RD Sharma Class 12 Exercise 13.1 Differentials, Errors and Approxim Solutions Maths - Download PDF Free Online

Differentials, errors and approximations exercise 13.1 question 1

Answer: $\mathrm{\Delta }y=0-$No change

Given: $y=sinx$ and $x$ changes from $\frac{\pi }{2}$ to $\frac{22}{14}$
Solution: Suppose $\chi =\frac{\pi }{2}$ Therefore $x+\mathrm{\Delta }x=\frac{22}{14}$
$\begin{array}{rl}& \to \frac{\pi }{2}+\mathrm{\Delta }x=\frac{22}{14}\\ \\ & \mathrm{\Delta }x=\frac{22}{14}-\frac{\pi }{2}\end{array}$

→ Differentiate y with respect to x

$\frac{dy}{dx}=\cos x$
⇒ As we know that $(\sin x)=\cos x$
$\therefore \frac{dy}{dx}=\cos x$
⇒ when,$x=\frac{\pi }{2}$ we have $\frac{dy}{dx}=\cos \left(\frac{\pi }{2}\right)$
$\begin{array}{rl}& \Rightarrow \left(\frac{dy}{dx}\right)x=\frac{\pi }{2}=0\\ \\ & \Rightarrow \mathrm{\Delta }y=\left(\frac{dy}{dx}\right)\mathrm{\Delta }x\end{array}$

Here $\frac{dy}{dx}=0$ and $\mathrm{\Delta }x=\frac{22}{14}-\frac{\pi }{2}$

$\mathrm{\Delta }y=(0)\left(\frac{22}{14}\right)-\left(\frac{\pi }{2}\right)$

$\Delta y=(0)$

Differentials, errors and approximations exercise 13.1 question 2

Answer: decrease $80\pi c{m}^2$
Hint: As we know, volume of a sphere of radius $x$ is given by $v=\left(\frac{4}{3}\right)\pi x^3$
Given: the radius of a sphere changes from 10cm to 9.8cm
Solution: Suppose $x$ be the radius of the sphere and $\mathrm{\Delta }x$ be the change in the value of $x$
$\Rightarrow$ Thus, we have $x=10$ and $x+\mathrm{\Delta }x=9.8$$\begin{array}{rl}& \Rightarrow 10+\mathrm{\Delta }x=9.8\\ \\ & \mathrm{\Delta }x=9.8-10\\ \\ & =-0.2\end{array}$

⇒ Differentiate $v$ with respect to $x$

$\begin{array}{rl}& \frac{dv}{dx}=\frac{d}{dx}\left(\left(\frac{4}{3}\right)\pi x^3\right)\\ \\ & \frac{dv}{dx}=\frac{4\pi }{3}\frac{d}{dx}\left(x^3\right)\\ \\ & \frac{d}{dx}\left(x^3\right)=n{x}^{n-1}\end{array}$

$\begin{array}{rl}& \frac{dv}{dx}=4\pi x^2\text{ when }\frac{dv}{dx}=4\pi \times 100=400\\ \\ & (x=10)\\ \\ & \Rightarrow \mathrm{\Delta }y=\left(\frac{dy}{dx}\right)\mathrm{\Delta }y\end{array}$

$\begin{array}{rl}& \frac{dv}{dx}=400\pi \text{ and }\mathrm{\Delta }x=-0.2\\ \\ & \mathrm{\Delta }x=(400\pi )(-0.2)=-80\pi \end{array}$

Differentials, errors and approximations exercise 13.1 question 3

Answer:$2k\pi c{m}^2$
Hint: Here, Area of a circular plate of radius $x$ is given by $A=\pi x^2$
Given: the radius of a circular plate initially is 10 cm and it increase by K%
Solution: Suppose $x$ be the radius of the plate, and $\mathrm{\Delta }x$ is the change in the value of $x$
Thus we have $x=10\text{ and }\mathrm{\Delta }x=\frac{K}{100}\times 10$
So, $\mathrm{\Delta }x=0.1K$
Differentiating A with respect to $x$
$\begin{array}{rl}& \frac{dA}{dx}=\frac{d}{dx}\left(\pi x^2\right)\\ \\ & \frac{dA}{dx}=\pi \frac{d}{dx}\left(x^2\right)=n{x}^{n-1}\\ \\ & \Rightarrow \frac{dA}{dx}=2\pi x\end{array}$
$\begin{array}{rl}& \Rightarrow \text{ when }x=10\text{ and }\frac{dA}{dx}=2\pi (10)\text{ so, }\frac{dA}{dx}=20\pi \\ \\ & \Rightarrow \mathrm{\Delta }y=\frac{dy}{dx}\mathrm{\Delta }x\end{array}$
$\begin{array}{rl}& \text{ Here, }\frac{dA}{dx}=20\pi \text{ and }\mathrm{\Delta }x=0.1K\\ \\ & \mathrm{\Delta }A=(20\pi )(0.1K)\\ \\ & \mathrm{\Delta }A=2K\pi \end{array}$

