Whatiknow:

Volume of Sphere = 4/3 pi r ^3

dv/dr = 4 pi r^2

The question

When the radius of a sphere is measured , the measurement has a 1 percent error. What is the resulting percentage error when the volume of sphere is computed using this inaccurate measurement of the radius?

Help!

I'm confused. I googled the answer, and there was work shown but i had no idea how the answer was 6%. Can someone please go through the steps? I don't understand how taking the derivative helps us find the percent error in volume.

I also know that the percent error is (measured value)/(actual value) * 100