posted by Hannah .
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?
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)
If r = 12, just plug the value to the last equation
How did you get 12?
Nevermind I meant how did you get 4/r but I figured it out. Thank You!!
Is this correct for the last part?
(8)(3.14)(12) X (1/(2)(3.14)(12^2) = 4/12
Ok do I actually have to do the calculations though?
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.
OK thank you!
Simplify the answer (4/12) into 1/3