IGNOU BMTC-132 Solved Assignment 2024 | B.Sc (G) CBCS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU BMTC-132 Assignment Question Paper 2024

bmtc-132-solved-assignment-2024-10cd7422-4b11-4531-9171-a6445826d855

bmtc-132-solved-assignment-2024-10cd7422-4b11-4531-9171-a6445826d855

BMTC-132 Solved Assignment 2024
1. State whether the following statements are true or false. Give a short proof or a counterexample in support of your Answer:
a) A real-valued function of three variables which is continuous everywhere is differentiable.
b) The function $f\left(x,y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$$f\left(x,y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$f(x,y)=ln((x+y)/(x))f(x, y)=\ln \left(\frac{x+y}{x}\right)$f\left(x,y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$ is not homogeneous function.
c) The cylindrical coordinates of the point whose spherical coordinates is $\left(8,\frac{\pi }{6},\frac{\pi }{2}\right)$$\left(8,\frac{\pi }{6},\frac{\pi }{2}\right)$(8,(pi)/(6),(pi)/(2))\left(8, \frac{\pi}{6}, \frac{\pi}{2}\right)$\left(8,\frac{\pi }{6},\frac{\pi }{2}\right)$ is $\left(8,\frac{\pi }{6},0\right)$$\left(8,\frac{\pi }{6},0\right)$(8,(pi)/(6),0)\left(8, \frac{\pi}{6}, 0\right)$\left(8,\frac{\pi }{6},0\right)$
d) The unique solution $y\left(x\right)$$y\left(x\right)$y(x)y(x)$y\left(x\right)$ of an ordinary differential equation:
$\frac{dy}{dx}=\left\{\begin{array}{ll}0,& \text{for}x<0\\ 1,& \text{for}x\ge 0\end{array}$$\frac{dy}{dx}=\left\{\begin{array}{l}0, \text{for}x<0\\ 1, \text{for}x\ge 0\end{array}\right\$(dy)/(dx)={[0″,”,” for “x < 0],[1″,”,” for “x >= 0]:}\frac{d y}{d x}= \begin{cases}0, & \text { for } x<0 \\ 1, & \text { for } x \geq 0\end{cases}$\frac{dy}{dx}=\left\{\begin{array}{ll}0,& \text{for}x<0\\ 1,& \text{for}x\ge 0\end{array}$
exists $\mathrm{\forall }x\in \mathbb{R}$$\mathrm{\forall }x\in \mathbb{R}$AA x inR\forall x \in \mathbb{R}$\mathrm{\forall }x\in \mathbb{R}$.
e) The differential equation:
${\left[1+{\left({y}^{\mathrm{\prime }}\right)}^{2}\right]}^{\frac{5}{3}}={y}^{\mathrm{\prime }\mathrm{\prime }}$${\left[1+{\left({y}^{\mathrm{\prime }}\right)}^{2}\right]}^{\frac{5}{3}}={y}^{\mathrm{\prime }\mathrm{\prime }}$[1+(y^(‘))^(2)]^((5)/(3))=y^(”)\left[1+\left(y^{\prime}\right)^2\right]^{\frac{5}{3}}=y^{\prime \prime}${\left[1+{\left({y}^{\mathrm{\prime }}\right)}^{2}\right]}^{\frac{5}{3}}={y}^{\mathrm{\prime }\mathrm{\prime }}$
is a second order differential equation of degree 3 .
f) Differential equation:
