anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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TuringTest
  • TuringTest
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)
anonymous
  • anonymous
am i finding the derivative for surface area or volume?
TuringTest
  • TuringTest
volume with respect to time dV/dt is what the question asks for

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

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