Exercise 8.8 - Chapter 8 Differentials and Partial Derivatives 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $8.8$
Choose the correct or the most suitable answer from the given four alternatives.
Question 1.
A circular template has a radius of $10 \mathrm{~cm}$. The measurement of radius has an approximate error of $0.02$ $\mathrm{cm}$. Then the percentage error in calculating area of this template is
(a) $0.2 \%$
(b) $0.4 \%$
(c) $0.04 \%$
(d) $0.08 \%$
Solution:
(b) $0.4 \%$
Hint:
$
\begin{aligned}
& \mathrm{r}=10 \mathrm{~cm} \\
& \mathrm{dr}=0.02 \\
& \mathrm{~A}=\pi r^2 \\
& d \mathrm{~A}=2 \pi r d r \\
& \frac{d \mathrm{~A}}{\mathrm{~A}} \times 100=\frac{2 \pi r d r}{\pi r^2} \times 100 \\
&=\frac{2}{10}(0.02) \times 100=0.4 \%
\end{aligned}
$
Question 2.
The percentage error of fifth root of 31 is approximately how many times the percentage error in 31 ?
(a) $\frac{1}{31}$
(b) $\frac{1}{5}$
(c) 5
(d) 31
Solution:
$\frac{1}{5}$
Hint:
We know that the percentage error in the «th root of a number is approximately $\frac{1}{n}$ times the percentage error in the number.
so $\frac{1}{5}$

Question 3.
If $u(x, y)=e^{x^2+y^2}$, then $\frac{\partial u}{\partial x}$ is equal to
(a) $e^{x^2+y^2}$
(b) $2 x u$
(c) $x^2 u$
(d) $y^2 u$
Solution:
(b) $2 \mathrm{xu}$
Hint:
$
\frac{\partial u}{\partial x}=2 x e^{x^2+y^2}=2 x u
$
Question 4.
If $v(x, y)=\log \left(e^x+e^y\right)$, then $\frac{\partial v}{\partial x}+\frac{\partial v}{\partial y}$ is equal to
(a) $e^x+e^y$
(b) $\frac{1}{e^x+e^y}$
(c) 2
(d) 1
Solution:
(d) 1
Hint:
$
\begin{aligned}
& \frac{\partial v}{\partial x}=\frac{e^x}{e^x+e^y} ; \frac{\partial v}{\partial y}=\frac{e^y}{e^x+e^y} \\
& \frac{\partial v}{\partial x}+\frac{\partial v}{\partial y}=\frac{e^x+e^y}{e^x+e^y}=1
\end{aligned}
$
Question 5.
If $\mathrm{w}(\mathrm{x}, \mathrm{y})=\mathrm{x}^{\mathrm{y}}, \mathrm{x}>0$, then $\frac{\partial w}{\partial x}$ is equal to ......
(a) $x^y \log x$
(b) $y \log x$
(c) $y x^{y-1}$
(d) $x \log y$
Solution:
(c) $y x^{y-1}$
Hint:
$
w(x, y)=x^y
$

$
\frac{\partial w}{\partial x}=y x^{y-1}
$
Question 6.
If $f(x, y)=e^{x y}$, then $\frac{\partial^2 f}{\partial x \partial y}$ is equal to
(a) $x y e^{x y}$
(b) $(1+x y) e^{x y}$
(c) $(1+y) e^{x y}$
(d) $(1+x) e^{x y}$
Solution:
(b) $(1+x y) e^{x y}$
Hint:
$
\begin{aligned}
f(x, y) & =e^{x y} \\
\frac{\partial f}{\partial y} & =x e^{x y} \\
\frac{\partial^2 f}{\partial x \partial y} & =x\left[y e^{x y}+e^{x y}\right] \\
& =e^{x y}[x y+1]
\end{aligned}
$

Question 7.
If we measure the side of a cube to be $4 \mathrm{~cm}$ with an error of $0.1 \mathrm{~cm}$, then the error in our calculation of the volume is ........
(a) $0.4 \mathrm{cu} . \mathrm{cm}$
(b) $0.45$ cu.cm
(c) $2 \mathrm{cu} . \mathrm{cm}$
(d) $4.8 \mathrm{culcm}$
Solution:
(d) $4.8 \mathrm{culcm}$
Hint.
$
\begin{aligned}
& a=4 \mathrm{~cm} \\
& \mathrm{da}=0.1 \mathrm{~cm} \\
& \mathrm{v}=\mathrm{a}^3 \\
& \mathrm{dv}=3 \mathrm{a}^2 \mathrm{da} \\
& =3(4)^2(0.1) \\
& =4.8 \mathrm{cu} . \mathrm{cm}
\end{aligned}
$
Question 8.
The change in the surface area $\mathrm{S}=6 \mathrm{x}^2$ of a cube when the edge length varies from $\mathrm{x}_0$ to $\mathrm{x}_0+\mathrm{dx}$ is $\ldots \ldots$
(a) $12 \mathrm{x}_0+\mathrm{dx}$
(b) $12 \mathrm{x}_0 \mathrm{dx}$
(c) $6 \mathrm{x}_0 \mathrm{dx}$
(d) $6 x_0+d x$
Solution:
(b) $12 \mathrm{x}_0 \mathrm{dx}$

