Exercise 8.2 - 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 8.2
Question 1.
Find differential dy for each of the following functions:
(i) $y=\frac{(1-2 x)^3}{3-4 x}$
(ii) $y=(3+\sin (2 x))^{2 / 3}$
(iii) $y=e^{x^2-5 x+7} \cos \left(x^2-1\right)$
Solution:
(i) $y=\frac{(1-2 x)^3}{3-4 x}$.
$
\begin{aligned}
d y \quad & =\left[\frac{(3-4 x) 3(1-2 x)^2(-2)-(1-2 x)^3(-4)}{(3-4 x)^2}\right] d x \\
\text { (i.e.,) } \quad d y & =\left\{\frac{(1-2 x)^2}{(3-4 x)^2}[-6(3-4 x)+4(1-2 x)]\right\} d x \\
\text { (i.e.,) } \quad d y & =\left\{\frac{(1-2 x)^2}{(3-4 x)^2}[16 x-14]\right\} d x=2 \frac{(8 x-7)(1-2 x)^2}{(3-4 x)^2} d x \\
d y & =\frac{2(8 x-7)(1-2 x)^2}{(3-4 x)^2} d x
\end{aligned}
$

(ii)
$
\begin{aligned}
y & =[3+\sin (2 x)]^{\frac{2}{3}} \\
d y & =\left\{\frac{2}{3}[3+\sin (2 x)]^{\frac{-1}{3}}[(\cos 2 x)(2)]\right\} d x \\
& =\left\{\frac{4}{3}(3+\sin 2 x)^{\frac{-1}{3}}(\cos 2 x)\right\} d x \\
& =\left\{\frac{4}{3} \frac{\cos 2 x}{(3+\sin 2 x)^{\frac{1}{3}}}\right\} d x
\end{aligned}
$
(iii)
$
\begin{aligned}
y & =e^{x^2-5 x+7} \cos \left(x^2-1\right) \\
d y & =\left\{e^{x^2-5 x+7}\left[-\sin \left(x^2-1\right)(2 x)\right]+\cos \left(x^2-1\right)\left[e^{x^2-5 x+7}(2 x-5)\right]\right\} d x \\
d y & =\left\{e^{x^2-5 x+7}\left[-2 x \sin \left(x^2-1\right)+(2 x-5) \cos \left(x^2-1\right)\right]\right\} d x \\
d y & =e^{x^2-5 x+7}\left\{(2 x-5) \cos \left(x^2-1\right)-2 x \sin \left(x^2-1\right)\right\} d x
\end{aligned}
$

Question 2 .
Find $\mathrm{df}$ for $f(x)=x^2+3 x$ and evaluate it for
(i) $\mathrm{x}=2$ and $\mathrm{dx}=0.1$
(ii) $\mathrm{x}=3$ and $\mathrm{dx}=0.02$
Solution:
Solution:
$\begin{aligned}
& y=f(x)=x^2+3 x \\
& d y=(2 x+3) d x
\end{aligned}$
(i) dy $\{$ when $x=2$ and ate $=0.1\}=[2(2)+3](0.1)$ $=7(0.1)=0.7$
(ii) dy $\{$ when $\mathrm{x}=3$ and $\mathrm{dx}=0.02\}=[2(3)+3](0.0 .2)$ $=9(0.02)=0.18$

Question 3.
Find $\Delta \mathrm{f}$ and $\mathrm{df}$ for the function $\mathrm{f}$ for the indicated values of $\mathrm{x}, \Delta \mathrm{x}$ and compare
(i) $f(x)=x^3-2 x^2 ; x=2, \Delta x=d x=0.5$
(ii) $f(x)=x^2+2 x+3 ; x=-0.5, \Delta x=d x=0.1$
Solution:
(i) $y=f(x)=x^3-2 x^2$
$d y=\left(3 x^2-4 x\right) d x$
dy $($ when $\mathrm{x}=2$ and $\mathrm{dx}=0.5)=\left[3\left(2^2\right)-4(2)\right](0.5)$
$=(12-8)(0.5)=4(0.5)=2$
(i.e.,) $\mathrm{df}=2$
Now $\Delta \mathrm{f}=\mathrm{f}(\mathrm{x}+\Delta \mathrm{x})-\mathrm{f}(\mathrm{x})$
Here $x=2$ and $\Delta x=0.5$
$f(x)=x^3-2 x^2$
So $f(x+\Delta x)=f(2+0.5)=f(2.5)=(2.5)^3-2(2.5)^2=(2.5)^2[2.5-2]=6.25(0.5)=3.125$ $f(x)=f(2)=2^3-2\left(2^2\right)=8-8=0$
So $\Delta \mathrm{f}=3.125-0=3.125$
(ii) $y=f(x)=x^2+2 x+3$
$d y=(2 x+2) d x$
$\mathrm{dy}($ when $\mathrm{x}=-0.5$ and $\mathrm{dx}=0.1)$
$=[2(-0.5)+2](0.1)$
$=(-1+2)(0.1)=(1)(0.1)=0.1$
(i.e.,) $\mathrm{df}=0.1$
Now $\Delta \mathrm{f}=\mathrm{f}(\mathrm{x}+\Delta \mathrm{x})-\mathrm{f}(\mathrm{x})$
Here $\mathrm{x}=-0.5$ and $\Delta \mathrm{x}=0.1$
$x^2+2 x+3$
$\mathrm{f}(\mathrm{x}+\Delta \mathrm{x})=\mathrm{f}(-0.5+0.1)=\mathrm{f}(-0.4)$
$=(-0.4)^2+2(-0.4)+3$
$=0.16-0.8+3=3.16-0.8=2.36$
$\mathrm{f}(\mathrm{x})=\mathrm{f}(-0.5)=(-0.5)^2+2(-0.5)+3$
$=0.25-1+3=3.25-1=2.25$
So $\Delta \mathrm{f}=\mathrm{f}(\mathrm{x}+\Delta \mathrm{x})-\mathrm{f}(\mathrm{x})=2.36-2.25=0.11$

