Expert: Alon Mandes - 9/30/2008

Question
A spherical balloon is inflated with gas at the rate of 800 cubic centimeters per minute. How fast is the radius of the balloon increasing at the instant the radius is 30 centimeters?

Answer
Let's first write the formula of the volume of the spherical balloon
V=(4/3)πR^3. This means that in each point of time , or in each
minute R(t)=[3V(t)/4π]^(1/3). We know that the the volume is increasing in the fashion of 800t. Hence, we can claim that :
R(t)=[600t/4π]^(1/3) or R(t)=5.75*t^(1/3). This the amount increase
of the radius of our spherical balloon. Very well, now we are interested in the rate of this, or how fast this radius grows, or
in other words, we are interested in finding the derivative of R(t)
at the point where r=30cm. Let's get to work :
Step 1 : Let's calculate the derivative : R'(t)=5.75*[t^(1/3)]'
        R'(t)=1.91*t^(-2/3). "Note that t^(1/3)'=(1/3)t^(-2/3)".
Step 2 : Let's find out, when does the radius gets 30cm ?
        To do so, we use the formula Ro=5.75*to^(1/3) -->
        30=5.75to^(1/3) --> 5.21=to^(1/3) --> to=142 minutes.
Step 3 : Let's calculate the R'(t){at t=to=142} :
        R'(142)=1.91*142^(-2/3)=0.07 cm/min

ALon.