$\mathrm{cos}x\frac{{d}^{2}y}{d{x}^{2}}+\frac{dy}{dx}+x{y}^{2}=0$$\mathrm{cos}x\frac{{d}^{2}y}{d{x}^{2}}+\frac{dy}{dx}+x{y}^{2}=0$cos x(d^(2)y)/(dx^(2))+(dy)/(dx)+xy^(2)=0\cos x \frac{d^2 y}{d x^2}+\frac{d y}{d x}+x y^2=0$\mathrm{cos}x\frac{{d}^{2}y}{d{x}^{2}}+\frac{dy}{dx}+x{y}^{2}=0$
in $\right]0,\pi \left[$$\right]0,\pi \left[$]0,pi[] 0, \pi[$\right]0,\pi \left[$ is a linear, homogeneous equation.
g) The total differential equation corresponding to the family of surfaces ${x}^{3}z+{x}^{2}y=c$${x}^{3}z+{x}^{2}y=c$x^(3)z+x^(2)y=cx^3 z+x^2 y=c${x}^{3}z+{x}^{2}y=c$, where $c$$c$cc$c$ is a parameter is $3{x}^{2}dz+x\left(ydx+xdy\right)=0$$3{x}^{2}dz+x\left(ydx+xdy\right)=0$3x^(2)dz+x(ydx+xdy)=03 x^2 d z+x(y d x+x d y)=0$3{x}^{2}dz+x\left(ydx+xdy\right)=0$.
h) Differential equation:
$5{x}^{2}{y}^{2}{z}^{2}=2p{x}^{2}{y}^{2}+5q{x}^{2}{y}^{3}+2p{z}^{2}+9{x}^{2}{y}^{2}$$5{x}^{2}{y}^{2}{z}^{2}=2p{x}^{2}{y}^{2}+5q{x}^{2}{y}^{3}+2p{z}^{2}+9{x}^{2}{y}^{2}$5x^(2)y^(2)z^(2)=2px^(2)y^(2)+5qx^(2)y^(3)+2pz^(2)+9x^(2)y^(2)5 x^2 y^2 z^2=2 p x^2 y^2+5 q x^2 y^3+2 p z^2+9 x^2 y^2$5{x}^{2}{y}^{2}{z}^{2}=2p{x}^{2}{y}^{2}+5q{x}^{2}{y}^{3}+2p{z}^{2}+9{x}^{2}{y}^{2}$
is a semi-linear partial differential equation of first order.
i) The differential equation:
$v\frac{du}{dv}={e}^{2v}+uv-u$$v\frac{du}{dv}={e}^{2v}+uv-u$v(du)/(dv)=e^(2v)+uv-uv \frac{d u}{d v}=e^{2 v}+u v-u$v\frac{du}{dv}={e}^{2v}+uv-u$
has an integrating factor $v\mathrm{exp}\left(-v\right)$$v\mathrm{exp}\left(-v\right)$v exp(-v)v \exp (-v)$v\mathrm{exp}\left(-v\right)$.
j) The simultaneous differential equation of simple harmonic motion of a particle in phase -plane is:
$\frac{dx}{y}=\frac{dy}{-{w}^{2}x}=dt\text{with}y\left({x}_{0}\right)={y}_{0}$$\frac{dx}{y}=\frac{dy}{-{w}^{2}x}=dt\text{with}y\left({x}_{0}\right)={y}_{0}$(dx)/(y)=(dy)/(-w^(2)x)=dt” with “y(x_(0))=y_(0)\frac{d x}{y}=\frac{d y}{-w^2 x}=d t \text { with } y\left(x_0\right)=y_0$\frac{dx}{y}=\frac{dy}{-{w}^{2}x}=dt\text{with}y\left({x}_{0}\right)={y}_{0}$
1. a) Solve the differential equation:
$\frac{dy}{dx}\frac{-\mathrm{tan}y}{\left(1+x\right)}=\left(1+x\right){e}^{x}\mathrm{sec}y$$\frac{dy}{dx}\frac{-\mathrm{tan}y}{\left(1+x\right)}=\left(1+x\right){e}^{x}\mathrm{sec}y$(dy)/(dx)(-tan y)/((1+x))=(1+x)e^(x)sec y\frac{d y}{d x} \frac{-\tan y}{(1+x)}=(1+x) e^x \sec y$\frac{dy}{dx}\frac{-\mathrm{tan}y}{\left(1+x\right)}=\left(1+x\right){e}^{x}\mathrm{sec}y$
b) Show that the differential equation:
$\left({y}^{2}+yx\right)dx+\left({z}^{2}+zx\right)dy+\left({y}^{2}-xy\right)dz=0$$\left({y}^{2}+yx\right)dx+\left({z}^{2}+zx\right)dy+\left({y}^{2}-xy\right)dz=0$(y^(2)+yx)dx+(z^(2)+zx)dy+(y^(2)-xy)dz=0\left(y^2+y x\right) d x+\left(z^2+z x\right) d y+\left(y^2-x y\right) d z=0$\left({y}^{2}+yx\right)dx+\left({z}^{2}+zx\right)dy+\left({y}^{2}-xy\right)dz=0$
