so confused. please help.

Question

anonymous · Answer

A high-altitude spherical weather balloon expands as it rises due to the drop in atmospheric pressure. Suppose that the radius r increases at the rate of 0.16 inches per second and that r = 38 inches at time t = 0. Determine the equation that models the volume V of the balloon at time t and find the volume when t = 280 seconds.

whpalmer4 · Answer

What class are you taking?

anonymous · Answer

precalc

whpalmer4 · Answer

Okay, then I won't use calculus to find the answer :-)

whpalmer4 · Answer

Do you know the formula for the volume of the sphere?

anonymous · Answer

no that is why I am having trouble because I cannot find the equation

whpalmer4 · Answer

Okay.  if memory fails, googling "formula volume of a sphere" should get you an answer, but I'll tell it to you now to save time.

$$V_{sphere} = \frac{4}{3}\pi r^3$$

whpalmer4 · Answer

Now we know that r = 38 inches at t = 0, and that r is increasing by 0.16 inches/second.  In calculus, we would write the latter quantity as $$\frac{dr}{dt} = 0.16 inch/s$$which just is a way of saying that the rate of change of r with respect to t is 0.16 inch/s.

whpalmer4 · Answer

So how would you write a formula for r as a function of t?  

t   r
0  38
10 38+(0.16*10)
20 38+(0.16*20)
etc.
?

anonymous · Answer

$$V=\frac{ 4 }{ 3 }\pi(38)^3$$

anonymous · Answer

no it would be $$V=\frac{ 4\pi(38+.16t)^3 }{ 3 }$$

whpalmer4 · Answer

Yeah, that looks better...

whpalmer4 · Answer

What do you get for your answer?

anonymous · Answer

so in fact it would be 6966.303506

whpalmer4 · Answer

Hmm, that's not what I get.  4/3 * pi is roughly 4.  .16+280 is roughly 45.  45 + 38 is 83, call it 85.  85^3 is a bit more than 600,000, so, my back of the envelope estimate is about 2.4 million.

anonymous · Answer

V(t) =2,377,823.53 in.3

whpalmer4 · Answer

Yep!

anonymous · Answer

2,377,823.53 in.^3

anonymous · Answer

thank you for your help!

whpalmer4 · Answer

a typical calculus problem would have a slightly different twist: you would know the rate at which the volume was changing, and be asked to find the rate of change of the radius at the time that it equalled some number.

whpalmer4 · Answer

something to look forward to :-)

whpalmer4 · Answer

something to look forward to :-)

whpalmer4 · Answer

or, you'd be asked to find the rate of change of the surface area of the sphere...

whpalmer4 · Answer

you'll have many opportunities to remember the formula for the volume and surface area, and you'll also learn how to derive again for when you forget :-)