+2x_2\ge 15$:
   Subtract surplus variable $s_2$$s_2$s_(2)s_2$s_2$ and add artificial variable $A_2$$A_2$A_(2)A_2$A_2$:
   $$x_1+2x_2-s_2+A_2=15.$$$$x_1+2x_2-s_2+A_2=15.$$x_(1)+2x_(2)-s_(2)+A_(2)=15.x_1 + 2x_2 – s_2 + A_2 = 15.$$x_1+2x_2-s_2+A_2=15.$$
3. For $2x_1-3x_2\le 9$$2x_1-3x_2\le 9$2x_(1)-3x_(2) <= 92x_1 – 3x_2 \leq 9$2x_1-3x_2\le 9$:
   Add slack variable $s_3$$s_3$s_(3)s_3$s_3$:
   $$2x_1-3x_2+s_3=9.$$$$2x_1-3x_2+s_3=9.$$2x_(1)-3x_(2)+s_(3)=9.2x_1 – 3x_2 + s_3 = 9.$$2x_1-3x_2+s_3=9.$$

Step 2: Formulate the Big M Objective Function

Step 3: Construct the Initial Simplex Tableau

Basis	$x_1$$x_1$x_(1)x_1$x_1$	$x_2$$x_2$x_(2)x_2$x_2$	$s_1$$s_1$s_(1)s_1$s_1$	$s_2$$s_2$s_(2)s_2$s_2$	$s_3$$s_3$s_(3)s_3$s_3$	$A_1$$A_1$A_(1)A_1$A_1$	$A_2$$A_2$A_(2)A_2$A_2$	RHS
$A_1$$A_1$A_(1)A_1$A_1$	1	1	-1	0	0	1	0	9
$A_2$$A_2$A_(2)A_2$A_2$	1	2	0	-1	0	0	1	15
$s_3$$s_3$s_(3)s_3$s_3$	2	-3	0	0	1	0	0	9
Z’	$2+2M$$2+2M$2+2M2 + 2M$2+2M$	$3+3M$$3+3M$3+3M3 + 3M$3+3M$	$-M$$-M$-M-M$-M$	$-M$$-M$-M-M$-M$	0	0	0	$24M$$24M$24 M24M$24M$

Step 4: Perform Simplex Iterations

• Entering Variable: $x_2$$x_2$x_(2)x_2$x_2$ (most positive coefficient in $Z^{\prime }$$Z^{\prime }$Z^(‘)Z’$Z^{\prime }$).
• Leaving Variable: $A_2$$A_2$A_(2)A_2$A_2$ (min ratio test: $min(9/1,15/2,\text{ignore}{s}_3)=7.5$$min(9/1,15/2,\text{ignore }{s}_3)=7.5$min(9//1,15//2,”ignore “s_(3))=7.5\min(9/1, 15/2, \text{ignore } s_3) = 7.5$min(9/1,15/2,\text{ignore }{s}_3)=7.5$).

Basis	$x_1$$x_1$x_(1)x_1$x_1$	$x_2$$x_2$x_(2)x_2$x_2$	$s_1$$s_1$s_(1)s_1$s_1$	$s_2$$s_2$s_(2)s_2$s_2$	$s_3$$s_3$s_(3)s_3$s_3$	$A_1$$A_1$A_(1)A_1$A_1$	$A_2$$A_2$A_(2)A_2$A_2$	RHS
$A_1$$A_1$A_(1)A_1$A_1$	0.5	0	-1	0.5	0	1	-0.5	1.5
$x_2$$x_2$x_(2)x_2$x_2$	0.5	1	0	-0.5	0	0	0.5	7.5
$s_3$$s_3$s_(3)s_3$s_3$	3.5	0	0	-1.5	1	0	1.5	31.5
Z’	$0.5+0.5M$$0.5+0.5M$0.5+0.5 M0.5 + 0.5M$0.5+0.5M$	0	$-M$$-M$-M-M$-M$	$1.5+1.5M$$1.5+1.5M$1.5+1.5 M1.5 + 1.5M$1.5+1.5M$	0	0	$-1.5-1.5M$$-1.5-1.5M$-1.5-1.5 M-1.5 – 1.5M$-1.5-1.5M$	$22.5-1.5M$$22.5-1.5M$22.5-1.5 M22.5 – 1.5M$22.5-1.5M$

• Entering Variable: $x_1$$x_1$x_(1)x_1$x_1$.
• Leaving Variable: $A_1$$A_1$A_(1)A_1$A_1$ (min ratio test: $min(1.5/0.5,7.5/0.5,31.5/3.5)=3$$min(1.5/0.5,7.5/0.5,31.5/3.5)=3$min(1.5//0.5,7.5//0.5,31.5//3.5)=3\min(1.5/0.5, 7.5/0.5, 31.5/3.5) = 3$min(1.5/0.5,7.5/0.5,31.5/3.5)=3$).

Basis	$x_1$$x_1$x_(1)x_1$x_1$	$x_2$$x_2$x_(2)x_2$x_2$	$s_1$$s_1$s_(1)s_1$s_1$	$s_2$$s_2$s_(2)s_2$s_2$	$s_3$$s_3$s_(3)s_3$s_3$	$A_1$$A_1$A_(1)A_1$A_1$	$A_2$$A_2$A_(2)A_2$A_2$	RHS
$x_1$$x_1$x_(1)x_1$x_1$	1	0	-2	1	0	2	-1	3
$x_2$$x_2$x_(2)x_2$x_2$	0	1	1	-1	0	-1	1	6
$s_3$$s_3$s_(3)s_3$s_3$	0	0	7	-5	1	-7	5	21
Z’	0	0	$1-2M$$1-2M$1-2M1 – 2M$1-2M$	$1+M$$1+M$1+M1 + M$1+M$	0	$-1+2M$$-1+2M$-1+2M-1 + 2M$-1+2M$	$-1-M$$-1-M$-1-M-1 – M$-1-M$	21

• Optimality Check: All coefficients in $Z^{\prime }$$Z^{\prime }$Z^(‘)Z’$Z^{\prime }$ are non-positive (since $M$$M$MM$M$ is large and positive).
• Artificial Variables: $A_1$$A_1$A_(1)A_1$A_1$ and $A_2$$A_2$A_(2)A_2$A_2$ are non-basic (coefficients are zero in $Z^{\prime }$$Z^{\prime }$Z^(‘)Z’$Z^{\prime }$), so they can be removed.

Step 5: Extract the Optimal Solution

Step 6: Check Uniqueness of the Optimal Solution

• Uniqueness Condition: If all non-basic variables in the final tableau have strictly negative coefficients in the $Z$$Z$ZZ$Z$-row, the solution is unique.
• Final $Z$$Z$ZZ$Z$-row Coefficients:$$\begin{gathered} \text{For}{s}_1:1-2M(\text{Negative for large}M), \\ \text{For}{s}_2:1+M(\text{Positive}), \\ \text{For}{A}_1:-1+2M(\text{Positive for large}M), \\ \text{For}{A}_2:-1-M(\text{Negative}). \end{gathered}$$$$\begin{gathered} \text{For }{s}_1:1-2M(\text{Negative for large }M), \\ \text{For }{s}_2:1+M(\text{Positive}), \\ \text{For }{A}_1:-1+2M(\text{Positive for large }M), \\ \text{For }{A}_2:-1-M(\text{Negative}). \end{gathered}$$“For “s_(1):1-2M quad(“Negative for large “M),”For “s_(2):1+M quad(“Positive”),”For “A_(1):-1+2M quad(“Positive for large “M),”For “A_(2):-1-M quad(“Negative”).\text{For } s_1: 1 – 2M \quad (\text{Negative for large } M), \\ \text{For } s_2: 1 + M \quad (\text{Positive}), \\ \text{For } A_1: -1 + 2M \quad (\text{Positive for large } M), \\ \text{For } A_2: -1 – M \quad (\text{Negative}).$$\begin{gathered} \text{For }{s}_1:1-2M(\text{Negative for large }M), \\ \text{For }{s}_2:1+M(\text{Positive}), \\ \text{For }{A}_1:-1+2M(\text{Positive for large }M), \\ \text{For }{A}_2:-1-M(\text{Negative}). \end{gathered}$$
• Conclusion: Since the coefficient for $s_2$$s_2$s_(2)s_2$s_2$ is positive, there exists an alternative optimal solution. Thus, the optimal solution is not unique.