Differentials, errors and approximations exercise 13.1 question 4

Answer: 2% Error
Hint: area of a cubical box of radius $x$ is given by $s=6x^2$
Given:$\mathrm{\Delta }x$ is the error in the value of $x\mathrm{\Delta }x=0.01x$
Solution: suppose $x$ be the of cubical box
So,$\mathrm{\Delta }x=0.01x$
Differentiate $s$ with respect to $x$
$\begin{array}{rl}& \frac{ds}{dx}=\frac{d}{dx}\left(6x^2\right)\\ \\ & \Rightarrow \frac{ds}{dx}=6\times \frac{d}{dx}\left(x^2\right)\\ \\ & \frac{d}{dx}\left(x^n\right)=n{x}^{n-1}\Rightarrow \frac{ds}{dx}=6x2x=12x\end{array}$
⇒ As we know that if $y=F(x)\mathrm{\Delta }x$ is a small increment in $x$ , then the corresponding increment in $y$
$\mathrm{\Delta }x=F(x+\mathrm{\Delta }y)-F(x)$ , is approximately given as $\mathrm{\Delta }y=\left(\frac{dy}{dx}\right)\mathrm{\Delta }x$
$\begin{array}{rl}& \Rightarrow \text{ Here, }\frac{ds}{dx}=12x\mathrm{\Delta }x=0.01x\\ \\ & \mathrm{\Delta }s=12x\times 0.01x\\ \\ & \mathrm{\Delta }s=0.12x^2\end{array}$
$\Rightarrow$ Percentage of error is
$\text{ Error }=\frac{0.12x^2}{6x^2}\times 100\mathrm{\%}=0.02\times 100\mathrm{\%}=2\mathrm{\%}$

Differentials, errors and approximations exercise 13.1 question 5

Answer: 0.3%
Hint: Surface volume of a sphere of radius $x$ is given by $v=\frac{4}{3}\pi x^3$
Given: let $x$ be the radius and $\mathrm{\Delta }x$ be the error in the value $x$
Solution: Suppose $x$ be the radius of the sphere and $\mathrm{\Delta }x$ be the error in the value $x$
⇒ Thus, we have $\mathrm{\Delta }x=\left(\frac{0.1}{100}\right)\times (x)$
So, $\mathrm{\Delta }x=0.001x$
⇒ Volume of a sphere, $v=\frac{4}{3}\pi x^3$
So, Differentiate v with respect $x$
$\begin{array}{rl}& \Rightarrow \frac{dv}{dx}=\frac{d}{dx}\left(\frac{4}{3}\pi x^3\right)\Rightarrow \frac{dv}{dx}=\frac{4\pi }{3}\frac{d}{dx}\left(x^3\right)\\ \\ \\ \\ & \Rightarrow \frac{dv}{dx}=\frac{4\pi }{3}\left(4x^2\right)=4\pi x^2\end{array}$
$\Rightarrow$ As we know, $y=f(x)$ and $\mathrm{\Delta }x$ is a smaller increment,
$\begin{array}{rl}& \mathrm{\Delta }y=\left(\frac{dv}{dx}\right)\mathrm{\Delta }x=0.001x\\ \\ & \mathrm{\Delta }x=0.004\pi x^3\\ \\ & \text{ Here, }\frac{dv}{dx}=4\pi x^2\text{ and }\mathrm{\Delta }x=0.001x\end{array}$

$\Rightarrow$ Percentage error $=\frac{0.004\pi x^3}{\frac{4}{3}\pi x^3}\times 100=0.003\times 100=0.3\mathrm{\%}$

