Exercise 8.1 - Chapter 8 - Differentials and Partial Derivatives - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Differentials and Partial Derivatives
$\operatorname{Ex} 8.1$
Question 1.
Let $\mathrm{f}(\mathrm{x})=\sqrt[3]{x}$. Find the linear approximation at $\mathrm{x}=27$. Use the linear approximation to $\sqrt[3]{27.2}$.

Solution:
$
\begin{aligned}
& f(x)=\sqrt[3]{x}=x^{\frac{1}{3}} \\
& f^{\prime}(x)=\frac{1}{3} x^{\frac{-2}{3}} f\left(x_0\right)=f(27)=(27)^{\frac{1}{3}}=3 \\
& x_0=27 \\
& \Delta x=0.2 \\
& x=27.2 \\
& f^{\prime}\left(x_0\right)=f^{\prime}(27)=\frac{1}{3}(27)^{\frac{-2}{3}} \\
& =\frac{1}{3}(3)^{-2}=\frac{1}{3}\left(\frac{1}{9}\right) \\
& =\frac{1}{27} \\
&
\end{aligned}
$
The required linear approximation $\mathrm{L}(x)=f\left(x_0\right)+f^{\prime}\left(x_0\right)\left(x-x_0\right)$
$
\begin{aligned}
\mathrm{L}(x) & =3+\frac{1}{27}(x-27) \\
\mathrm{L}(27.2) & =3+\frac{1}{27}(27.2-27) \\
& =3+\frac{1}{27}(0.2) \\
& =3+0.0074=3.0074 \text { (approximately) }
\end{aligned}
$
Question 2.
Use the linear approximation to find approximate values of
(i) $(123)^{\frac{2}{3}}$
(ii) $\sqrt[4]{15}$
(iii) $\sqrt[3]{26}$

Solution:
(i) $(123)^{\frac{2}{3}}=(125-2)^{2 / 3}$

we have,
$
\begin{aligned}
x & =123 \\
x_0 & =125 \\
\Delta x & =-2
\end{aligned}
$
Let $f(x)=x^{2 / 3}$
$
\begin{aligned}
& f^{\prime}(x)=\frac{2}{3} x^{\frac{-1}{3}} \\
& f\left(x_0\right)=f(125)=(125)^{\frac{2}{3}}=25 \\
& f^{\prime}\left(x_0\right)=\frac{2}{3}(125)^{\frac{-1}{3}}=\frac{2}{3}\left(\frac{1}{5}\right)=\frac{2}{15}
\end{aligned}
$
$
\begin{aligned}
x & =123 \\
x_0 & =125 \\
\Delta x & =-2
\end{aligned}
$
The required linear approximation $\mathrm{L}(x)=f\left(x_0\right)+f^{\prime}\left(x_0\right)\left(x-x_0\right)$
$
\begin{aligned}
\mathrm{L}(x) & =25+\frac{2}{15}(x-125) \\
\mathrm{L}(123) & =25+\frac{2}{15}(123-125) \\
& =25+\frac{2}{15}(-2)=25-\frac{4}{15} \\
& =25-0.2667 \\
& =24.7333 \text { (approximately) }
\end{aligned}
$

(ii) $\sqrt[4]{15}$ consider $(15)^{\frac{1}{4}}=(16-1)^{\frac{1}{4}}$

we have,
$
\begin{aligned}
x & =15 \\
x_0 & =16 \\
\Delta x & =-1 .
\end{aligned}
$
Let $f(x)=x^{\frac{1}{4}}$ we have,
$
\begin{aligned}
x & =15 \\
x_0 & =16 \\
\Delta x & =-1 .
\end{aligned}
$
$
f^{\prime}(x)=\frac{1}{4} x^{\frac{-3}{4}} f\left(x_0\right)=f(16)=(16)^{\frac{1}{4}}=2
$

$
f^{\prime}\left(x_0\right)=f^{\prime}(16)=\frac{1}{4}(16)^{\frac{-3}{4}}=\frac{1}{4}\left(\frac{1}{8}\right)=\frac{1}{32}
$
The required linear approximation $\mathrm{L}(x)=f\left(x_0\right)+f^{\prime}\left(x_0\right)\left(x-x_0\right)$
$
\begin{aligned}
\mathrm{L}(x) & =2+\frac{1}{32}(x-16) \\
\mathrm{L}(15) & =2+\frac{1}{32}(15-16) \\
& =2-\frac{1}{32} \\
& =2-0.03125 \\
& =1.96875 \text { (approximately) }
\end{aligned}
$