Final Answer

• Optimal Solution: $x_1=3$$x_1=3$x_(1)=3x_1 = 3$x_1=3$, $x_2=6$$x_2=6$x_(2)=6x_2 = 6$x_2=6$, with $Z=24$$Z=24$Z=24Z = 24$Z=24$.
• Uniqueness: The optimal solution is not unique because the coefficient of $s_2$$s_2$s_(2)s_2$s_2$ in the final $Z$$Z$ZZ$Z$-row is positive.

(a) Prove that the oscillation of a real-valued bounded function $f$$f$ff$f$ defined on $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$ is the supremum of the set $\{|f(x_1)-f(x_2)|:x_1,x_2\in [a,b]\}$$\{|f(x_1)-f(x_2)|:x_1,x_2\in [a,b]\}${|f(x_(1))-f(x_(2))|:x_(1),x_(2)in[a,b]}\{ |f(x_1) – f(x_2)| : x_1, x_2 \in [a, b] \}$\{|f(x_1)-f(x_2)|:x_1,x_2\in [a,b]\}$.

Answer:

Proof: Oscillation of a Bounded Function on $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$

Definitions:

1. Oscillation of $f$$f$ff$f$ on $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$:
   $$\text{osc}(f,[a,b])=\underset{x\in [a,b]}{sup}f(x)-\underset{x\in [a,b]}{inf}f(x).$$$$\text{osc}(f,[a,b])=\underset{x\in [a,b]}{sup} f(x)-\underset{x\in [a,b]}{inf} f(x).$$“osc”(f,[a,b])=s u p_(x in[a,b])f(x)-i n f_(x in[a,b])f(x).\text{osc}(f, [a, b]) = \sup_{x \in [a, b]} f(x) – \inf_{x \in [a, b]} f(x).$$\text{osc}(f,[a,b])=\underset{x\in [a,b]}{sup}f(x)-\underset{x\in [a,b]}{inf}f(x).$$
2. Set of Absolute Differences:
   $$S=\{|f(x_1)-f(x_2)|:x_1,x_2\in [a,b]\}.$$$$S=\{|f(x_1)-f(x_2)|:x_1,x_2\in [a,b]\}.$$S={|f(x_(1))-f(x_(2))|:x_(1),x_(2)in[a,b]}.S = \{ |f(x_1) – f(x_2)| : x_1, x_2 \in [a, b] \}.$$S=\{|f(x_1)-f(x_2)|:x_1,x_2\in [a,b]\}.$$

Goal:

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Step 1: Show $\text{osc}(f,[a,b])\ge supS$$\text{osc}(f,[a,b])\ge supS$“osc”(f,[a,b]) >= s u p S\text{osc}(f, [a, b]) \geq \sup S$\text{osc}(f,[a,b])\ge supS$

• By definition of supremum and infimum, for any $x_1,x_2\in [a,b]$$x_1,x_2\in [a,b]$x_(1),x_(2)in[a,b]x_1, x_2 \in [a, b]$x_1,x_2\in [a,b]$:$$f(x_1)\le \underset{x\in [a,b]}{sup}f(x),f(x_2)\ge \underset{x\in [a,b]}{inf}f(x).$$$$f(x_1)\le \underset{x\in [a,b]}{sup} f(x),f(x_2)\ge \underset{x\in [a,b]}{inf} f(x).$$f(x_(1)) <= s u p_(x in[a,b])f(x),quad f(x_(2)) >= i n f_(x in[a,b])f(x).f(x_1) \leq \sup_{x \in [a, b]} f(x), \quad f(x_2) \geq \inf_{x \in [a, b]} f(x).$$f(x_1)\le \underset{x\in [a,b]}{sup}f(x),f(x_2)\ge \underset{x\in [a,b]}{inf}f(x).$$
• Therefore:$$|f(x_1)-f(x_2)|\le supf-inff=\text{osc}(f,[a,b]).$$$$|f(x_1)-f(x_2)|\le supf-inff=\text{osc}(f,[a,b]).$$|f(x_(1))-f(x_(2))| <= s u p f-i n f f=”osc”(f,[a,b]).|f(x_1) – f(x_2)| \leq \sup f – \inf f = \text{osc}(f, [a, b]).$$|f(x_1)-f(x_2)|\le supf-inff=\text{osc}(f,[a,b]).$$
• Since this holds for all $x_1,x_2$$x_1,x_2$x_(1),x_(2)x_1, x_2$x_1,x_2$, it follows that:$$supS\le \text{osc}(f,[a,b]).$$$$supS\le \text{osc}(f,[a,b]).$$s u p S <= “osc”(f,[a,b]).\sup S \leq \text{osc}(f, [a, b]).$$supS\le \text{osc}(f,[a,b]).$$

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Step 2: Show $\text{osc}(f,[a,b])\le supS$$\text{osc}(f,[a,b])\le supS$“osc”(f,[a,b]) <= s u p S\text{osc}(f, [a, b]) \leq \sup S$\text{osc}(f,[a,b])\le supS$

• Let $supf=f(x^{\ast })$$supf=f(x^{\ast })$s u p f=f(x^(**))\sup f = f(x^*)$supf=f(x^{\ast })$ and $inff=f(x_{\ast })$$inff=f(x_{\ast })$i n f f=f(x_(**))\inf f = f(x_*)$inff=f(x_{\ast })$ for some $x^{\ast },x_{\ast }\in [a,b]$$x^{\ast },x_{\ast }\in [a,b]$x^(**),x_(**)in[a,b]x^*, x_* \in [a, b]$x^{\ast },x_{\ast }\in [a,b]$ (since $f$$f$ff$f$ is bounded and $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$ is compact, the supremum and infimum are attained).
• Then:$$\text{osc}(f,[a,b])=f(x^{\ast })-f(x_{\ast })=|f(x^{\ast })-f(x_{\ast })|\in S.$$$$\text{osc}(f,[a,b])=f(x^{\ast })-f(x_{\ast })=|f(x^{\ast })-f(x_{\ast })|\in S.$$“osc”(f,[a,b])=f(x^(**))-f(x_(**))=|f(x^(**))-f(x_(**))|in S.\text{osc}(f, [a, b]) = f(x^*) – f(x_*) = |f(x^*) – f(x_*)| \in S.$$\text{osc}(f,[a,b])=f(x^{\ast })-f(x_{\ast })=|f(x^{\ast })-f(x_{\ast })|\in S.$$
• Since $supS$$supS$s u p S\sup S$supS$ is the least upper bound of $S$$S$SS$S$, and $\text{osc}(f,[a,b])\in S$$\text{osc}(f,[a,b])\in S$“osc”(f,[a,b])in S\text{osc}(f, [a, b]) \in S$\text{osc}(f,[a,b])\in S$, we have:$$\text{osc}(f,[a,b])\le supS.$$$$\text{osc}(f,[a,b])\le supS.$$“osc”(f,[a,b]) <= s u p S.\text{osc}(f, [a, b]) \leq \sup S.$$\text{osc}(f,[a,b])\le supS.$$

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Step 3: Conclusion

Final Answer

Classify the singular point $z=0$$z=0$z=0z = 0$z=0$ of the function $f(z)=\frac{e^z}{z-\sin z}$$f(z)=\frac{e^z}{z-\sin z}$f(z)=(e^(z))/(z-sin z)f(z) = \frac{e^z}{z – \sin z}$f(z)=\frac{e^z}{z-\sin z}$ and obtain the principal part of its Laurent series expansion.

