The surface area S (in square meters) of a hot-air balloon is given by
S(r) = 4π r2
where ris the radius of the balloon (in meters). If the radiusris increasing with time t(in seconds)
according to the formula
r(t) = (2/3)t^3
, t ≥ 0, find the surface area S of the balloon as a function of the
time t.

Question

The surface area S (in square meters) of a hot-air balloon is given by
S(r) = 4π r2
where ris the radius of the balloon (in meters). If the radiusris increasing with time t(in seconds) 
according to the formula
r(t) = (2/3)t^3
, t ≥ 0, find the surface area S of the balloon as a function of the 
time t.

e.mccormick · Answer

OK, how far have you gotten with this?

anonymous · Answer

nowhere, I have no clue what to do

e.mccormick · Answer

OK, had any function composition?  Like $(f\circ g)(x)$ stuff?

anonymous · Answer

Nope, this is the first assignment from this chapter. I take math online so im pretty freaking confused.

e.mccormick · Answer

AAARGH!  I had it mostly typed and then it booted me.  Welp, time to try again.

anonymous · Answer

aw thank you!

e.mccormick · Answer

OK.  I was going to go over composition really fast.

Lets say I have $f(x)=x-1$ and $g(x)=(x+4)^2$.  Then $(f\circ g)(x)$ means I plug the equation for g into the equation for f.  It becomes this:$$f(g(x))=((x+4)^2)-1$$Your question is similar to this.

e.mccormick · Answer

Typed it in another window...  just in case.

anonymous · Answer

Oh ok I vaguely remember doing that in high school

e.mccormick · Answer

Now you have:
$S(r) = 4\pi r^2$ and are told r is radius.
Then they do a second formula for radius:
$r(t) = (2/3)t^3$

What they want is $(S\circ r)(t)$ which is also called $S(r(t))$

anonymous · Answer

would it look like 4pi ((2/3)t^3))^2 ?

e.mccormick · Answer

Yes, and you might do a little distribution. $$(x^m)^n=x^{m\times n}$$