(iii)

we have,
$
\begin{aligned}
x & =26 \\
x_0 & =27 \\
\Delta x & =-1
\end{aligned}
$
$
\begin{aligned}
& \sqrt[3]{26}=(26)^{\frac{1}{3}}=(27-1)^{\frac{1}{3}} \\
& \text { consider } f(x)=x^{\frac{1}{3}} \\
& f^{\prime}(x)=\frac{1}{3} x^{\frac{-2}{3}} \\
& x=26 \\
& x_0=27 \\
& \Delta x=-1 \\
& f\left(x_0\right)=f(27)=(27)^{\frac{1}{3}}=3 \\
& f^{\prime}\left(x_0\right)=f^{\prime}(27)=\frac{1}{3}(27)^{\frac{-2}{3}}=\frac{1}{3}\left(\frac{1}{9}\right)=\frac{1}{27} \\
&
\end{aligned}
$
The réquired linear approximation $\mathrm{L}(x)=f\left(x_0\right)+f^{\prime}\left(x_0\right)\left(x-x_0\right)$
$
\begin{aligned}
\mathrm{L}(x) & =3+\frac{1}{27}(x-27) \\
\mathrm{L}(26) & =3+\frac{1}{27}(26-27)=3+\frac{1}{27}(-1) \\
& =3-\frac{1}{27} \\
& =3-0.037 \\
& =2.963 \text { (approximately) }
\end{aligned}
$

Question 3.
Find a linear approximation for the following functions at the indicated points.
(i) $f(x)=x^3-5 x+12, x_0=2$
(ii) $\mathrm{g}(\mathrm{x})=\sqrt{x^2+9}+\mathrm{x}_0=-4$
(iii) $\mathrm{h}(\mathrm{x})=\frac{x}{x+1}, \mathrm{x}_0=1$
Solution:
(i) $
\begin{aligned}
& f(x)=x^3-5 x+12 \\
& f(x)=3 x^2-5 \\
& f\left(x_0\right)=f(2)=(2)^3-5(2)+12=8-10+12=10 \\
& f^{\prime}\left(x_0\right)=f^{\prime}(2)=3(2)^2-5=12-5=7
\end{aligned}
$
The required linear approximation $L(x)=f\left(x_0\right)+f^{\prime}\left(x_0\right)\left(x-x_0\right)$
$
\begin{aligned}
& =10+7(x-2) \\
& =10+7 x-14 \\
& =7 x-4
\end{aligned}
$

(ii)
$
\begin{aligned}
g(x) & =\sqrt{x^2+9} ; x_0=-4 \\
g^{\prime}(x) & =\frac{1}{2 \sqrt{x^2+9}}(2 x)=\frac{x}{\sqrt{x^2+9}} \\
g\left(x_0\right) & =g(-4)=\sqrt{16+9}=\sqrt{25}=5 \\
g^{\prime}\left(x_0\right) & =\frac{-4}{\sqrt{25}}=\frac{-4}{5}
\end{aligned}
$
The required linear approximation $\mathrm{L}(x)=g\left(x_0\right)+g^{\prime}\left(x_0\right)\left(x-x_0\right)$
$
\begin{aligned}
& =5-\frac{4}{5}(x+4) \\
& =5-\frac{4 x}{5}-\frac{16}{5}=\frac{9}{5}-\frac{4 x}{5} \\
& =\frac{9-4 x}{5}
\end{aligned}
$

(iii)
$
\begin{aligned}
h(x) & =\frac{x}{x+1} ; x_0=1 \\
h^{\prime}(x) & =\frac{(x+1)(1)-x(1)}{(x+1)^2} \\
h^{\prime}(x) & =\frac{1}{(x+1)^2} \\
h\left(x_0\right)=h(1) & =\frac{1}{1+1}=\frac{1}{2} \\
h^{\prime}\left(x_0\right)=h^{\prime}(1) & =\frac{1}{(1+1)^2}=\frac{1}{4}
\end{aligned}
$
The required linear approximation $\mathrm{L}(x)=h\left(x_0\right)+h^{\prime}\left(x_0\right)\left(x-x_0\right)$
$
\begin{aligned}
& =\frac{1}{2}+\frac{1}{4}(x-1) \\
& =\frac{1}{2}+\frac{x}{4}-\frac{1}{4}=\frac{x}{4}+\frac{1}{4} \\
& =\frac{x+1}{4}
\end{aligned}
$
Question 4.
The radius of a circular plate is measured as $12.65 \mathrm{~cm}$ instead of the actual length $12.5 \mathrm{~cm}$. find the following in calculating the area of the circular plate:
(i) Absolute error
(ii) Relative error
(iii) Percentage error
Solution:

