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\large 
\underline
{MA1161 \hspace{5mm}Calculus I with Technology \hspace{50mm} Ch. 11-6, ex. 16}


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Class,  

Here is the correct solution to problem 16 I attempted to try in class.  Serves me right for not drawing a picture.

The questions gives us the information that the change in the volume with respect to time is proportional to the surface area.  Therefore, we have the equation:

$$\frac{dV}{dt}=-k(4\pi r^2)$$

We also know from implicit differentiation that the following equation is true:

$$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$$

This means the following equation is true:

$$\frac{dr}{dt}=-k$$

We know that $k$ equals $\frac{1}{2}$ $cm/month$ based on the information given in the problem. Now we have a differential equation for the radius with respect to time, and solving the equation with a starting value of 1:

$$r=1-\frac{t}{2}$$

Solving this equation with the value $r=.2$ gives us the solution of $t=1.6$.
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