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anonymous

  • 5 years ago

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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  1. TuringTest
    • 5 years ago
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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)

  2. anonymous
    • 5 years ago
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    am i finding the derivative for surface area or volume?

  3. TuringTest
    • 5 years ago
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    volume with respect to time dV/dt is what the question asks for

  4. anonymous
    • 5 years ago
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    oh thats right.

  5. anonymous
    • 5 years ago
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    so would it be 4pi(r)^2dr/dt?

  6. TuringTest
    • 5 years ago
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    the chain rule is going to give exactly now what does the first part of that formula look like?

  7. TuringTest
    • 5 years ago
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    [4pi(r)^2]dr/dt ^^^^^^^ familiar?

  8. anonymous
    • 5 years ago
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    yes surface area formula!

  9. TuringTest
    • 5 years ago
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    so we can just plug in A=3 for that and dr/dt=1/5 and we're done :)

  10. TuringTest
    • 5 years ago
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    plus now you know that for a sphere A=dV/dr which is a nice thing to know

  11. anonymous
    • 5 years ago
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    3=4pi(r)^2(1/5)

  12. TuringTest
    • 5 years ago
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    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

  13. anonymous
    • 5 years ago
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    3/5

  14. TuringTest
    • 5 years ago
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    yup

  15. anonymous
    • 5 years ago
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    oh wow so it would 3/5 cubic meters per minute

  16. anonymous
    • 5 years ago
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    thank you!

  17. TuringTest
    • 5 years ago
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    anytime!

  18. anonymous
    • 5 years ago
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    i have another problem i need help with if you can help me im putting it up :)

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