Answer:

Step 1: Analyze the Denominator $z-\sin z$$z-\sin z$z-sin zz – \sin z$z-\sin z$ Near $z=0$$z=0$z=0z = 0$z=0$

Step 2: Analyze the Numerator $e^z$$e^z$e^(z)e^z$e^z$ Near $z=0$$z=0$z=0z = 0$z=0$

Step 3: Determine the Nature of the Singularity

Step 4: Find the Laurent Series Expansion Near $z=0$$z=0$z=0z = 0$z=0$

Step 5: Identify the Principal Part

Final Answer

• Classification: $z=0$$z=0$z=0z = 0$z=0$ is a pole of order 3.
• Principal Part of Laurent Series:

A department head has 5 subordinates and 5 jobs to be performed. The time (in hours) that each subordinate will take to perform each job is given in the matrix below:

	A	B	C	D	E
I	4	9	4	12	4
II	15	11	20	5	8
III	17	7	15	12	18
IV	9	13	11	9	14
V	6	11	12	9	14

Answer:

Evaluate, using binary arithmetic, the following numbers in their given system:

Answer:

Solution:

(i) $(634.235{)}_8-(132.223{)}_8$$(634.235{)}_8-(132.223{)}_8$(634.235)_(8)-(132.223)_(8)(634.235)_8 – (132.223)_8$(634.235{)}_8-(132.223{)}_8$)

• $(634.235{)}_8$$(634.235{)}_8$(634.235)_(8)(634.235)_8$(634.235{)}_8$ to binary:
  $$6\to 110,3\to 011,4\to 100,.2\to 010,3\to 011,5\to 101$$$$6\to 110,3\to 011,4\to 100,.2\to 010,3\to 011,5\to 101$$6rarr110,quad3rarr011,quad4rarr100,quad.quad2rarr010,quad3rarr011,quad5rarr1016 \rightarrow 110, \quad 3 \rightarrow 011, \quad 4 \rightarrow 100, \quad . \quad 2 \rightarrow 010, \quad 3 \rightarrow 011, \quad 5 \rightarrow 101$$6\to 110,3\to 011,4\to 100,.2\to 010,3\to 011,5\to 101$$
  $$(634.235{)}_8=(110011100.010011101{)}_2$$$$(634.235{)}_8=(110011100.010011101{)}_2$$(634.235)_(8)=(110011100.010011101)_(2)(634.235)_8 = (110\,011\,100.010\,011\,101)_2$$(634.235{)}_8=(110011100.010011101{)}_2$$
• $(132.223{)}_8$$(132.223{)}_8$(132.223)_(8)(132.223)_8$(132.223{)}_8$ to binary:
  $$1\to 001,3\to 011,2\to 010,.2\to 010,2\to 010,3\to 011$$$$1\to 001,3\to 011,2\to 010,.2\to 010,2\to 010,3\to 011$$1rarr001,quad3rarr011,quad2rarr010,quad.quad2rarr010,quad2rarr010,quad3rarr0111 \rightarrow 001, \quad 3 \rightarrow 011, \quad 2 \rightarrow 010, \quad . \quad 2 \rightarrow 010, \quad 2 \rightarrow 010, \quad 3 \rightarrow 011$$1\to 001,3\to 011,2\to 010,.2\to 010,2\to 010,3\to 011$$
  $$(132.223{)}_8=(001011010.010010011{)}_2$$$$(132.223{)}_8=(001011010.010010011{)}_2$$(132.223)_(8)=(001011010.010010011)_(2)(132.223)_8 = (001\,011\,010.010\,010\,011)_2$$(132.223{)}_8=(001011010.010010011{)}_2$$

• Subtraction process (using two’s complement for negative results):$$110011100.010011101-001011010.010010011=101000010.000001010$$$$110011100.010011101-001011010.010010011=101000010.000001010$$110011100.010011101-001011010.010010011=101000010.000001010110\,011\,100.010\,011\,101 – 001\,011\,010.010\,010\,011 = 101\,000\,010.000\,001\,010$$110011100.010011101-001011010.010010011=101000010.000001010$$

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(ii) $(7AB.432{)}_{16}-(5CA.D61{)}_{16}$$(7AB.432{)}_{16}-(5CA.D61{)}_{16}$(7AB.432)_(16)-(5CA.D 61)_(16)(7AB.432)_{16} – (5CA.D61)_{16}$(7AB.432{)}_{16}-(5CA.D61{)}_{16}$)

• $(7AB.432{)}_{16}$$(7AB.432{)}_{16}$(7AB.432)_(16)(7AB.432)_{16}$(7AB.432{)}_{16}$ to binary:
  $$7\to 0111,A\to 1010,B\to 1011,.4\to 0100,3\to 0011,2\to 0010$$$$7\to 0111,A\to 1010,B\to 1011,.4\to 0100,3\to 0011,2\to 0010$$7rarr0111,quad A rarr1010,quad B rarr1011,quad.quad4rarr0100,quad3rarr0011,quad2rarr00107 \rightarrow 0111, \quad A \rightarrow 1010, \quad B \rightarrow 1011, \quad . \quad 4 \rightarrow 0100, \quad 3 \rightarrow 0011, \quad 2 \rightarrow 0010$$7\to 0111,A\to 1010,B\to 1011,.4\to 0100,3\to 0011,2\to 0010$$
  $$(7AB.432{)}_{16}=(011110101011.010000110010{)}_2$$$$(7AB.432{)}_{16}=(011110101011.010000110010{)}_2$$(7AB.432)_(16)=(011110101011.010000110010)_(2)(7AB.432)_{16} = (0111\,1010\,1011.0100\,0011\,0010)_2$$(7AB.432{)}_{16}=(011110101011.010000110010{)}_2$$
• $(5CA.D61{)}_{16}$$(5CA.D61{)}_{16}$(5CA.D 61)_(16)(5CA.D61)_{16}$(5CA.D61{)}_{16}$ to binary:
  $$5\to 0101,C\to 1100,A\to 1010,.D\to 1101,6\to 0110,1\to 0001$$$$5\to 0101,C\to 1100,A\to 1010,.D\to 1101,6\to 0110,1\to 0001$$5rarr0101,quad C rarr1100,quad A rarr1010,quad.quad D rarr1101,quad6rarr0110,quad1rarr00015 \rightarrow 0101, \quad C \rightarrow 1100, \quad A \rightarrow 1010, \quad . \quad D \rightarrow 1101, \quad 6 \rightarrow 0110, \quad 1 \rightarrow 0001$$5\to 0101,C\to 1100,A\to 1010,.D\to 1101,6\to 0110,1\to 0001$$
  $$(5CA.D61{)}_{16}=(010111001010.110101100001{)}_2$$$$(5CA.D61{)}_{16}=(010111001010.110101100001{)}_2$$(5CA.D 61)_(16)=(010111001010.110101100001)_(2)(5CA.D61)_{16} = (0101\,1100\,1010.1101\,0110\,0001)_2$$(5CA.D61{)}_{16}=(010111001010.110101100001{)}_2$$

• Subtraction process (using two’s complement for negative results):$$011110101011.010000110010-010111001010.110101100001=000111011100.011011010001$$$$011110101011.010000110010-010111001010.110101100001=000111011100.011011010001$$011110101011.010000110010-010111001010.110101100001=000111011100.0110110100010111\,1010\,1011.0100\,0011\,0010 – 0101\,1100\,1010.1101\,0110\,0001 = 0001\,1101\,1100.0110\,1101\,0001$$011110101011.010000110010-010111001010.110101100001=000111011100.011011010001$$

A planet of mass $m$$m$mm$m$ is revolving around the sun of mass $M$$M$MM$M$. The kinetic energy $T$$T$TT$T$ and the potential energy $V$$V$VV$V$ of the planet are given by $T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)$$T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)$T=(1)/(2)m(r^(˙)^(2)+r^(2)theta^(˙)^(2))T = \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\theta}^2)$T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)$ and $V=GMm(\frac{1}{2a}-\frac{1}{r})$$V=GMm\left(\frac{1}{2a}-\frac{1}{r}\right)$V=GMm((1)/(2a)-(1)/(r))V = G M m \left( \frac{1}{2a} – \frac{1}{r} \right)$V=GMm(\frac{1}{2a}-\frac{1}{r})$, where $(r,\theta )$$(r,\theta )$(r,theta)(r, \theta)$(r,\theta )$ are the polar coordinates of the planet at time $t$$t$tt$t$, $G$$G$GG$G$ is the gravitational constant, and $2a$$2a$2a2a$2a$ is the major axis of the ellipse (the path of the planet). Find the Hamiltonian and the Hamilton equations of the planet’s motion.