is integrable and find its integral.
1. a) Solve the differential equation:
${x}^{2}\frac{{d}^{2}y}{d{x}^{2}}-2x\frac{dy}{dx}-4y={x}^{2}+2\mathrm{ln}x$${x}^{2}\frac{{d}^{2}y}{d{x}^{2}}-2x\frac{dy}{dx}-4y={x}^{2}+2\mathrm{ln}x$x^(2)(d^(2)y)/(dx^(2))-2x(dy)/(dx)-4y=x^(2)+2ln xx^2 \frac{d^2 y}{d x^2}-2 x \frac{d y}{d x}-4 y=x^2+2 \ln x${x}^{2}\frac{{d}^{2}y}{d{x}^{2}}-2x\frac{dy}{dx}-4y={x}^{2}+2\mathrm{ln}x$
b) Using Charpit’s method, find the complete integral of the differential equation:
$p\left(1+{q}^{2}\right)+\left(b-z\right)q=0,$$p\left(1+{q}^{2}\right)+\left(b-z\right)q=0,$p(1+q^(2))+(b-z)q=0,p\left(1+q^2\right)+(b-z) q=0,$p\left(1+{q}^{2}\right)+\left(b-z\right)q=0,$
$b$$b$bb$b$ being a constant.
1. a) Find all the first order partial derivatives of the following function:
$h\left(x,y,t\right)={e}^{x-t}\mathrm{cos}\left(y+t\right)$$h\left(x,y,t\right)={e}^{x-t}\mathrm{cos}\left(y+t\right)$h(x,y,t)=e^(x-t)cos(y+t)h(x, y, t)=e^{x-t} \cos (y+t)$h\left(x,y,t\right)={e}^{x-t}\mathrm{cos}\left(y+t\right)$
What is the value of $\frac{\mathrm{\partial }h}{\mathrm{\partial }t}$$\frac{\mathrm{\partial }h}{\mathrm{\partial }t}$(del h)/(del t)\frac{\partial h}{\partial t}$\frac{\mathrm{\partial }h}{\mathrm{\partial }t}$ at $\left(0,\frac{\pi }{2},0\right)$$\left(0,\frac{\pi }{2},0\right)$(0,(pi)/(2),0)\left(0, \frac{\pi}{2}, 0\right)$\left(0,\frac{\pi }{2},0\right)$ ?
b) Find the limit of:
$f\left(x,y\right)=\frac{{x}^{2}-{y}^{2}}{{x}^{2}+{y}^{2}}$$f\left(x,y\right)=\frac{{x}^{2}-{y}^{2}}{{x}^{2}+{y}^{2}}$f(x,y)=(x^(2)-y^(2))/(x^(2)+y^(2))f(x, y)=\frac{x^2-y^2}{x^2+y^2}$f\left(x,y\right)=\frac{{x}^{2}-{y}^{2}}{{x}^{2}+{y}^{2}}$
as $\left(x,y\right)\to \left(0,0\right)$$\left(x,y\right)\to \left(0,0\right)$(x,y)rarr(0,0)(x, y) \rightarrow(0,0)$\left(x,y\right)\to \left(0,0\right)$ along:
i) $y=3x$$y=3x$y=3xy=3 x$y=3x$
ii) $y=5x$$y=5x$y=5xy=5 x$y=5x$
What can you conclude about $\underset{\left(x,y\right)\to \left(0,0\right)}{lim}f\left(x,y\right)$$\underset{\left(x,y\right)\to \left(0,0\right)}{lim} f\left(x,y\right)$lim_((x,y)rarr(0,0))f(x,y)\lim _{(x, y) \rightarrow(0,0)} f(x, y)$\underset{\left(x,y\right)\to \left(0,0\right)}{lim}f\left(x,y\right)$ ? Justify your answer.
c) If $\mathrm{cos}\alpha ,\mathrm{cos}\beta$$\mathrm{cos}\alpha ,\mathrm{cos}\beta$cos alpha,cos beta\cos \alpha, \cos \beta$\mathrm{cos}\alpha ,\mathrm{cos}\beta$ and $\mathrm{cos}\gamma$$\mathrm{cos}\gamma$cos gamma\cos \gamma$\mathrm{cos}\gamma$ are the direction cosines of a line, then show that:
