Explain solution RD Sharma class 12 chapter Differentials Errors and Approximations exercise 13.1 question 8 maths

Hint: volume of a sphere of radius $x$ is given by $v=\frac{4}{3} \pi x^{3}$

Given: we have $\Delta x=0.01 k x$

Solution: Suppose the error in measuring the radius of a sphere be $k$

⇒ Suppose $x$ be the radius of the sphere and $\Delta x$ be the error in the value of $x$

⇒ Thus, we have $\Delta x=\left(\frac{k}{100}\right) \times(x)$

So, $\Delta x=0.01 k x$

⇒ Differentiate v with respect to $x$

\begin{aligned} &\frac{d v}{d x}=\frac{d}{d x}\left(\frac{4}{3}\right) \pi x^{3} \\\\ &\Rightarrow \frac{d v}{d x}=\frac{4 \pi}{3} \frac{d}{d x}\left(x^{3}\right) \Rightarrow \frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \end{aligned}

\begin{aligned} &\frac{d v}{d x}=\frac{4 \pi}{3}\left(3 x^{2}\right) \frac{d v}{d x}=4 \pi x^{2} \\\\ &\Rightarrow \Delta y=\frac{d y}{d x} \Delta x, \text { Here } \frac{d v}{d x}=4 \pi x^{2} \Delta x=0.01 k x \end{aligned}

$\therefore \Delta x=0.04 k \pi x^{3}$

⇒ percentage of error is,

\begin{aligned} &\text { Error }=\frac{0.04 k \pi x^{3}}{\frac{4}{3} \pi x^{3}} \times 100 \% \\\\ \end{aligned}

$=\frac{0.04 k \pi x^{3}}{4 \pi x^{3}} \times 100 \times 3=3 k \%$