Answer:

1. Lagrangian of the System

• Kinetic Energy (T):$$T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)$$$$T=\frac{1}{2}m\left({\dot{r}}^2+r^2{\dot{\theta }}^2\right)$$T=(1)/(2)m(r^(˙)^(2)+r^(2)theta^(˙)^(2))T = \frac{1}{2} m \left( \dot{r}^2 + r^2 \dot{\theta}^2 \right)$$T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)$$
• Potential Energy (V):$$V=GMm(\frac{1}{2a}-\frac{1}{r})$$$$V=GMm\left(\frac{1}{2a}-\frac{1}{r}\right)$$V=GMm((1)/(2a)-(1)/(r))V = G M m \left( \frac{1}{2a} – \frac{1}{r} \right)$$V=GMm(\frac{1}{2a}-\frac{1}{r})$$

2. Generalized Momenta

3. Hamiltonian

4. Hamilton’s Equations

(a) For $r$$r$rr$r$ and $p_r$$p_r$p_(r)p_r$p_r$:

(b) For $\theta$$\theta$theta\theta$\theta$ and $p_{\theta }$$p_{\theta }$p_( theta)p_\theta$p_{\theta }$:

Final Answer:

• Hamiltonian:
  $$H=\frac{p_r^2}{2m}+\frac{p_{\theta }^2}{2m{r}^2}+GMm(\frac{1}{2a}-\frac{1}{r})$$$$H=\frac{p_r^2}{2m}+\frac{p_{\theta }^2}{2m{r}^2}+GMm\left(\frac{1}{2a}-\frac{1}{r}\right)$$H=(p_(r)^(2))/(2m)+(p_( theta)^(2))/(2mr^(2))+GMm((1)/(2a)-(1)/(r))H = \frac{p_r^2}{2m} + \frac{p_\theta^2}{2m r^2} + G M m \left( \frac{1}{2a} – \frac{1}{r} \right)$$H=\frac{p_r^2}{2m}+\frac{p_{\theta }^2}{2m{r}^2}+GMm(\frac{1}{2a}-\frac{1}{r})$$
• Hamilton’s Equations:
  $$\{\begin{array}{l}\dot{r}={\displaystyle \frac{p_r}{m}},\\ {\dot{p}}_r={\displaystyle \frac{p_{\theta }^2}{m{r}^3}}-{\displaystyle \frac{GMm}{r^2}},\\ \dot{\theta }={\displaystyle \frac{p_{\theta }}{m{r}^2}},\\ {\dot{p}}_{\theta }=0(\text{angular momentum is conserved}).\end{array}$$$$\left\{\begin{array}{l}\dot{r}={\displaystyle \frac{p_r}{m}},\\ {\dot{p}}_r={\displaystyle \frac{p_{\theta }^2}{m{r}^3}}-{\displaystyle \frac{GMm}{r^2}},\\ \dot{\theta }={\displaystyle \frac{p_{\theta }}{m{r}^2}},\\ {\dot{p}}_{\theta }=0(\text{angular momentum is conserved}).\end{array}\right.$${[r^(˙)=(p_(r))/(m)”,”],[p^(˙)_(r)=(p_( theta)^(2))/(mr^(3))-(GMm)/(r^(2))”,”],[theta^(˙)=(p_( theta))/(mr^(2))”,”],[p^(˙)_(theta)=0quad(“angular momentum is conserved”).]:}\begin{cases} \dot{r} = \dfrac{p_r}{m}, \\ \dot{p}_r = \dfrac{p_\theta^2}{m r^3} – \dfrac{G M m}{r^2}, \\ \dot{\theta} = \dfrac{p_\theta}{m r^2}, \\ \dot{p}_\theta = 0 \quad (\text{angular momentum is conserved}). \end{cases}$$\{\begin{array}{l}\dot{r}={\displaystyle \frac{p_r}{m}},\\ {\dot{p}}_r={\displaystyle \frac{p_{\theta }^2}{m{r}^3}}-{\displaystyle \frac{GMm}{r^2}},\\ \dot{\theta }={\displaystyle \frac{p_{\theta }}{m{r}^2}},\\ {\dot{p}}_{\theta }=0(\text{angular momentum is conserved}).\end{array}$$

In a fluid motion, there is a source of strength $2m$$2m$2m2m$2m$ placed at $z=2$$z=2$z=2z = 2$z=2$ and two sinks of strength $m$$m$mm$m$ placed at $z=2+i$$z=2+i$z=2+iz = 2 + i$z=2+i$ and $z=2-i$$z=2-i$z=2-iz = 2 – i$z=2-i$. Find the streamlines.

Answer:

1. The Complex Potential

• A source of strength $k$$k$kk$k$ at $z_0$$z_0$z_(0)z_0$z_0$ has a complex potential $k\ln (z-z_0)$$k\ln (z-z_0)$k ln(z-z_(0))k \ln(z – z_0)$k\ln (z-z_0)$.
• A sink of strength $k$$k$kk$k$ at $z_0$$z_0$z_(0)z_0$z_0$ has a complex potential $-k\ln (z-z_0)$$-k\ln (z-z_0)$-k ln(z-z_(0))-k \ln(z – z_0)$-k\ln (z-z_0)$.

• Source of strength $2m$$2m$2m2m$2m$ at $z=2$$z=2$z=2z = 2$z=2$.
• Sink of strength $m$$m$mm$m$ at $z=2+i$$z=2+i$z=2+iz = 2 + i$z=2+i$.
• Sink of strength $m$$m$mm$m$ at $z=2-i$$z=2-i$z=2-iz = 2 – i$z=2-i$.

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2. The Stream Function

• ${\theta }_1=\arg (z-2)=\arg ((x-2)+iy)=\arctan \left(\frac{y}{x-2}\right)$${\theta }_1=\arg (z-2)=\arg ((x-2)+iy)=\arctan \left(\frac{y}{x-2}\right)$theta_(1)=arg(z-2)=arg((x-2)+iy)=arctan((y)/(x-2))\theta_1 = \arg(z-2) = \arg((x-2) + iy) = \arctan\left(\frac{y}{x-2}\right)${\theta }_1=\arg (z-2)=\arg ((x-2)+iy)=\arctan \left(\frac{y}{x-2}\right)$
• ${\theta }_2=\arg (z-2-i)=\arg ((x-2)+i(y-1))=\arctan \left(\frac{y-1}{x-2}\right)$${\theta }_2=\arg (z-2-i)=\arg ((x-2)+i(y-1))=\arctan \left(\frac{y-1}{x-2}\right)$theta_(2)=arg(z-2-i)=arg((x-2)+i(y-1))=arctan((y-1)/(x-2))\theta_2 = \arg(z-2-i) = \arg((x-2) + i(y-1)) = \arctan\left(\frac{y-1}{x-2}\right)${\theta }_2=\arg (z-2-i)=\arg ((x-2)+i(y-1))=\arctan \left(\frac{y-1}{x-2}\right)$
• ${\theta }_3=\arg (z-2+i)=\arg ((x-2)+i(y+1))=\arctan \left(\frac{y+1}{x-2}\right)$${\theta }_3=\arg (z-2+i)=\arg ((x-2)+i(y+1))=\arctan \left(\frac{y+1}{x-2}\right)$theta_(3)=arg(z-2+i)=arg((x-2)+i(y+1))=arctan((y+1)/(x-2))\theta_3 = \arg(z-2+i) = \arg((x-2) + i(y+1)) = \arctan\left(\frac{y+1}{x-2}\right)${\theta }_3=\arg (z-2+i)=\arg ((x-2)+i(y+1))=\arctan \left(\frac{y+1}{x-2}\right)$