Differentials, errors and approximations exercise 13.1 question 6

Answer: 0.7%

Hint: Here we use the basic concept of area and logarithm
Given:$p{v}^{1.4}$ =constant
Solution: Given as $p{v}^{1.4}$=constant and the decrease in $v$ is $\frac{1}{2}$
$\Rightarrow$ Thus, we have $\mathrm{\Delta }v=\left(\frac{-1}{2}\right)100xv$
So, $\mathrm{\Delta }v=-0.005v$
$\Rightarrow$ Now, $p{v}^{1.4}$ =constant
taking log on both sides,
$\log \left(p{v}^{1.4}\right)=\log (\text{ constant })$
$\Rightarrow$ Differentiate both sides with respect to v
$\begin{array}{rl}& \frac{d}{dp}(\log p)x\frac{dp}{dv}+\frac{d}{dv}(1.4\log v)=0\\ \\ & \Rightarrow \frac{d}{dp}(\log p)x\frac{dp}{dv}+1.4\frac{d}{dv}(\log v)=0\end{array}$
$\begin{array}{rl}& \Rightarrow \frac{d}{dx}(\log x)=\frac{1}{x}\\ \\ & \Rightarrow \frac{d}{dv}=\frac{-1.4}{v}P\end{array}$
$\begin{array}{rl}& \mathrm{\Delta }p=\left(\frac{-1.4}{v}p\right)(-0.005x)\\ \\ & \mathrm{\Delta }p=0.007p\end{array}$
$\Rightarrow$ Percentage of error is
$\text{ Error }=\frac{0.007p}{p}\times 100=0.7\mathrm{\%}$

Differentials, errors and approximations exercise 13.1 question 7

Answer: 2k%
Hint: total surface area of the cone is given by $s=\pi r^2+\pi rl$
Given: $\mathrm{\Delta }x=\left(\frac{k}{100}\right)\times (x)$
$\mathrm{\Delta }x=0.01kx$
Solution: Suppose $x$ be the height of the cone and $\mathrm{\Delta }x$ be the change in the value of $x$
⇒ Thus we have $\mathrm{\Delta }x=\left(\frac{k}{100}\right)\times x$
$\mathrm{\Delta }x=0.01kx$
⇒ suppose us assume that radius, the semi height and semi vertical angle of the come to be $r$, 1 and respectively as shown

⇒ from above figure, using trigonometry

$\tan d=\frac{OB}{OA}=\frac{r}{x}$
⇒ we have also
$\cos x=\frac{OA}{AB}=\frac{x}{l}$
$1=\frac{x}{\cos x}$
⇒ Now the total surface area of the cone is given by
$s=\pi x^2+\pi rl$
⇒ from the above, we have $r=x\tan x\text{ and }l=x\sec x$
$\Rightarrow s=\pi x^2+\pi (x\tan x\cdot x\sec x)$
⇒ Differentiate s with respect to $x$
$\begin{array}{rl}& \frac{ds}{dx}=\frac{d}{dx}[\pi x^2\tan x(\tan x+\sec x)]\\ \\ & \Rightarrow \frac{ds}{dx}=\pi \tan x(\tan x+\sec x)\frac{1}{dx}\left(x^2\right)\end{array}$
$\begin{array}{rl}& \frac{d}{dx}\left(x^n\right)=n{x}^{n-1}\\ \\ & \Rightarrow \frac{ds}{dx}=2\pi x\tan x(\tan x+\sec x)\end{array}$
$\begin{array}{rl}& \Rightarrow \text{ so, }\mathrm{\Delta }x=\left(\frac{dy}{dx}\right)\mathrm{\Delta }x\\ \\ & \Rightarrow \text{ Here }\frac{ds}{dx}=2\pi x(\tan x+\sec x)\text{ and }\mathrm{\Delta }x=0.01kx\\ \\ & \Rightarrow \mathrm{\Delta }s=(2\pi x\tan (x)[\tan (x)+\sec (x)](0.01kx)\end{array}$
⇒ Percentage of increase in S
$\begin{array}{rl}& \text{ Increase }=\frac{\mathrm{\Delta }s}{s}\times 100y\\ \\ & =100+\times 0..03k\pi x^2\tan x[\tan x+\sec x]\\ \\ & \pi x^2\tan x(\tan x+\sec x)\end{array}$
$\begin{array}{rl}& \text{ Increase }=0.02k\times 100\\ \\ & =2k\mathrm{\%}\end{array}$

Differentials, errors and approximations exercise 13.1 question 8

Answer: 3k %
Hint: volume of a sphere of radius $x$ is given by $v=\frac{4}{3}\pi x^3$
Given: we have $\mathrm{\Delta }x=0.01kx$
Solution: Suppose the error in measuring the radius of a sphere be $k$
⇒ Suppose $x$ be the radius of the sphere and $\mathrm{\Delta }x$ be the error in the value of $x$
⇒ Thus, we have $\mathrm{\Delta }x=\left(\frac{k}{100}\right)\times (x)$
So, $\mathrm{\Delta }x=0.01kx$
⇒ Differentiate v with respect to $x$
$\begin{array}{rl}& \frac{dv}{dx}=\frac{d}{dx}\left(\frac{4}{3}\right)\pi x^3\\ \\ & \Rightarrow \frac{dv}{dx}=\frac{4\pi }{3}\frac{d}{dx}\left(x^3\right)\Rightarrow \frac{d}{dx}\left(x^n\right)=n{x}^{n-1}\end{array}$
$\begin{array}{rl}& \frac{dv}{dx}=\frac{4\pi }{3}\left(3x^2\right)\frac{dv}{dx}=4\pi x^2\\ \\ & \Rightarrow \mathrm{\Delta }y=\frac{dy}{dx}\mathrm{\Delta }x,\text{ Here }\frac{dv}{dx}=4\pi x^2\mathrm{\Delta }x=0.01kx\end{array}$
$\therefore \mathrm{\Delta }x=0.04k\pi x^3$
⇒ percentage of error is,
$\begin{array}{rl}& \text{ Error }=\frac{0.04k\pi x^3}{\frac{4}{3}\pi x^3}\times 100\mathrm{\%}\\ \end{array}$
$=\frac{0.04k\pi x^3}{4\pi x^3}\times 100\times 3=3k\mathrm{\%}$