${\mathrm{sin}}^{2}\alpha +{\mathrm{sin}}^{2}\beta +{\mathrm{sin}}^{2}\gamma =2$${\mathrm{sin}}^{2}\alpha +{\mathrm{sin}}^{2}\beta +{\mathrm{sin}}^{2}\gamma =2$sin^(2)alpha+sin^(2)beta+sin^(2)gamma=2\sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma=2${\mathrm{sin}}^{2}\alpha +{\mathrm{sin}}^{2}\beta +{\mathrm{sin}}^{2}\gamma =2$
1. a) Solve the differential equation:
$2x\frac{dy}{dx}+y\left(6{y}^{2}-x-1\right)=0$$2x\frac{dy}{dx}+y\left(6{y}^{2}-x-1\right)=0$2x(dy)/(dx)+y(6y^(2)-x-1)=02 x \frac{d y}{d x}+y\left(6 y^2-x-1\right)=0$2x\frac{dy}{dx}+y\left(6{y}^{2}-x-1\right)=0$
b) The rate of change of the price of a commodity is proportional to the difference between the demand $D$$D$DD$D$ and the supply $\mathrm{S}$$\mathrm{S}$S\mathrm{S}$\mathrm{S}$. If $D=\alpha -bP$$D=\alpha -bP$D=alpha-bPD=\alpha-b P$D=\alpha -bP$ and $S=c\mathrm{sin}\beta t$$S=c\mathrm{sin}\beta t$S=c sin beta tS=c \sin \beta t$S=c\mathrm{sin}\beta t$, where $a,b,c$$a,b,c$a,b,ca, b, c$a,b,c$ and $\beta$$\beta$beta\beta$\beta$ are constants, determine $P\left(t\right)$$P\left(t\right)$P(t)P(t)$P\left(t\right)$. It is given that at $t=0,P={P}_{0}$$t=0,P={P}_{0}$t=0,P=P_(0)t=0, P=P_0$t=0,P={P}_{0}$.
1. a) Find the differential equations of the space curve in which the two families of surfaces:
$u={x}^{2}+{y}^{2}+3z={c}_{1}$$u={x}^{2}+{y}^{2}+3z={c}_{1}$u=x^(2)+y^(2)+3z=c_(1)u=x^2+y^2+3 z=c_1$u={x}^{2}+{y}^{2}+3z={c}_{1}$
and $v=zx+{x}^{2}+{y}^{2}={c}_{2}$$v=zx+{x}^{2}+{y}^{2}={c}_{2}$v=zx+x^(2)+y^(2)=c_(2)v=z x+x^2+y^2=c_2$v=zx+{x}^{2}+{y}^{2}={c}_{2}$
intersect.
b) Transform the given equation to Clairaut’s form and hence find its general solution:
$xy\left(y-px\right)=x-py$$xy\left(y-px\right)=x-py$xy(y-px)=x-pyx y(y-p x)=x-p y$xy\left(y-px\right)=x-py$
Also find its singular solution, if it exists.
1. a) Find the envelope and the characteristic curves of the family of curves:
${x}^{2}\left(y-c{\right)}^{2}+{z}^{2}={c}^{2}{\mathrm{cos}}^{2}\alpha ,$${x}^{2}\left(y-c{\right)}^{2}+{z}^{2}={c}^{2}{\mathrm{cos}}^{2}\alpha ,$x^(2)(y-c)^(2)+z^(2)=c^(2)cos^(2)alpha,x^2(y-c)^2+z^2=c^2 \cos ^2 \alpha,${x}^{2}\left(y-c{\right)}^{2}+{z}^{2}={c}^{2}{\mathrm{cos}}^{2}\alpha ,$
$c$$c$cc$c$ and $\alpha$$\alpha$alpha\alpha$\alpha$ are constants.
b) Using Lagrange’s method, solve the differential equation:
$\left({x}^{2}-{y}^{2}-{z}^{2}\right)p+2xq=2xz$$\left({x}^{2}-{y}^{2}-{z}^{2}\right)p+2xq=2xz$(x^(2)-y^(2)-z^(2))p+2xq=2xz\left(x^2-y^2-z^2\right) p+2 x q=2 x z$\left({x}^{2}-{y}^{2}-{z}^{2}\right)p+2xq=2xz$
1. a) Using the method of variation of parameters, solve the differential equation:
