Mathematics: The radius of a sphere increases at a constant rate of 2 com/min at the time when the volume of the sphere is 40 cm^3ç what is the rate of increase of the volume in cm^3/min?

The radius of a sphere increases at a constant

orlando19882000

14.01.2022

$\frac{dV}{dt}=525 cm^{3}/min$

Step-by-step explanation:

The volume of a sphere is:

$V=\frac{4}{3}\pi r^{3}$ (1)

We know that:

dr/dt = 2 cm/min (increasing rate of radius)V = 40 cm³

If we take the derivative of (1) with respect of time t, we ca n find the rate of increase of the volume.

$\frac{dV}{dt}=\frac{4}{3}\pi 3r^{2}\frac{dr}{dt}=4\pi r^{2}\frac{dr}{dt}$ (2)

We also know that the volume is 40 cm³, then using the (1) we can get the radius at this value.

Solving (1) for r, we have:

$r=\left(\frac{3\cdot V}{4\pi}\right)^{1/3}=\left(\frac{3\cdot 400}{4\pi}\right)^{1/3}=4.57 cm$

Finally dV/dt will be:

$\frac{dV}{dt}=4\pi (4.57)^{2}\cdot 2=525 cm^{3}/min$

I hope it helps you!

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