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provide solution for RD Sharma maths class 12 chapter Differentials Errors and Approximations exercise  13.1 question 2

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Answer: decrease 80\pi \; cm^{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 \Delta x be the change in the value of x

\Rightarrow Thus, we have x=10 and x+ \Delta x=9.8

\begin{aligned} &\Rightarrow 10+\Delta x=9.8 \\\\ &\Delta x=9.8-10 \\\\ &=-0.2 \end{aligned}

⇒ Differentiate v with respect to x

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

\begin{aligned} &\frac{d v}{d x}=4 \pi x^{2} \text { when } \frac{d v}{d x}=4 \pi \times 100=400 \\\\ &(x=10) \\\\ &\Rightarrow \Delta y=\left(\frac{d y}{d x}\right) \Delta y \end{aligned}

\begin{aligned} &\frac{d v}{d x}=400 \pi \text { and } \Delta x=-0.2 \\\\ &\Delta x=(400 \pi)(-0.2)=-80 \pi \end{aligned}

 

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