$\frac{{d}^{2}y}{d{x}^{2}}+y={\mathrm{sec}}^{3}x$$\frac{{d}^{2}y}{d{x}^{2}}+y={\mathrm{sec}}^{3}x$(d^(2)y)/(dx^(2))+y=sec^(3)x\frac{d^2 y}{d x^2}+y=\sec ^3 x$\frac{{d}^{2}y}{d{x}^{2}}+y={\mathrm{sec}}^{3}x$
b) Solve the differential equation:
$\left(2x{e}^{y}{y}^{4}+2x{y}^{3}y\right)dx+\left({x}^{2}{y}^{4}{e}^{y}-{x}^{2}{y}^{2}-3x\right)dy=0$$\left(2x{e}^{y}{y}^{4}+2x{y}^{3}y\right)dx+\left({x}^{2}{y}^{4}{e}^{y}-{x}^{2}{y}^{2}-3x\right)dy=0$(2xe^(y)y^(4)+2xy^(3)y)dx+(x^(2)y^(4)e^(y)-x^(2)y^(2)-3x)dy=0\left(2 x e^y y^4+2 x y^3 y\right) d x+\left(x^2 y^4 e^y-x^2 y^2-3 x\right) d y=0$\left(2x{e}^{y}{y}^{4}+2x{y}^{3}y\right)dx+\left({x}^{2}{y}^{4}{e}^{y}-{x}^{2}{y}^{2}-3x\right)dy=0$
1. a) Show that the limit of the function $f\left(x,y\right)$$f\left(x,y\right)$f(x,y)f(x, y)$f\left(x,y\right)$ exists at the origin, where:
$f\left(x,y\right)=\left\{\begin{array}{cc}x\mathrm{cos}\frac{1}{y}+y\mathrm{cos}\frac{1}{x},& \left(x,y\right)\ne \left(0,0\right)\\ 0,& \left(x,y\right)=\left(0,0\right)\end{array}$$f\left(x,y\right)=\left\{\begin{array}{cc}x\mathrm{cos}\frac{1}{y}+y\mathrm{cos}\frac{1}{x},& \left(x,y\right)\ne \left(0,0\right)\\ 0,& \left(x,y\right)=\left(0,0\right)\end{array}\right\$f(x,y)={[x cos (1)/(y)+y cos (1)/(x)”,”,(x”,”y)!=(0″,”0)],[0″,”,(x”,”y)=(0″,”0)]:}f(x, y)=\left\{\begin{array}{cc} x \cos \frac{1}{y}+y \cos \frac{1}{x}, & (x, y) \neq(0,0) \\ 0, & (x, y)=(0,0) \end{array}\right.$f\left(x,y\right)=\left\{\begin{array}{cc}x\mathrm{cos}\frac{1}{y}+y\mathrm{cos}\frac{1}{x},& \left(x,y\right)\ne \left(0,0\right)\\ 0,& \left(x,y\right)=\left(0,0\right)\end{array}$
Do the repeated limits of $f\left(x,y\right)$$f\left(x,y\right)$f(x,y)f(x, y)$f\left(x,y\right)$ exist? Justify your answer.
b) For the function:
$f\left(x,y\right)={x}^{3}+xy-2{y}^{2},$$f\left(x,y\right)={x}^{3}+xy-2{y}^{2},$f(x,y)=x^(3)+xy-2y^(2),f(x, y)=x^3+x y-2 y^2,$f\left(x,y\right)={x}^{3}+xy-2{y}^{2},$
Find the polynomial given by:
${f}_{xx}\left(1,2\right)\left(x-2{\right)}^{2}+{f}_{xy}\left(1,2\right)\left(x-2\right)\left(y-1\right)+{f}_{yy}\left(1,2\right)\left(y-1{\right)}^{2}$${f}_{xx}\left(1,2\right)\left(x-2{\right)}^{2}+{f}_{xy}\left(1,2\right)\left(x-2\right)\left(y-1\right)+{f}_{yy}\left(1,2\right)\left(y-1{\right)}^{2}$f_(xx)(1,2)(x-2)^(2)+f_(xy)(1,2)(x-2)(y-1)+f_(yy)(1,2)(y-1)^(2)f_{x x}(1,2)(x-2)^2+f_{x y}(1,2)(x-2)(y-1)+f_{y y}(1,2)(y-1)^2${f}_{xx}\left(1,2\right)\left(x-2{\right)}^{2}+{f}_{xy}\left(1,2\right)\left(x-2\right)\left(y-1\right)+{f}_{yy}\left(1,2\right)\left(y-1{\right)}^{2}$
$$a^2=b^2+c^2-2bc\:Cos\left(A\right)$$

