# Related rate

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Use differentials to find the approximate change in the surface area when the radius of the sphere is increased from 10cm to 10.2cm.

I know A=4pir^2 and A' is 8pir but I'm not sure how to set it up from here. Can someone please show me all the steps involved? (This isn't HW, just a textbook question that I can't figure out, thanks!)

• Related rate -

Ha! I figured it out.

dA=8pir dr
r=10 , dr=10.2

dA=8(10)(.2)pi
dA=16pi (approx)

check:
A(10)=4(10)^2pi
A(10)=400pi
A(10.2)=4(10.2)^2pi
A(10.2)=416.16

A(10.2)-A(10)=16.16pi (exact)

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