Differentials, errors and approximations exercise 13.1 question 9 (i)

Answer: 5.03
Hint: Here we use the formula
$\mathrm{\Delta }y=f(x+\mathrm{\Delta }x)-f(x)$
Given: $\sqrt{25.3}$
Solution: let $y=\sqrt{x}$
$\begin{array}{rl}& x=25\mathrm{\&}\mathrm{\Delta }x=0.3\\ \\ & \Rightarrow \text{ Since }y=\sqrt{x}\\ \\ & \frac{dy}{dx}=\frac{d\sqrt{x}}{dx}=\frac{1}{2\sqrt{x}}\end{array}$
⇒ Now,
$\begin{array}{rl}& \mathrm{\Delta }y=\frac{dy}{dx}\mathrm{\Delta }x\\ \\ & =\frac{1}{2\sqrt{x}}\times 0.3\\ \\ & =\frac{1}{2\sqrt{x}}\times 0.3=0.03\end{array}$
$\begin{array}{rl}& \Rightarrow \text{ Also, }\mathrm{\Delta }y=\frac{f(x+\mathrm{\Delta }x)-f(x)}{}=\sqrt{25+0.3}-\sqrt{25}\\ \\ & \sqrt{25.3}=5.03\end{array}$

Differentials, errors and approximations exercise 13.1 question 9 (ii)

Answer: 0.208
Hint: here we use the formula
Given:$(0.009{)}^{\frac{1}{3}}$
Solution: consider $y=x^{\frac{1}{3}}\text{ Let }x=0.009\text{ and }\mathrm{\Delta }x=0.001$$\mathrm{\Delta }x=(x+\mathrm{\Delta }x{)}^{\frac{1}{3}}=(0.009{)}^{\frac{1}{3}}-(0.008{)}^{\frac{1}{3}}(0.009{)}^{\frac{1}{3}}=0.2+\mathrm{\Delta }y$

⇒ Now, $dy$ is approximately equal to $\mathrm{\Delta }y$ and is given by,

$\begin{array}{rl}& dy=\left(\frac{dy}{dx}\right)\mathrm{\Delta }x=\frac{1}{3(x{)}^{\frac{2}{3}}}(\mathrm{\Delta }x)\\ \\ & =\frac{3}{3\times 0.04}(0.001)=\frac{0.001}{0.12}\\ \\ & =0.008\end{array}$
Hence, the approximate value of $(0.004{)}^{\frac{1}{3}}i\text{ is }0.2+0.08=0.208$

Differentials, errors and approximations exercise 13.1 question 9 (iii)

Answer: 0.191667

Hint: Here we use the formula
$\mathrm{\Delta }y=f(x+\mathrm{\Delta }x)-f(x)$
Given:$(0.007{)}^{\frac{1}{3}}$
Solution: consider the function $f(x)=3\sqrt{x}$
Let:
$\begin{array}{rl}& x=0.008\\ \\ & x+\mathrm{\Delta }x=0.007\end{array}$
Then,
$\begin{array}{rl}& \mathrm{\Delta }x=-0.001\\ \\ & \text{ For }x=0.008,\end{array}$
$\begin{array}{rl}& y=\sqrt{0.008}=0.2\\ \\ & \text{ Let, }dx=\mathrm{\Delta }x=0.001\\ \\ & \text{ Now }y=3\sqrt{x}\end{array}$
$\begin{array}{rl}& \frac{dy}{dx}=\frac{1}{3(x{)}^{\frac{2}{3}}}\\ \\ & \left(\frac{dy}{dx}\right)x=0.008=\frac{1}{0.12}\times 0.001=\frac{1}{120}\end{array}$$\begin{array}{rl}& \Rightarrow \mathrm{\Delta }y=\frac{1}{120}=0.008333\\ \\ & (0.007{)}^{\frac{1}{3}}=y+\mathrm{\Delta }y=0.191667\end{array}$