BMTC-132 Sample Solution 2024

bmtc-132-solved-assignment-2024-ss-10cd7422-4b11-4531-9171-a6445826d855

bmtc-132-solved-assignment-2024-ss-10cd7422-4b11-4531-9171-a6445826d855

BMTC-132 Solved Assignment 2024
1. State whether the following statements are true or false. Give a short proof or a counterexample in support of your answer:
a) A real-valued function of three variables which is continuous everywhere is differentiable.
The statement "A real-valued function of three variables which is continuous everywhere is differentiable" is false.

Justification

Continuity and differentiability are related but distinct properties of functions. While differentiability implies continuity, the converse is not necessarily true. A function can be continuous at a point but not differentiable at that point. This concept holds for functions of any number of variables, including functions of three variables.

Counterexample

A classic counterexample in the context of functions of a single variable is the absolute value function, $f\left(x\right)=|x|$$f\left(x\right)=|x|$f(x)=|x|f(x) = |x|$f\left(x\right)=|x|$, which is continuous everywhere but not differentiable at $x=0$$x=0$x=0x = 0$x=0$.
For a function of three variables, we can construct a similar example. Consider the function
$f\left(x,y,z\right)=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$$f\left(x,y,z\right)=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$f(x,y,z)=sqrt(x^(2)+y^(2)+z^(2))f(x, y, z) = \sqrt{x^2 + y^2 + z^2}$f\left(x,y,z\right)=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$
This function represents the distance from the origin in three-dimensional space and is continuous everywhere. However, it is not differentiable at the origin $\left(0,0,0\right)$$\left(0,0,0\right)$(0,0,0)(0, 0, 0)$\left(0,0,0\right)$.
To see why, consider the partial derivative with respect to $x$$x$xx$x$ at the origin. The partial derivative is defined as the limit of the difference quotient:
$\underset{h\to 0}{lim}\frac{f\left(h,0,0\right)-f\left(0,0,0\right)}{h}$$\underset{h\to 0}{lim} \frac{f\left(h,0,0\right)-f\left(0,0,0\right)}{h}$lim_(h rarr0)(f(h,0,0)-f(0,0,0))/(h)\lim_{h \to 0} \frac{f(h, 0, 0) – f(0, 0, 0)}{h}$\underset{h\to 0}{lim}\frac{f\left(h,0,0\right)-f\left(0,0,0\right)}{h}$
Substituting the function, we get:
$\underset{h\to 0}{lim}\frac{\sqrt{{h}^{2}}}{h}$$\underset{h\to 0}{lim} \frac{\sqrt{{h}^{2}}}{h}$lim_(h rarr0)(sqrt(h^(2)))/(h)\lim_{h \to 0} \frac{\sqrt{h^2}}{h}$\underset{h\to 0}{lim}\frac{\sqrt{{h}^{2}}}{h}$
This simplifies to:
$\underset{h\to 0}{lim}\frac{|h|}{h}$$\underset{h\to 0}{lim} \frac{|h|}{h}$lim_(h rarr0)(|h|)/(h)\lim_{h \to 0} \frac{|h|}{h}$\underset{h\to 0}{lim}\frac{|h|}{h}$
This limit does not exist because it approaches $1$$1$11$1$ as $h$$h$hh$h$ approaches $0$$0$00$0$ from the right and $-1$$-1$-1-1$-1$ as $h$$h$hh$h$ approaches $0$$0$00$0$ from the left. Therefore, the function $f\left(x,y,z\right)=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$$f\left(x,y,z\right)=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$f(x,y,z)=sqrt(x^(2)+y^(2)+z^(2))f(x, y, z) = \sqrt{x^2 + y^2 + z^2}$f\left(x,y,z\right)=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$ is not differentiable at the origin, even though it is continuous everywhere.