We know that Area of the circular plate $\mathrm{A}(\mathrm{r})=\pi \mathrm{r}^2, \mathrm{~A}^{\prime}(\mathrm{r})=2 \pi \mathrm{r}$
Change in Area $=\mathrm{A}^{\prime}(12.5)(0.15)=3.75 \pi \mathrm{cm}^2$
Exact calculation of the change in Area $=\mathrm{A}(12.65)-\mathrm{A}(12.5)$
$
\begin{aligned}
& =160.0225 \pi-156.25 \pi \\
& =3.7725 \pi \mathrm{cm}^2
\end{aligned}
$
(i) Absolute error $=$ Actual value $-$ Approximate value
$
=3.7725 \pi-3.75 \pi
$
$
=0.0225 \pi \mathrm{cm}^2
$
(ii)
$
\begin{aligned}
\text { Relative error } & =\frac{\text { Actual value }-\text { Approximate value }}{\text { Actual value }} \\
& =\frac{3.7725 \pi-3.75 \pi}{3.7725 \pi} \\
& =\frac{0.0225 \pi}{3.7725 \pi} \\
& =0.006 \mathrm{~cm}^2
\end{aligned}
$
(iii)
$
\begin{aligned}
\text { Percentage error } & =\text { Relative error } \times 100 \\
& =0.006 \times 100 \\
& =0.6 \%
\end{aligned}
$
Question 5.
A sphere is made of ice having radius $10 \mathrm{~cm}$. Its radius decreases from $10 \mathrm{~cm}$ to $9.8 \mathrm{~cm}$. Find approximations for the following:
(i) change in the volume
(ii) change in the surface area
Solution:
(i) We know that Volume of sphere

$
\begin{aligned}
v(r) & =\frac{4}{3} \pi r^3 \\
v^{\prime}(r) & =\frac{4}{3} \pi\left(3 r^2\right) \\
v^{\prime}(r) & =4 \pi r^2
\end{aligned}
$
we have
$
r=10 \mathrm{~cm}
$
Change in volume at $r=10$ is
$
\begin{aligned}
& =v^{\prime}(r)[10-9.8] \\
& =4 \pi(10)^2(0.2) \\
& =8 \pi \mathrm{cm}^3
\end{aligned}
$
$\therefore$ Volume decreases by $80 \pi \mathrm{cm}^3$
(ii) Surface area of the sphere
$
\begin{aligned}
& \mathrm{S}(\mathrm{r})=4 \pi \mathrm{r}^2 \\
& \mathrm{~S}^{\prime}(\mathrm{r})=8 \pi \mathrm{r}
\end{aligned}
$
Change in surface area at $r=10$ is
$
=\mathrm{S}^{\prime}(\mathrm{r})[10-9.8]
$
$
=8 \pi(10)(0.2)=16 \pi \mathrm{cm}^2
$
$\therefore$ Surface Area decreases by $16 \pi \mathrm{cm}^2$
Question 6.
The time $\mathrm{T}$, taken for a complete oscillation of a single pendulum with length 1 , is given by the equation $\mathrm{T}=2 \pi \sqrt{\frac{l}{g}}$, where $\mathrm{g}$ is a constant. Find the approximate percentage error in the calculated value of $T$ corresponding to an error of 2 percent in the value of 1

Solution:
$
\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{l}{g}} \\
& \mathrm{~T}=2 \pi\left(\frac{l}{g}\right)^{\frac{1}{2}} \\
& \log \mathrm{T}=\log 2 \pi+\frac{1}{2} \log \left(\frac{l}{g}\right) \\
& \log \mathrm{T}=\log 2 \pi+\frac{1}{2}[\log l-\log g] \\
& \text { D.w.r to ' } l \\
& \frac{1}{\mathrm{~T}} \frac{d \mathrm{~T}}{d l}=0+\frac{1}{2}\left[\frac{1}{l}-0\right] \\
& \text { Percentage error } \frac{d \mathrm{~T}}{\mathrm{~T}}=\frac{1}{2} \cdot \frac{1}{l} d l \\
& \frac{\Delta \mathrm{T}}{\mathrm{T}} \times 100=\frac{1}{2} \cdot \frac{1}{l} d l \times 100 \\
& \text { (But } d l=0.02 l)=\frac{1}{2}\left(\frac{1}{l}\right) \cdot(0.02 l) \times 100 \\
& =1 \% \\
&
\end{aligned}
$

Question 7.
Show that the percentage error in the $\mathrm{n}^{\text {th }}$ root of a number is approximately $\frac{1}{n}$ times the percentage error in the number.
Solution:
Let $\mathrm{x}$ be the number
$
\begin{aligned}
\therefore y & =f(x)=(x)^{\frac{1}{n}} \\
\log y & =\frac{1}{n} \log x \\
\text { D.w.r to ' } x^{\prime} & \frac{1}{y} \cdot \frac{1}{x} \\
\frac{1}{y} \frac{d y}{d x} & =\frac{1}{n} \times \frac{1}{x} d x
\end{aligned}
$