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3. Equation of the Streamlines

(a) Find the surface passing through the two lines $z=x=0$$z=x=0$z=x=0z = x = 0$z=x=0$ and $z-1=x-y=0$$z-1=x-y=0$z-1=x-y=0z – 1 = x – y = 0$z-1=x-y=0$, and satisfying the partial differential equation $\frac{{\partial }^{2}z}{\partial x^2}-4\frac{{\partial }^{2}z}{\partial x\partial y}+4\frac{{\partial }^{2}z}{\partial y^2}=0$$\frac{{\partial }^{2}z}{\partial x^2}-4\frac{{\partial }^{2}z}{\partial x\partial y}+4\frac{{\partial }^{2}z}{\partial y^2}=0$(del^(2)z)/(delx^(2))-4(del^(2)z)/(del x del y)+4(del^(2)z)/(dely^(2))=0\frac{\partial^2 z}{\partial x^2} – 4 \frac{\partial^2 z}{\partial x \partial y} + 4 \frac{\partial^2 z}{\partial y^2} = 0$\frac{{\partial }^{2}z}{\partial x^2}-4\frac{{\partial }^{2}z}{\partial x\partial y}+4\frac{{\partial }^{2}z}{\partial y^2}=0$.

Answer:

Given:

1. Lines:
   • $L_1:z=x=0$$L_1:z=x=0$L_(1):z=x=0L_1: z = x = 0$L_1:z=x=0$ (the $y$$y$yy$y$-axis),
   • $L_2:z-1=x-y=0$$L_2:z-1=x-y=0$L_(2):z-1=x-y=0L_2: z – 1 = x – y = 0$L_2:z-1=x-y=0$ (the line $x=y$$x=y$x=yx = y$x=y$ at height $z=1$$z=1$z=1z = 1$z=1$).
2. PDE:
   $$\frac{{\partial }^{2}z}{\partial x^2}-4\frac{{\partial }^{2}z}{\partial x\partial y}+4\frac{{\partial }^{2}z}{\partial y^2}=0.$$$$\frac{{\partial }^{2}z}{\partial x^2}-4\frac{{\partial }^{2}z}{\partial x\partial y}+4\frac{{\partial }^{2}z}{\partial y^2}=0.$$(del^(2)z)/(delx^(2))-4(del^(2)z)/(del x del y)+4(del^(2)z)/(dely^(2))=0.\frac{\partial^2 z}{\partial x^2} – 4 \frac{\partial^2 z}{\partial x \partial y} + 4 \frac{\partial^2 z}{\partial y^2} = 0.$$\frac{{\partial }^{2}z}{\partial x^2}-4\frac{{\partial }^{2}z}{\partial x\partial y}+4\frac{{\partial }^{2}z}{\partial y^2}=0.$$

Step 1: Solve the PDE

Step 2: Apply Boundary Conditions

1. On $L_1$$L_1$L_(1)L_1$L_1$: $x=0$$x=0$x=0x = 0$x=0$, $z=0$$z=0$z=0z = 0$z=0$.
   Substituting $x=0$$x=0$x=0x = 0$x=0$ into the general solution:
   $$z(0,y)=f(y)+0\cdot g(y)=f(y)=0.$$$$z(0,y)=f(y)+0\cdot g(y)=f(y)=0.$$z(0,y)=f(y)+0*g(y)=f(y)=0.z(0, y) = f(y) + 0 \cdot g(y) = f(y) = 0.$$z(0,y)=f(y)+0\cdot g(y)=f(y)=0.$$
   Thus, $f(y)=0$$f(y)=0$f(y)=0f(y) = 0$f(y)=0$, and the solution simplifies to:
   $$z(x,y)=xg(y+2x).$$$$z(x,y)=xg(y+2x).$$z(x,y)=xg(y+2x).z(x, y) = x g(y + 2x).$$z(x,y)=xg(y+2x).$$
2. On $L_2$$L_2$L_(2)L_2$L_2$: $x=y$$x=y$x=yx = y$x=y$, $z=1$$z=1$z=1z = 1$z=1$.
   Substituting $x=y$$x=y$x=yx = y$x=y$ into the simplified solution:
   $$z(y,y)=yg(y+2y)=yg(3y)=1.$$$$z(y,y)=yg(y+2y)=yg(3y)=1.$$z(y,y)=yg(y+2y)=yg(3y)=1.z(y, y) = y g(y + 2y) = y g(3y) = 1.$$z(y,y)=yg(y+2y)=yg(3y)=1.$$
   Thus, $g(3y)=\frac{1}{y}$$g(3y)=\frac{1}{y}$g(3y)=(1)/(y)g(3y) = \frac{1}{y}$g(3y)=\frac{1}{y}$. Let $\xi =3y$$\xi =3y$xi=3y\xi = 3y$\xi =3y$, so $y=\frac{\xi }{3}$$y=\frac{\xi }{3}$y=(xi)/(3)y = \frac{\xi}{3}$y=\frac{\xi }{3}$, and:
   $$g(\xi )=\frac{3}{\xi }.$$$$g(\xi )=\frac{3}{\xi }.$$g(xi)=(3)/(xi).g(\xi) = \frac{3}{\xi}.$$g(\xi )=\frac{3}{\xi }.$$
   Therefore, $g(y+2x)=\frac{3}{y+2x}$$g(y+2x)=\frac{3}{y+2x}$g(y+2x)=(3)/(y+2x)g(y + 2x) = \frac{3}{y + 2x}$g(y+2x)=\frac{3}{y+2x}$.

Step 3: Final Solution

Verification:

