Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
0 votes

If the radius of a sphere is measured as \(7 \: m\) with an error of \(0.02 \: m\), then find the approximate error in calculating its volume.

$\begin{array}{1 1} (A)\;3.90 \pi m^3 \\ (B)\;3.92\;\pi m^3 \\(C)\;3.94\; \pi\; m^3 \\ (D)\;3.9\;\pi m^3 \end{array} $

Can you answer this question?

1 Answer

0 votes
  • Let $y=f(x)$
  • $\Delta y=f(x+\Delta x)-f(x)$
  • $dy=\big(\large\frac{dy}{dx}\big)\Delta x$
  • Volume of the sphere $V=\large\frac{4}{3}$$\pi r^3$
Step 1:
Radius of the sphere=$7m$
Error in measurement of radius =$\Delta r$
Volume of the sphere $V=\large\frac{4}{3}$$\pi r^3$
$\Delta V=\large\frac{dV}{dr}\times \Delta r$------(1)
$V=\large\frac{4}{3}\pi r^3$
$\large\frac{dv}{dx}=\frac{4}{3}$$\times 3r^2\times \pi$[Differentiating with respect to x]
Step 2:
Substitute the value of $\large\frac{dv}{dr}$ and $\Delta r$ in equation (1)
$\Delta V=\large\frac{4}{3}$$\times 3r^2\times \pi\times 0.02$
$\quad\quad=4\pi r^2\times 0.02$
$\quad\quad=4\pi 7^2\times 0.02$
$\quad\quad=4\pi\times 49\times 0.02$
$\quad\quad=392\pi \times 0.01$
$\quad\quad=3.92\pi m^3$
answered Aug 6, 2013 by sharmaaparna1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App