AP Calc AB question Here is the question: The lollipop is in the shape of a sphere where the radius is in centimeters. The equation of the circle is x^2 + y^2 = 9. When the circle is rotated about the x-axis, a sphere is formed. The volume of the lollipop decreases at .05 mm^3/lick and the average rate of licking is 10 licks/minute. 


*

*At what rate is the radius of the sphere decreasing after 5000 licks? 10000 licks? 

*After 30 mins, what is the rate of change of the volume? 

*After how many licks is the rate of change of the Surface Area = rate of change of the Volume?

 A: Let $V(t)$ denote the volume of the portion of the lollipop remaining after $t$ minutes given in cubic centimeters. Since the lollipop is licked $10$ times per minute, we can approximate $V(t)$ by
$$
    V(t) = [\text{Initial Volume}] - .005t = 36\pi - (.005)(10)t = 36\pi - .05t.
$$
(Why does this make sense?)
For part (1), you want to determine $\frac{dr}{dt}$ at $t = 500$ and $t = 1000$. By the formula $V = (4/3)\pi r^3$, show that
$$
    \frac{dr}{dt} = \frac{1}{4 \pi r^2}\frac{dV}{dt}.
$$
You can approximate $dV/dt$ using the change in volume per lick and average rate of licking they provided. (Be sure to approximate it with change in volume per minute not per lick and that you're consistently using cm or mm.) Note that the problem seems to be implicitly assuming that the lollipop remains a sphere after each lick (so that the notion of a radius $r$ even makes sense). Hence after $t$ minutes, we can assume the lollipop is still spherical with radius $r_t$ cm and volume
$$
    \frac{4}{3}\pi r_t^3 = 36\pi - (.005)(10)t
$$
Thus
$$
    r_t = \left(\frac{108\pi - .05t}{4}\right)^{1/3}.
$$
So to finish (1), you need to figure out what $t$ corresponds to $500$ licks and $1000$ licks, compute $r_t$ at these values, and plug into the above formula for $dr/dt$.
(2) is the quickest of the three parts. You're actually given the answer, but there's something you need to check first: is there any lollipop leftover after 30 minutes of licking?
For (3), use related rates in a similar fashion to what you did for part (1) together with the formula for the surface area of a sphere with radius $r$, i.e., the formula $SA(r) = 4\pi r^2$.
