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A spherical balloon is inflated with helium at the rate of 100pi ft^3/min
how fast is the balloon's radius increasing at the instant the radius is 5ft?
how fast is the surface are increasing?
 
alena said:
A spherical balloon is inflated with helium at the rate of 100pi ft^3/min. how fast is the balloon's radius increasing at the instant the radius is 5ft? how fast is the surface are increasing?
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Sphere!! What do you know about spheres?

Volume = (4/3)[pi]r<sup>3</sup>
Surface Area = 4[pi]r<sup>2</sup>

That should do.

Define volume as a function of the radius.

V(r) = (4/3)[pi]r<sup>3</sup>
Find its derivative, dV/dr (or V'(r))
dV/dr = 4[pi]r<sup>2</sup> -- Never mind that this looks like the formula for the surface area. That is NOT a coincidence, but you can ponder it later. Let's stay on task.

dV = 4[pi]r<sup>2</sup>*dr

The problem statement says:
dV = 100pi ft^3/min
r = 5ft
This leave a little algebra to solve for dr.

You tell me how to do the Surface Area part.
 
so i kno that dv/dt =100ft^3/min
so do i plug in the 5 into the original equation or should i take the derivative of the whole formula first then plug in 5 and 100??
 
wow ok i got the first part!! yay
um disregard my question.. let me go on to the next one
THNAK U!!!
 
so when u get the surface area u plug it into 4pi r^2
then u take the derivative of that which is 8pi r dr
so u get 40 pi dr?
 
Wasn't that sweet to make you get dr from the first part of the problem?

I think you have it.
 
sweet thanks
could you help me with this one??
a baseball diamond is a square 90ft an each side. a player runs from first base to second at a rate of 16ft/sec
a) at what rate is the player's distnace form thrid base changing when the player is 30ft from first base
b) at what rates are angles theta and theta 2 changing at that time
 
Come one...draw a right triangle and figure it out. Squares are easier than spheres. :)

Recommendation: Start a new thread when posting an unrelated question.
 
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