Conclusion

Thus, the statement is false. A real-valued function of three variables that is continuous everywhere is not necessarily differentiable everywhere.
b) The function $f\left(x,y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$$f\left(x,y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$f(x,y)=ln((x+y)/(x))f(x, y)=\ln \left(\frac{x+y}{x}\right)$f\left(x,y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$ is not homogeneous function.
To determine the truth of the statement "The function $f\left(x,y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$$f\left(x,y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$f(x,y)=ln((x+y)/(x))f(x, y) = \ln\left(\frac{x+y}{x}\right)$f\left(x,y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$ is not a homogeneous function", we need to verify if the function is homogeneous of any degree. A function $f\left(x,y\right)$$f\left(x,y\right)$f(x,y)f(x, y)$f\left(x,y\right)$ is homogeneous of degree $n$$n$nn$n$ if it satisfies:
$f\left(\lambda x,\lambda y\right)={\lambda }^{n}f\left(x,y\right)$$f\left(\lambda x,\lambda y\right)={\lambda }^{n}f\left(x,y\right)$f(lambda x,lambda y)=lambda ^(n)f(x,y)f(\lambda x, \lambda y) = \lambda^n f(x, y)$f\left(\lambda x,\lambda y\right)={\lambda }^{n}f\left(x,y\right)$
for all scalars $\lambda$$\lambda$lambda\lambda$\lambda$. If a function satisfies this condition for some degree $n$$n$nn$n$, it is homogeneous; otherwise, it is not.
For the given function $f\left(x,y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$$f\left(x,y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$f(x,y)=ln((x+y)/(x))f(x, y) = \ln\left(\frac{x+y}{x}\right)$f\left(x,y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$, let’s substitute $\lambda x$$\lambda x$lambda x\lambda x$\lambda x$ and $\lambda y$$\lambda y$lambda y\lambda y$\lambda y$ and simplify:
$f\left(\lambda x,\lambda y\right)=\mathrm{ln}\left(\frac{\lambda x+\lambda y}{\lambda x}\right)$$f\left(\lambda x,\lambda y\right)=\mathrm{ln}\left(\frac{\lambda x+\lambda y}{\lambda x}\right)$f(lambda x,lambda y)=ln((lambda x+lambda y)/(lambda x))f(\lambda x, \lambda y) = \ln\left(\frac{\lambda x + \lambda y}{\lambda x}\right)$f\left(\lambda x,\lambda y\right)=\mathrm{ln}\left(\frac{\lambda x+\lambda y}{\lambda x}\right)$
Simplifying this expression will help us compare it with ${\lambda }^{n}f\left(x,y\right)$${\lambda }^{n}f\left(x,y\right)$lambda ^(n)f(x,y)\lambda^n f(x, y)${\lambda }^{n}f\left(x,y\right)$ to determine if the function is homogeneous and, if so, of which degree. Let’s perform this calculation.
Upon simplifying the function $f\left(\lambda x,\lambda y\right)=\mathrm{ln}\left(\frac{\lambda x+\lambda y}{\lambda x}\right)$$f\left(\lambda x,\lambda y\right)=\mathrm{ln}\left(\frac{\lambda x+\lambda y}{\lambda x}\right)$f(lambda x,lambda y)=ln((lambda x+lambda y)/(lambda x))f(\lambda x, \lambda y) = \ln\left(\frac{\lambda x + \lambda y}{\lambda x}\right)$f\left(\lambda x,\lambda y\right)=\mathrm{ln}\left(\frac{\lambda x+\lambda y}{\lambda x}\right)$, we find that it simplifies to:
$f\left(\lambda x,\lambda y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$$f\left(\lambda x,\lambda y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$f(lambda x,lambda y)=ln((x+y)/(x))f(\lambda x, \lambda y) = \ln\left(\frac{x + y}{x}\right)$f\left(\lambda x,\lambda y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$
This is identical to the original function $f\left(x,y\right)$$f\left(x,y\right)$f(x,y)f(x, y)$f\left(x,y\right)$. Therefore, the function satisfies the condition:
$f\left(\lambda x,\lambda y\right)=f\left(x,y\right)$$f\left(\lambda x,\lambda y\right)=f\left(x,y\right)$f(lambda x,lambda y)=f(x,y)f(\lambda x, \lambda y) = f(x, y)$f\left(\lambda x,\lambda y\right)=f\left(x,y\right)$
which is the same as:
$f\left(\lambda x,\lambda y\right)={\lambda }^{0}f\left(x,y\right)$$f\left(\lambda x,\lambda y\right)={\lambda }^{0}f\left(x,y\right)$f(lambda x,lambda y)=lambda^(0)f(x,y)f(\lambda x, \lambda y) = \lambda^0 f(x, y)$f\left(\lambda x,\lambda y\right)={\lambda }^{0}f\left(x,y\right)$
since ${\lambda }^{0}=1$${\lambda }^{0}=1$lambda^(0)=1\lambda^0 = 1${\lambda }^{0}=1$ for any $\lambda$$\lambda$lambda\lambda$\lambda$. This means that the function $f\left(x,y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$$f\left(x,y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$f(x,y)=ln((x+y)/(x))f(x, y) = \ln\left(\frac{x+y}{x}\right)$f\left(x,y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$ is indeed homogeneous, but of degree $n=0$$n=0$n=0n = 0$n=0$.
Thus, the statement "The function $f\left(x,y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$$f\left(x,y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$f(x,y)=ln((x+y)/(x))f(x, y)=\ln \left(\frac{x+y}{x}\right)$f\left(x,y\right)=\mathrm{ln}\left(\frac{x+y}{x}\right)$ is not a homogeneous function" is false, as the function is homogeneous of degree 0.
c) The cylindrical coordinates of the point whose spherical coordinates is $\left(8,\frac{\pi }{6},\frac{\pi }{2}\right)$$\left(8,\frac{\pi }{6},\frac{\pi }{2}\right)$(8,(pi)/(6),(pi)/(2))\left(8, \frac{\pi}{6}, \frac{\pi}{2}\right)$\left(8,\frac{\pi }{6},\frac{\pi }{2}\right)$ is $\left(8,\frac{\pi }{6},0\right)$$\left(8,\frac{\pi }{6},0\right)$(8,(pi)/(6),0)\left(8, \frac{\pi}{6}, 0\right)$\left(8,\frac{\pi }{6},0\right)$
The statement "The cylindrical coordinates of the point whose spherical coordinates are $\left(8,\frac{\pi }{6},\frac{\pi }{2}\right)$$\left(8,\frac{\pi }{6},\frac{\pi }{2}\right)$(8,(pi)/(6),(pi)/(2))\left(8, \frac{\pi}{6}, \frac{\pi}{2}\right)$\left(8,\frac{\pi }{6},\frac{\pi }{2}\right)$ is $\left(8,\frac{\pi }{6},0\right)$$\left(8,\frac{\pi }{6},0\right)$(8,(pi)/(6),0)\left(8, \frac{\pi}{6}, 0\right)$\left(8,\frac{\pi }{6},0\right)$" is false. To demonstrate this, let’s convert the given spherical coordinates to cylindrical coordinates and compare.

