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March 25, 2017

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Gas is escaping from a spherical balloon at the rate of 2ft^3/min. How fast is the surface area shrinking (ds/dt) when the radius is 12ft? (A sphere of radius r has volume v=4/3 pi r^3 and surface area S=4pi r^2.)

Remember that ds/dt = ds/dr X dr/dt

Step 1: Find ds/dr

I have no clue where to start. I was going to set 2 equal to 4 pi r^2 and then take the derivative but I really have no idea.

Step 2: Find dr/dr. (HInt dv/dt= dv/dr X dr/dt

Would I set 2 equal to volume and then take derivative?

Step 3: Find ds/dt?

Step 4: Evaluate ds/dt when the radius is 12ft.
After I find my equation I would just plug in 12 correct?

  • Math(Please help) - ,

    S = 4pi*r^2
    dS/dr = 4pi*2r = 8pi*r

    V = (4/3)pi*r^3
    dV/dr = 4pi*r^2
    dV/dt = dV/dr * dr/dt
    2 = 4pi*r^2 * dr/dt
    dr/dt = 1/(2pi*r^2)

    dS/dt = dS/dr * dr/dt
    = 8pi*r * 1/(2pi*r^2)
    = 4/r
    If r = 12, just plug the value to the last equation

  • Math(Please help) - ,

    How did you get 12?

  • Math(Please help) - ,

    Nevermind I meant how did you get 4/r but I figured it out. Thank You!!

  • Math(Please help) - ,

    Is this correct for the last part?

    (8)(3.14)(12) X (1/(2)(3.14)(12^2) = 4/12

  • Math(Please help) - ,

    Yes

  • Math(Please help) - ,

    Ok do I actually have to do the calculations though?

  • Math(Please help) - ,

    If you are not given the value, that means you have to write the steps I told you until the very last equation.
    If you are given the value (of r), you don't have to put the value in every equation. Just follow my steps until the last equation, then plug the value after that. It's easier that way.

  • Math(Please help) - ,

    OK thank you!

  • Math(Please help) - ,

    Simplify the answer (4/12) into 1/3

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