At what approximate rate (in cubic meters per minute) is the volume of a sphere changing at the instant when its surface area is 3 square meters and the radius is increasing at the rate of 1/5 meters per minute?

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At what approximate rate (in cubic meters per minute) is the volume of a sphere changing at the instant when its surface area is 3 square meters and the radius is increasing at the rate of 1/5 meters per minute?

Mathematics
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V=(4/3)pi*r^3 A=4pi*r^2=3 and we are given that dr/dt=1/5 now compute dV/dt to find the rate change of volume and watch what formula you get (remember to use the chain rule because this is the derivative w/respect to t)
am i finding the derivative for surface area or volume?
volume with respect to time dV/dt is what the question asks for

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Other answers:

oh thats right.
so would it be 4pi(r)^2dr/dt?
the chain rule is going to give exactly now what does the first part of that formula look like?
[4pi(r)^2]dr/dt ^^^^^^^ familiar?
yes surface area formula!
so we can just plug in A=3 for that and dr/dt=1/5 and we're done :)
plus now you know that for a sphere A=dV/dr which is a nice thing to know
3=4pi(r)^2(1/5)
you put the 3 where dV/dt should be dV/dt=4pi(r)^2(dr/dt)=A(dr/dt) ...now plug in for A and dr/dt
3/5
yup
oh wow so it would 3/5 cubic meters per minute
thank you!
anytime!
i have another problem i need help with if you can help me im putting it up :)

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