Spherical Coordinates to Cylindrical Coordinates Conversion

The spherical coordinates $\left(r,\theta ,\varphi \right)$$\left(r,\theta ,\varphi \right)$(r,theta,phi)(r, \theta, \phi)$\left(r,\theta ,\varphi \right)$ are related to cylindrical coordinates $\left(\rho ,\varphi ,z\right)$$\left(\rho ,\varphi ,z\right)$(rho,phi,z)(\rho, \phi, z)$\left(\rho ,\varphi ,z\right)$ by the following relationships:
• $\rho =r\mathrm{sin}\left(\varphi \right)$$\rho =r\mathrm{sin}\left(\varphi \right)$rho=r sin(phi)\rho = r \sin(\phi)$\rho =r\mathrm{sin}\left(\varphi \right)$
• $z=r\mathrm{cos}\left(\varphi \right)$$z=r\mathrm{cos}\left(\varphi \right)$z=r cos(phi)z = r \cos(\phi)$z=r\mathrm{cos}\left(\varphi \right)$
• $\varphi$$\varphi$phi\phi$\varphi$ (the angle in the xy-plane from the x-axis) remains the same in both coordinate systems.
Given the spherical coordinates $\left(8,\frac{\pi }{6},\frac{\pi }{2}\right)$$\left(8,\frac{\pi }{6},\frac{\pi }{2}\right)$(8,(pi)/(6),(pi)/(2))\left(8, \frac{\pi}{6}, \frac{\pi}{2}\right)$\left(8,\frac{\pi }{6},\frac{\pi }{2}\right)$, where $r=8$$r=8$r=8r = 8$r=8$, $\theta =\frac{\pi }{6}$$\theta =\frac{\pi }{6}$theta=(pi)/(6)\theta = \frac{\pi}{6}$\theta =\frac{\pi }{6}$, and $\varphi =\frac{\pi }{2}$$\varphi =\frac{\pi }{2}$phi=(pi)/(2)\phi = \frac{\pi}{2}$\varphi =\frac{\pi }{2}$, we can convert them to cylindrical coordinates:
• $\rho =8\mathrm{sin}\left(\frac{\pi }{2}\right)=8$$\rho =8\mathrm{sin}\left(\frac{\pi }{2}\right)=8$rho=8sin((pi)/(2))=8\rho = 8 \sin\left(\frac{\pi}{2}\right) = 8