1. Check PDE:
   Compute the second derivatives:
   $$z_x=\frac{3(y+2x)-3x\cdot 2}{(2x+y{)}^2}=\frac{3y}{(2x+y{)}^2},$$$$z_x=\frac{3(y+2x)-3x\cdot 2}{(2x+y{)}^2}=\frac{3y}{(2x+y{)}^2},$$z_(x)=(3(y+2x)-3x*2)/((2x+y)^(2))=(3y)/((2x+y)^(2)),z_x = \frac{3(y + 2x) – 3x \cdot 2}{(2x + y)^2} = \frac{3y}{(2x + y)^2},$$z_x=\frac{3(y+2x)-3x\cdot 2}{(2x+y{)}^2}=\frac{3y}{(2x+y{)}^2},$$
   $$z_{xx}=\frac{-6y\cdot 2}{(2x+y{)}^3}=\frac{-12y}{(2x+y{)}^3},$$$$z_{xx}=\frac{-6y\cdot 2}{(2x+y{)}^3}=\frac{-12y}{(2x+y{)}^3},$$z_(xx)=(-6y*2)/((2x+y)^(3))=(-12 y)/((2x+y)^(3)),z_{xx} = \frac{-6y \cdot 2}{(2x + y)^3} = \frac{-12y}{(2x + y)^3},$$z_{xx}=\frac{-6y\cdot 2}{(2x+y{)}^3}=\frac{-12y}{(2x+y{)}^3},$$
   $$z_y=\frac{-3x}{(2x+y{)}^2},$$$$z_y=\frac{-3x}{(2x+y{)}^2},$$z_(y)=(-3x)/((2x+y)^(2)),z_y = \frac{-3x}{(2x + y)^2},$$z_y=\frac{-3x}{(2x+y{)}^2},$$
   $$z_{yy}=\frac{6x}{(2x+y{)}^3},$$$$z_{yy}=\frac{6x}{(2x+y{)}^3},$$z_(yy)=(6x)/((2x+y)^(3)),z_{yy} = \frac{6x}{(2x + y)^3},$$z_{yy}=\frac{6x}{(2x+y{)}^3},$$
   $$z_{xy}=\frac{3(2x+y{)}^2-3y\cdot 2(2x+y)}{(2x+y{)}^4}=\frac{6x+3y-6y}{(2x+y{)}^3}=\frac{6x-3y}{(2x+y{)}^3}.$$$$z_{xy}=\frac{3(2x+y{)}^2-3y\cdot 2(2x+y)}{(2x+y{)}^4}=\frac{6x+3y-6y}{(2x+y{)}^3}=\frac{6x-3y}{(2x+y{)}^3}.$$z_(xy)=(3(2x+y)^(2)-3y*2(2x+y))/((2x+y)^(4))=(6x+3y-6y)/((2x+y)^(3))=(6x-3y)/((2x+y)^(3)).z_{xy} = \frac{3(2x + y)^2 – 3y \cdot 2(2x + y)}{(2x + y)^4} = \frac{6x + 3y – 6y}{(2x + y)^3} = \frac{6x – 3y}{(2x + y)^3}.$$z_{xy}=\frac{3(2x+y{)}^2-3y\cdot 2(2x+y)}{(2x+y{)}^4}=\frac{6x+3y-6y}{(2x+y{)}^3}=\frac{6x-3y}{(2x+y{)}^3}.$$
   Substituting into the PDE:
   $$z_{xx}-4z_{xy}+4z_{yy}=\frac{-12y}{(2x+y{)}^3}-4\cdot \frac{6x-3y}{(2x+y{)}^3}+4\cdot \frac{6x}{(2x+y{)}^3}=0.$$$$z_{xx}-4z_{xy}+4z_{yy}=\frac{-12y}{(2x+y{)}^3}-4\cdot \frac{6x-3y}{(2x+y{)}^3}+4\cdot \frac{6x}{(2x+y{)}^3}=0.$$z_(xx)-4z_(xy)+4z_(yy)=(-12 y)/((2x+y)^(3))-4*(6x-3y)/((2x+y)^(3))+4*(6x)/((2x+y)^(3))=0.z_{xx} – 4 z_{xy} + 4 z_{yy} = \frac{-12y}{(2x + y)^3} – 4 \cdot \frac{6x – 3y}{(2x + y)^3} + 4 \cdot \frac{6x}{(2x + y)^3} = 0.$$z_{xx}-4z_{xy}+4z_{yy}=\frac{-12y}{(2x+y{)}^3}-4\cdot \frac{6x-3y}{(2x+y{)}^3}+4\cdot \frac{6x}{(2x+y{)}^3}=0.$$
   Simplifying:
   $$-12y-24x+12y+24x=0.$$$$-12y-24x+12y+24x=0.$$-12 y-24 x+12 y+24 x=0.-12y – 24x + 12y + 24x = 0.$$-12y-24x+12y+24x=0.$$
   The PDE is satisfied.
2. Check Boundary Conditions:
   • On $L_1$$L_1$L_(1)L_1$L_1$: $x=0$$x=0$x=0x = 0$x=0$, $z=\frac{0}{0+y}=0$$z=\frac{0}{0+y}=0$z=(0)/(0+y)=0z = \frac{0}{0 + y} = 0$z=\frac{0}{0+y}=0$.
   • On $L_2$$L_2$L_(2)L_2$L_2$: $x=y$$x=y$x=yx = y$x=y$, $z=\frac{3y}{2y+y}=1$$z=\frac{3y}{2y+y}=1$z=(3y)/(2y+y)=1z = \frac{3y}{2y + y} = 1$z=\frac{3y}{2y+y}=1$.

Final Answer:

Solve the system of linear equations

Answer:

Gauss-Seidel Method

1. From $7x_1-x_2+2x_3=11$$7x_1-x_2+2x_3=11$7x_(1)-x_(2)+2x_(3)=117x_1 – x_2 + 2x_3 = 11$7x_1-x_2+2x_3=11$:

2. From $2x_1+8x_2-x_3=9$$2x_1+8x_2-x_3=9$2x_(1)+8x_(2)-x_(3)=92x_1 + 8x_2 – x_3 = 9$2x_1+8x_2-x_3=9$:

3. From $x_1-2x_2+9x_3=7$$x_1-2x_2+9x_3=7$x_(1)-2x_(2)+9x_(3)=7x_1 – 2x_2 + 9x_3 = 7$x_1-2x_2+9x_3=7$:

Iteration Process

• Compute $x_1^{(1)}$$x_1^{(1)}$x_(1)^((1))x_1^{(1)}$x_1^{(1)}$:

• Compute $x_2^{(1)}$$x_2^{(1)}$x_(2)^((1))x_2^{(1)}$x_2^{(1)}$, using $x_1^{(1)}$$x_1^{(1)}$x_(1)^((1))x_1^{(1)}$x_1^{(1)}$:

• Compute $x_3^{(1)}$$x_3^{(1)}$x_(3)^((1))x_3^{(1)}$x_3^{(1)}$, using $x_1^{(1)}$$x_1^{(1)}$x_(1)^((1))x_1^{(1)}$x_1^{(1)}$, $x_2^{(1)}$$x_2^{(1)}$x_(2)^((1))x_2^{(1)}$x_2^{(1)}$:

• Compute $x_1^{(2)}$$x_1^{(2)}$x_(1)^((2))x_1^{(2)}$x_1^{(2)}$:

• Compute $x_2^{(2)}$$x_2^{(2)}$x_(2)^((2))x_2^{(2)}$x_2^{(2)}$:

• Compute $x_3^{(2)}$$x_3^{(2)}$x_(3)^((2))x_3^{(2)}$x_3^{(2)}$:

• Compute $x_1^{(3)}$$x_1^{(3)}$x_(1)^((3))x_1^{(3)}$x_1^{(3)}$:

• Compute $x_2^{(3)}$$x_2^{(3)}$x_(2)^((3))x_2^{(3)}$x_2^{(3)}$:

• Compute $x_3^{(3)}$$x_3^{(3)}$x_(3)^((3))x_3^{(3)}$x_3^{(3)}$:

• Compute $x_1^{(4)}$$x_1^{(4)}$x_(1)^((4))x_1^{(4)}$x_1^{(4)}$:

• Compute $x_2^{(4)}$$x_2^{(4)}$x_(2)^((4))x_2^{(4)}$x_2^{(4)}$:

• Compute $x_3^{(4)}$$x_3^{(4)}$x_(3)^((4))x_3^{(4)}$x_3^{(4)}$:

• Compute $x_1^{(5)}$$x_1^{(5)}$x_(1)^((5))x_1^{(5)}$x_1^{(5)}$:

• Compute $x_2^{(5)}$$x_2^{(5)}$x_(2)^((5))x_2^{(5)}$x_2^{(5)}$:

• Compute $x_3^{(5)}$$x_3^{(5)}$x_(3)^((5))x_3^{(5)}$x_3^{(5)}$:

