Question

The radius of sphere is given by r=3t.The rate of change of volume at t=1 is equal to?

Answer 1

Answer:

$\frac{dV}{dt}_{at\:t=1}=339.12$

Explanation:

Formula used:

Volume of sphere of radius 'r' $= \frac{4}{3}\pi\:r^3$ cubic units

Given:

$r=3t$

Then volum of the sphere given

$V=\frac{4}{3}\pi\:r^3$

$V=\frac{4}{3}\pi\:(3t)^3$

$V=\frac{4}{3}\pi\:(27)t^3$

$V=4\pi\:(9)t^3$

$V=36\pi\:t^3$

Differentiate with respect to t

$\frac{dV}{dt}=36\pi\:3t^2$

$\frac{dV}{dt}=108\pi\:t^2$

$\frac{dV}{dt}_{at\:t=1}=108\pi\:(1)^2$

$\frac{dV}{dt}_{at\:t=1}=108\pi\:$

$\frac{dV}{dt}_{at\:t=1}=108(3.14)$

$\frac{dV}{dt}_{at\:t=1}=339.12$

Answer 2

Answer:

Answer will be 108 pi

Explanation:

Just replace r=3t in V=4/3pir^3 and differentiate it.