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grigory [225]
2 years ago
5

The radius r of a sphere is increasing at a constant rate of 0.04 centimeters per second. (Note: The volume of a sphere with rad

ius r is V=4/3 pi r³.)
At the time when the radius of the sphere is 10 centimeters, what is the rate of increase of its volume?
​
Mathematics
1 answer:
-Dominant- [34]2 years ago
3 0

Answer:

The rate of the volume increase will be \frac{dV}{dt}=50.27 cm^{3}/s

Step-by-step explanation:

Let's take the derivative with respect to time on each side of the volume equation.

\frac{dV}{dt}=4\pi R^{2}\frac{dR}{dt}

Now, we just need to put all the values on the rate equation.

We know that:

dR/dt is 0.04 cm/s  

And we need to know what is dV/dt when R = 10 cm.

Therefore using the equation of the volume rate:

\frac{dV}{dt}=4\pi 10^{2}0.04

\frac{dV}{dt}=50.27 cm^{3}/s

I hope it helps you!

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