Final Answer

1. $7x_1-x_2+2x_3=7\cdot 1.464-0.8598+2\cdot 0.8062=10.248-0.8598+1.6124=11.0006\approx 11$$7x_1-x_2+2x_3=7\cdot 1.464-0.8598+2\cdot 0.8062=10.248-0.8598+1.6124=11.0006\approx 11$7x_(1)-x_(2)+2x_(3)=7*1.464-0.8598+2*0.8062=10.248-0.8598+1.6124=11.0006~~117x_1 – x_2 + 2x_3 = 7 \cdot 1.464 – 0.8598 + 2 \cdot 0.8062 = 10.248 – 0.8598 + 1.6124 = 11.0006 \approx 11$7x_1-x_2+2x_3=7\cdot 1.464-0.8598+2\cdot 0.8062=10.248-0.8598+1.6124=11.0006\approx 11$
2. $2x_1+8x_2-x_3=2\cdot 1.464+8\cdot 0.8598-0.8062=2.928+6.8784-0.8062=9.0002\approx 9$$2x_1+8x_2-x_3=2\cdot 1.464+8\cdot 0.8598-0.8062=2.928+6.8784-0.8062=9.0002\approx 9$2x_(1)+8x_(2)-x_(3)=2*1.464+8*0.8598-0.8062=2.928+6.8784-0.8062=9.0002~~92x_1 + 8x_2 – x_3 = 2 \cdot 1.464 + 8 \cdot 0.8598 – 0.8062 = 2.928 + 6.8784 – 0.8062 = 9.0002 \approx 9$2x_1+8x_2-x_3=2\cdot 1.464+8\cdot 0.8598-0.8062=2.928+6.8784-0.8062=9.0002\approx 9$
3. $x_1-2x_2+9x_3=1.464-2\cdot 0.8598+9\cdot 0.8062=1.464-1.7196+7.2558=7.0002\approx 7$$x_1-2x_2+9x_3=1.464-2\cdot 0.8598+9\cdot 0.8062=1.464-1.7196+7.2558=7.0002\approx 7$x_(1)-2x_(2)+9x_(3)=1.464-2*0.8598+9*0.8062=1.464-1.7196+7.2558=7.0002~~7x_1 – 2x_2 + 9x_3 = 1.464 – 2 \cdot 0.8598 + 9 \cdot 0.8062 = 1.464 – 1.7196 + 7.2558 = 7.0002 \approx 7$x_1-2x_2+9x_3=1.464-2\cdot 0.8598+9\cdot 0.8062=1.464-1.7196+7.2558=7.0002\approx 7$

A mechanical system with 2 degrees of freedom has the Lagrangian

Answer:

Given Lagrangian:

Transformation:

Step 1: Express ${\dot{x}}^2+{\dot{y}}^2$${\dot{x}}^2+{\dot{y}}^2$x^(˙)^(2)+y^(˙)^(2)\dot{x}^2 + \dot{y}^2${\dot{x}}^2+{\dot{y}}^2$ in terms of ${\dot{q}}_1,{\dot{q}}_2$${\dot{q}}_1,{\dot{q}}_2$q^(˙)_(1),q^(˙)_(2)\dot{q}_1, \dot{q}_2${\dot{q}}_1,{\dot{q}}_2$:

Step 2: Express the potential terms in terms of $q_1,q_2$$q_1,q_2$q_(1),q_(2)q_1, q_2$q_1,q_2$:

Step 3: Combine all terms in the Lagrangian:

Step 4: Eliminate the $q_1{q}_2$$q_1{q}_2$q_(1)q_(2)q_1 q_2$q_1{q}_2$ term:

Step 5: Lagrange’s Equations for $q_1$$q_1$q_(1)q_1$q_1$ and $q_2$$q_2$q_(2)q_2$q_2$:

Final Answer:

1. The parameter $\theta$$\theta$theta\theta$\theta$ is:
   $$\boxed{{\displaystyle \theta =\frac{1}{2}\arctan \left(\frac{2k}{m(w_2^2-w_1^2)}\right)}}$$$$\boxed{{\displaystyle \theta =\frac{1}{2}\arctan \left(\frac{2k}{m(w_2^2-w_1^2)}\right)}}$$theta=(1)/(2)arctan((2k)/(m(w_(2)^(2)-w_(1)^(2))))\boxed{\theta = \frac{1}{2} \arctan \left( \frac{2k}{m (w_2^2 – w_1^2)} \right)}$$\boxed{{\displaystyle \theta =\frac{1}{2}\arctan \left(\frac{2k}{m(w_2^2-w_1^2)}\right)}}$$
2. The Lagrange’s equations for $q_1$$q_1$q_(1)q_1$q_1$ and $q_2$$q_2$q_(2)q_2$q_2$ are:
   $$\boxed{{\displaystyle \begin{array}{rl}{\ddot{q}}_1+{\mathrm{\Omega }}_1^2{q}_1& =0,\\ {\ddot{q}}_2+{\mathrm{\Omega }}_2^2{q}_2& =0,\end{array}}}$$$$\boxed{{\displaystyle \begin{array}{r}{\ddot{q}}_1+{\mathrm{\Omega }}_1^2{q}_1=0,\\ {\ddot{q}}_2+{\mathrm{\Omega }}_2^2{q}_2=0,\end{array}}}$$[q^(¨)_(1)+Omega_(1)^(2)q_(1)=0″,”],[q^(¨)_(2)+Omega_(2)^(2)q_(2)=0″,”]\boxed{ \begin{aligned} \ddot{q}_1 + \Omega_1^2 q_1 &= 0, \\ \ddot{q}_2 + \Omega_2^2 q_2 &= 0, \end{aligned} }$$\boxed{{\displaystyle \begin{array}{rl}{\ddot{q}}_1+{\mathrm{\Omega }}_1^2{q}_1& =0,\\ {\ddot{q}}_2+{\mathrm{\Omega }}_2^2{q}_2& =0,\end{array}}}$$
   where ${\mathrm{\Omega }}_1^2$${\mathrm{\Omega }}_1^2$Omega_(1)^(2)\Omega_1^2${\mathrm{\Omega }}_1^2$ and ${\mathrm{\Omega }}_2^2$${\mathrm{\Omega }}_2^2$Omega_(2)^(2)\Omega_2^2${\mathrm{\Omega }}_2^2$ are the transformed frequencies (constants determined by $\theta$$\theta$theta\theta$\theta$).

A perfectly rough ball is at rest within a hollow cylindrical roller. The roller is drawn along a level path with uniform velocity $V$$V$VV$V$. Let $a$$a$aa$a$ and $b$$b$bb$b$ be the radii of the ball and the roller respectively. If $V^2>\frac{27}{7}g(b-a)$$V^2>\frac{27}{7}g(b-a)$V^(2) > (27)/(7)g(b-a)V^2 > \frac{27}{7} g (b – a)$V^2>\frac{27}{7}g(b-a)$, then show that the ball will roll completely round the inside of the roller.

Answer:

• Potential energy: $P{E}_{\text{bottom}}=-mgR$$P{E}_{\text{bottom}}=-mgR$PE_(“bottom”)=-mgRPE_{\text{bottom}} = -m g R$P{E}_{\text{bottom}}=-mgR$
• Kinetic energy: $K{E}_{\text{bottom}}=\frac{7}{10}m{v}^2$$K{E}_{\text{bottom}}=\frac{7}{10}m{v}^2$KE_(“bottom”)=(7)/(10)mv^(2)KE_{\text{bottom}} = \frac{7}{10} m v^2$K{E}_{\text{bottom}}=\frac{7}{10}m{v}^2$, where $v$$v$vv$v$ is the speed at the bottom.

• Potential energy: $P{E}_{\text{top}}=-mgR\cos {180}^{\circ }=mgR$$P{E}_{\text{top}}=-mgR\cos {180}^{\circ }=mgR$PE_(“top”)=-mgR cos 180^(@)=mgRPE_{\text{top}} = -m g R \cos 180^\circ = m g R$P{E}_{\text{top}}=-mgR\cos {180}^{\circ }=mgR$
• Kinetic energy: $K{E}_{\text{top}}=\frac{7}{10}m{v}_{\text{top}}^2$$K{E}_{\text{top}}=\frac{7}{10}m{v}_{\text{top}}^2$KE_(“top”)=(7)/(10)mv_(“top”)^(2)KE_{\text{top}} = \frac{7}{10} m v_{\text{top}}^2$K{E}_{\text{top}}=\frac{7}{10}m{v}_{\text{top}}^2$

Compute a root of the equation ${\log }_{10}(2x+1)-x^2+3=0$${\log }_{10}(2x+1)-x^2+3=0$log_(10)(2x+1)-x^(2)+3=0\log_{10}(2x + 1) – x^2 + 3 = 0${\log }_{10}(2x+1)-x^2+3=0$, in the interval $[0,3]$$[0,3]$[0,3][0, 3]$[0,3]$, by the Regula-Falsi method, correct to 6 decimal places.

