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?
1 answer:
Answer:
The rate of the volume increase will be ![\frac{dV}{dt}=50.27 cm^{3}/s](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D50.27%20cm%5E%7B3%7D%2Fs)
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}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D4%5Cpi%20R%5E%7B2%7D%5Cfrac%7BdR%7D%7Bdt%7D)
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](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D4%5Cpi%2010%5E%7B2%7D0.04)
![\frac{dV}{dt}=50.27 cm^{3}/s](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D50.27%20cm%5E%7B3%7D%2Fs)
I hope it helps you!
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