Question 4.
Assuming $\log _{10} \mathrm{e}=0.4343$, find an approximate value of $\log _{10} 1003$.
Solution:
To find $\log 1003$
$1003=1000+3$ and
$\log 1000=3$
Let $y=\log x$
$d y=\frac{1}{x} d x$
Here $x=1000$ and $d x=3=3 \log _{10} e$
So $d y=\frac{1}{1000}[3 \times 0.4343]=0.0013029$
Now $y+d y=\log 1000+d y$
$=3+0.0013029$
$=3.0013029$

Question 5.
The trunk of a tree has diameter $30 \mathrm{~cm}$. During the following year, the circumference grew $6 \mathrm{~cm}$.
(i) Approximately, how much did the tree's diameter grow?
(ii) What is the percentage increase in area of the tree's cross-section?
Solution:
(i) Given $\mathrm{r}=15 \mathrm{~cm}$ and rate of change of perimeter $=6 \mathrm{~cm}$
To find the rate of change of diameter
Now perimeter $=p=2 \pi r$
So $\mathrm{dp}=2 \pi \mathrm{dr}$

Here $\mathrm{dp}=6 \mathrm{~cm}$ (given)
$
\begin{aligned}
\Rightarrow 6 & =2 \pi d r \\
\Rightarrow \frac{6}{2 \pi} & =d r\left(\text { i.e.,) } d r=\frac{3}{\pi}\right.
\end{aligned}
$
(i.e.,) rate of change in radius $=d r=\frac{3}{\pi}$
$\Rightarrow \quad$ so rate of change in diameter $=d(2 r)=2 d r=2 \times \frac{3}{\pi}=\frac{6}{\pi} \mathrm{cm}$

(ii) $\quad \mathrm{A}=\pi r^2$
$
\begin{aligned}
\mathrm{A} & =\pi r^2 \\
\text { Now } d \mathrm{~A} & =\pi(2 r) d r \\
& =2 \pi r d r \\
\text { so } \frac{d \mathrm{~A}}{\mathrm{~A}} & =\frac{2 \pi r d r}{\pi r^2}=\frac{2 d r}{r} \\
& =\frac{2\left(\frac{3}{\pi}\right)}{r(=15)}=\frac{6}{15 \pi}=\frac{2}{5 \pi} \\
\text { so } \% \text { error } & =\frac{\mathrm{dA}}{\mathrm{A}} \times 100 \\
& =\frac{2}{5 \pi} \times 100=\frac{40}{\pi}
\end{aligned}
$

Question 6.
An egg of a particular bird is very nearly spherical. If the radius to the inside of the shell is $5 \mathrm{~mm}$ and radius to the outside of the shell is $5.3 \mathrm{~mm}$, find the volume of the shell approximately.
Solution:
$
v=\frac{4}{3} \pi r^3
$
Here $r=5 \mathrm{~cm}$ and $d r=0.3$
$
\text { so } \begin{aligned}
d v & =\frac{4}{3} \pi(25) \frac{3}{10} \\
& =30 \pi \mathrm{mm}^3
\end{aligned}
$

Question 7.
Assume that the cross section of the artery of human is circular. A drug is given to a patient to dilate his arteries. If the radius of an artery is increased from $2 \mathrm{~mm}$ to $2.1 \mathrm{~mm}$, how much is cross-sectional area increased approximately?
Solution:
Area of circle $=\mathrm{A}=\pi \mathrm{r}^2$.
$\begin{aligned} \text { So } d \mathrm{~A} & =\pi(2 r) d r \\ d \mathrm{~A} & =2 \pi r d r \\ \text { (i.e.,) } r=2 \mathrm{~mm} \text { and } d r=0.1 \mathrm{~mm}) & \\ \text { sore } d \mathrm{~A} & =2 \pi(2)(0.1)=0.4 \pi\end{aligned}$