Question 9.
the approximate change in the volume: $\mathrm{V}$ of a cube of side $\mathrm{x}$ mentres caused by increasing the side by $1 \%$ is
(a) $0.3 \mathrm{xdx} \mathrm{m}^3$
(b) $0.03 \mathrm{x} \mathrm{m}^3$
(c) $0.03 \mathrm{x}^2 \mathrm{~m}^3$
(d) $0.03 \mathrm{x}^3 \mathrm{~m}^3$
Solution:
(d) $0.03 \mathrm{x}^3 \mathrm{~m}^3$
Hint:
Let the side of the cube be $x$ units
$
\mathrm{v}=\mathrm{x}^3
$
when $\mathrm{dx}=0.01 \mathrm{x}$
$
\begin{aligned}
& d v=3 x^2 d x \\
& =3 x^2(0.01 \mathrm{x}) \\
& =0.03 \mathrm{x}^3 \mathrm{~m}^3
\end{aligned}
$
Question 10.
If $\mathrm{g}(\mathrm{x}, \mathrm{y})=3 \mathrm{x}^2-5 \mathrm{y}+2 \mathrm{y}^2, \mathrm{x}(\mathrm{t})=\mathrm{e}^{\mathrm{t}}$ and $\mathrm{y}(\mathrm{t})=\cos \mathrm{t}$, then $\frac{d g}{d t}$ is equal to $\ldots \ldots .$.
(a) $6 \mathrm{e}^{2 \mathrm{t}}+5 \sin \mathrm{t}-4 \cos \mathrm{t} \sin \mathrm{t}$
(b) $6 \mathrm{e}^{2 \mathrm{t}}-5 \sin \mathrm{t}+4 \cos \mathrm{t} \sin \mathrm{t}$
(c) $3 \mathrm{e}^{2 \mathrm{t}}+5 \sin \mathrm{t}+4 \cos \mathrm{t} \sin \mathrm{t}$
(d) $3 \mathrm{e}^{2 \mathrm{t}}-5 \sin \mathrm{t}+4 \cos \mathrm{t} \sin \mathrm{t}$
Solution:
(a) $6 \mathrm{e}^{2 \mathrm{t}}+5 \sin \mathrm{t}-4 \cos \mathrm{t} \sin \mathrm{t}$

Question 11.
If $f(x)=\frac{x}{x+1}$ then its differential is given by
(a) $\frac{-1}{(x+1)^2} d x$
(b) $\frac{1}{(x+1)^2} d x$
(c) $\frac{1}{x+1} d x$
(d) $\frac{-1}{x+1} d x$
Solution:
(b) $\frac{1}{(x+1)^2} d x$
Hint:
$
\begin{aligned}
& d f=\frac{(x+1)(1)-x(1+0)}{(x+1)^2} d x \\
& d f=\frac{1}{(x+1)^2} d x
\end{aligned}
$

Question 12.
If $u(x, y)=x^2+3 x y+y-2019$, then $\left.\frac{\partial u}{\partial x}\right|_{(4,-5)}$ is equal to
(a) $-4$
(b) $-3$
(c) $-7$
(d) 13
Solution:
(c) -7
Hint:
$
\begin{aligned}
\frac{\partial u}{\partial x} & =2 x+3 y \\
\left(\frac{\partial u}{\partial x}\right)_{(4,-5)} & =8-15=-7
\end{aligned}
$

Question 13.
Linear approximation for $g(x)=\cos x$ at $x=\frac{\pi}{2}$ is
(a) $x+\frac{\pi}{2}$
(b) $-x+\frac{\pi}{2}$
(c) $x-\frac{\pi}{2}$
$(d)-x-\frac{\pi}{2}$
Solution:
(b) $-x+\frac{\pi}{2}$
Hint:
$\begin{aligned} g(x) & =\cos x ; x_0=\frac{\pi}{2} \\ g^{\prime}(x) & =-\sin x \\ g\left(\frac{\pi}{2}\right) & =\cos \left(\frac{\pi}{2}\right)=0 \\ g^{\prime}\left(\frac{\pi}{2}\right) & =-\sin \left(\frac{\pi}{2}\right)=-1 \\ \text { Linear approximation } \quad \mathrm{L}(x) & =g\left(x_0\right)+g^{\prime}\left(x_0\right)\left(x-x_0\right) \\ & =0+(-1)\left(x-\frac{\pi}{2}\right) \\ & =-x+\frac{\pi}{2}\end{aligned}$

Question 14.
If $w(x, y, z)=x^2(y-z)+y^2(z-x)+z^2(x-y)$, then (a) $x y+y z+z x$
(a) $x y+y z+z x$
(b) $x(y+z)$
(c) $y(z+x)$
(d) 0
Solution:
(d) 0
Hint:
$
\begin{aligned}
\frac{\partial w}{\partial x} & =2 x(y-z)-y^2+z^2=2 x y-2 x z-y^2+z^2 \\
\frac{\partial w}{\partial y} & =x^2+2 y z-z^2 \\
\frac{\partial w}{\partial z} & =-x^2+y^2+2 z x-2 z y \\
\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y}+\frac{\partial w}{\partial z} & =0
\end{aligned}
$
Question 15.
If $f(x, y, z)=x y+y z+z x$, then $f_x-f_z$ is equal to
(a) $z-x$
(b) $y-z$
(c) $x-z$
(d) $y-x$
Solution:
(a) $z-x$
Hint:
$
\begin{aligned}
& f_x=y+z \\
& f_z=y+x \\
& f_x-f_z=y+z-y-x=z-x
\end{aligned}
$