Answer:

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Step 1: Define the Function

Step 2: Check the Endpoints

Step 3: Apply the Regula-Falsi Method

Iteration 1:

• $a=0$$a=0$a=0a = 0$a=0$, $f(a)=3$$f(a)=3$f(a)=3f(a) = 3$f(a)=3$
• $b=3$$b=3$b=3b = 3$b=3$, $f(b)\approx -5.1549$$f(b)\approx -5.1549$f(b)~~-5.1549f(b) \approx -5.1549$f(b)\approx -5.1549$
• Compute $x_1$$x_1$x_(1)x_1$x_1$:

• Evaluate $f(x_1)$$f(x_1)$f(x_(1))f(x_1)$f(x_1)$:

• Since $f(x_1)>0$$f(x_1)>0$f(x_(1)) > 0f(x_1) > 0$f(x_1)>0$, the root lies in $[1.1036,3]$$[1.1036,3]$[1.1036,3][1.1036, 3]$[1.1036,3]$. Update $a=1.1036$$a=1.1036$a=1.1036a = 1.1036$a=1.1036$.

Iteration 2:

• $a=1.1036$$a=1.1036$a=1.1036a = 1.1036$a=1.1036$, $f(a)\approx 2.2881$$f(a)\approx 2.2881$f(a)~~2.2881f(a) \approx 2.2881$f(a)\approx 2.2881$
• $b=3$$b=3$b=3b = 3$b=3$, $f(b)\approx -5.1549$$f(b)\approx -5.1549$f(b)~~-5.1549f(b) \approx -5.1549$f(b)\approx -5.1549$
• Compute $x_2$$x_2$x_(2)x_2$x_2$:

• Evaluate $f(x_2)$$f(x_2)$f(x_(2))f(x_2)$f(x_2)$:

• Since $f(x_2)>0$$f(x_2)>0$f(x_(2)) > 0f(x_2) > 0$f(x_2)>0$, the root lies in $[1.6868,3]$$[1.6868,3]$[1.6868,3][1.6868, 3]$[1.6868,3]$. Update $a=1.6868$$a=1.6868$a=1.6868a = 1.6868$a=1.6868$.

Iteration 3:

• $a=1.6868$$a=1.6868$a=1.6868a = 1.6868$a=1.6868$, $f(a)\approx 0.7955$$f(a)\approx 0.7955$f(a)~~0.7955f(a) \approx 0.7955$f(a)\approx 0.7955$
• $b=3$$b=3$b=3b = 3$b=3$, $f(b)\approx -5.1549$$f(b)\approx -5.1549$f(b)~~-5.1549f(b) \approx -5.1549$f(b)\approx -5.1549$
• Compute $x_3$$x_3$x_(3)x_3$x_3$:

• Evaluate $f(x_3)$$f(x_3)$f(x_(3))f(x_3)$f(x_3)$:

• Since $f(x_3)>0$$f(x_3)>0$f(x_(3)) > 0f(x_3) > 0$f(x_3)>0$, the root lies in $[1.8624,3]$$[1.8624,3]$[1.8624,3][1.8624, 3]$[1.8624,3]$. Update $a=1.8624$$a=1.8624$a=1.8624a = 1.8624$a=1.8624$.

Continue Iterating Until Convergence

Final Root (Correct to 6 Decimal Places):

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Verification:

Determine under what conditions the velocity field $u=c(x^2-y^2),v=-2cxy,w=0$$u=c(x^2-y^2),v=-2cxy,w=0$u=c(x^(2)-y^(2)),v=-2cxy,w=0u = c(x^2 – y^2), v = -2cxy, w = 0$u=c(x^2-y^2),v=-2cxy,w=0$ is a solution to the Navier-Stokes momentum equations. Assuming that the conditions are met, determine the resulting pressure distribution, when $z$$z$zz$z$ is up and the external body forces are $B_x=0=B_y,B_z=-g$$B_x=0=B_y,B_z=-g$B_(x)=0=B_(y),B_(z)=-gB_x = 0 = B_y, B_z = -g$B_x=0=B_y,B_z=-g$.

Answer:

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Step 1: Write the Navier-Stokes Equations

• $\mathbf{u}=(u,v,w)$$\mathbf{u}=(u,v,w)$u=(u,v,w)\mathbf{u} = (u, v, w)$\mathbf{u}=(u,v,w)$ is the velocity field,
• $p$$p$pp$p$ is the pressure,
• $\mathbf{B}=(B_x,B_y,B_z)$$\mathbf{B}=(B_x,B_y,B_z)$B=(B_(x),B_(y),B_(z))\mathbf{B} = (B_x, B_y, B_z)$\mathbf{B}=(B_x,B_y,B_z)$ is the body force per unit volume.

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Step 2: Check the Continuity Equation

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Step 3: Compute the Convective Term $(\mathbf{u}\cdot \mathrm{\nabla })\mathbf{u}$$(\mathbf{u}\cdot \mathrm{\nabla })\mathbf{u}$(u*grad)u(\mathbf{u} \cdot \nabla) \mathbf{u}$(\mathbf{u}\cdot \mathrm{\nabla })\mathbf{u}$

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Step 4: Compute the Viscous Term $\mu {\mathrm{\nabla }}^2\mathbf{u}$$\mu {\mathrm{\nabla }}^2\mathbf{u}$mugrad^(2)u\mu \nabla^2 \mathbf{u}$\mu {\mathrm{\nabla }}^2\mathbf{u}$

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Step 5: Substitute into the Navier-Stokes Equations

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Step 6: Solve for the Pressure Gradient

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Step 7: Integrate to Find the Pressure

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Conditions for Validit

1. Steady Flow: $\frac{\partial \mathbf{u}}{\partial t}=0$$\frac{\partial \mathbf{u}}{\partial t}=0$(delu)/(del t)=0\frac{\partial \mathbf{u}}{\partial t} = 0$\frac{\partial \mathbf{u}}{\partial t}=0$.
2. Incompressibility: $\mathrm{\nabla }\cdot \mathbf{u}=0$$\mathrm{\nabla }\cdot \mathbf{u}=0$grad*u=0\nabla \cdot \mathbf{u} = 0$\mathrm{\nabla }\cdot \mathbf{u}=0$ (satisfied).
3. Viscous Term: $\mu {\mathrm{\nabla }}^2\mathbf{u}=\mathbf{0}$$\mu {\mathrm{\nabla }}^2\mathbf{u}=\mathbf{0}$mugrad^(2)u=0\mu \nabla^2 \mathbf{u} = \mathbf{0}$\mu {\mathrm{\nabla }}^2\mathbf{u}=\mathbf{0}$ (satisfied).
4. Body Force: $\mathbf{B}=(0,0,-\rho g)$$\mathbf{B}=(0,0,-\rho g)$B=(0,0,-rho g)\mathbf{B} = (0, 0, -\rho g)$\mathbf{B}=(0,0,-\rho g)$, conservative with potential $\varphi =\rho gz$$\varphi =\rho gz$phi=rho gz\phi = \rho g z$\varphi =\rho gz$.
5. Pressure: $p(x,y,z)=-\frac{\rho c^2}{2}(x^2+y^2{)}^2-\rho gz+C$$p(x,y,z)=-\frac{\rho c^2}{2}(x^2+y^2{)}^2-\rho gz+C$p(x,y,z)=-(rhoc^(2))/(2)(x^(2)+y^(2))^(2)-rho gz+Cp(x, y, z) = -\frac{\rho c^2}{2} (x^2 + y^2)^2 – \rho g z + C$p(x,y,z)=-\frac{\rho c^2}{2}(x^2+y^2{)}^2-\rho gz+C$.

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Conclusion