Question 8.
In a newly developed city, it is estimated that the voting population (in thousands) will increase according to $V(t)=30+12 t^2-t^3, 0 \leq t \leq 8$ where $t$ is the time in years. Find the approximate change in voters for the time change from 4 to $4 \frac{1}{6}$ year.
Solution:
$
v=30+12 t^2-t^3 \quad 0 \leq t \leq 8
$
So $d v=\left(24 t-3 t^2\right) d t$
Here $\quad t=4$ and $d t=\frac{1}{6}$
So
$
\begin{aligned}
d v & =\left[24(4)-3\left(4^2\right)\right] \frac{1}{6} \\
& \left.=(96-48) \frac{1}{6}=\frac{48}{6}=8 \text { (in thousand }\right)=8000
\end{aligned}
$
Question 9.
The relation between the number of words $\mathrm{y}$ a person learns in $\mathrm{x}$ hours is given by $\mathrm{y}=52 \sqrt{x}, 0 \leq$ $\mathrm{x} \leq 9$. What is the approximate number of words learned when $\mathrm{x}$ changes from
(i) 1 to $1.1$ hour?
(ii) 4 to $4.1$ hour?

Solution:
$
\begin{aligned}
y & =52 \sqrt{x} \\
d y & =52\left(\frac{1}{2 \sqrt{x}} d x\right)=\frac{26}{\sqrt{x}} d x
\end{aligned}
$
(i) $d y$ (when $x=1$ and $d x=0.1 \mathrm{hr}$ )
$
=\frac{26}{\sqrt{1}}(0.1)=2.6 \sim 3 \text { words }
$
(ii)
$
\begin{aligned}
d y(\text { when } x=4 \text { and } d x & =0.1 \mathrm{hr}) \\
& =\frac{26}{\sqrt{4}}(0.1)=\frac{26}{2}(0.1) \\
& =13(0.1)=1.3 \sim 1 \text { word }
\end{aligned}
$

Question 10.
A circular plate expands uniformly under the influence of heat. If it's radius increases from $10.5$ $\mathrm{cm}$ to $10.75 \mathrm{~cm}$, then find an approximate change in the area and the approximate percentage change in the area.
Solution:
Here radius is changing from $10.5 \mathrm{~cm}$ to $10.75 \mathrm{~cm}$ $\Rightarrow \mathrm{r}=10.5 \mathrm{~cm}$ and $\mathrm{dr}=0.25 \mathrm{~cm}$
Now area $=\mathrm{A}=\pi \mathrm{r}^2$ $\Rightarrow \mathrm{dA}=\pi(2 \mathrm{r}) \mathrm{dr}$
(i) So dA
(when $\mathrm{r}=10.5 \mathrm{~cm}$ and $\mathrm{dr}=0.25 \mathrm{~cm}$ )
$=\pi(2 \times 10.5)(0.25)$
$=5.25 \pi$
(ii) Percentage error in Area $=\frac{d \mathrm{~A}}{\mathrm{~A}} \times 100$
(i.e.,) $\begin{aligned} \text { Here } d \mathrm{~A} & =2 \pi r d r \\ \mathbf{A} & =\pi r^2\end{aligned}$
So $\frac{d \mathrm{~A}}{\mathrm{~A}}=\frac{2 \pi r d r}{\pi r^2}=\frac{2 d r}{r}$
$\frac{d \mathrm{~A}}{\mathrm{~A}} \times 100($ when $r=10.5$ and $d r=0.25 \mathrm{~cm})$
$=\frac{2(0.25)}{10.5} \times 100$
$=4.76 \%$

Question 11.
A coat of paint of thickness $0.2 \mathrm{~cm}$ is applied to the faces of a cube whose edge is $10 \mathrm{~cm}$. Use the differentials to find approximately how many cubic centimeters of paint is used to paint this cube. Also calculate the exact amount of paint used to paint this cube.
Solution:
(i) $v=a^3$
so $\mathrm{dv}=\mathrm{a}^2$ da
$\mathrm{dv}$ (when) $\mathrm{a}=10 \mathrm{~cm}$ and $\mathrm{da}=0.20 \mathrm{~cm}$
$=3\left(10^2\right)(0.2)$
$300 \times 0.2=60 \mathrm{~cm}^3$
Actual paint used $=\mathrm{v}$ at $\mathrm{x}+\Delta \mathrm{x}=10.2$ and $\mathrm{x}=10 \mathrm{~cm}$
$=\mathrm{a}^3$ at $\mathrm{x}+\Delta \mathrm{x}=10.2$ and $\mathrm{x}=10$
$=(10.2)^3-(10)=61.2 \mathrm{~cm}^3$