$(a)$ Consider $an$ investment that doubles $in$ value $if$ the interest rate $is\frac{?}{(5\frac{?}{?}\%\frac{?}{?})}$ compounded continuously. Let $\frac{?}{(A_0\frac{?}{?})}$ represent the amount $of$ the initial investment and let $\frac{?}{(t\frac{?}{?})}$ represent time $in$ years. Write $an$ equation that can $be$ solved for $\frac{?}{(t\frac{?}{?})}to$ determine how long $it$ will take the investment $to$ double $in$ value. How many years will $it$ take $an$ investment $to$ double $in$ value $if$ the interest rate $is\frac{5}{\%}$ compounded continuously? $(Round$ your answer $to$ two decimal places.$)(b)$ What $is$ the equivalent annual interest rate $(in$ percent$)?(Round$ your answer $to$ two decimal places.$)\frac{?}{?}\%$

Suppose that the volume $Vof$ a rolling snowball increases $so$ that $\frac{dV}{dt}is$ proportional $to$ the surface area $of$ the snowball $at$ time $t.$ Show that the radius $r$ increases $at$ a constant

rate, that $is,\frac{dr}{dt}is$ constant.

The volume $of$ the snowball $is$ given $byV=\frac{4}{3}{r}^3,so\frac{dV}{dt}=\frac{4}{3}*()\frac{dr}{dt}=4r^2\frac{dr}{dt}.$

Since the change $in$ volume $is$ proportional $to$ the surface area $S,$ with $S=4r^2,we$ also have the following for some constant $k.$

$\frac{dV}{dt}=k*r^3$

$\frac{dV}{dt}=r*4k^2$

$\frac{dV}{dk}=k*2r$

$\frac{dV}{dt}=k*4r^2$

$\frac{dV}{dt}=r*k^3$

Equating the two expressions for $\frac{dV}{dt}$ gives $4r^2\frac{dr}{dt}=k*()=>\frac{dr}{dt}=k_r$ that $is,\frac{dr}{dt}is$ constant.