Question 1165072
<font face="Times New Roman" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  A(r)\ =\ 4\pi{r^2}]


But *[tex \Large r] is a function of time, *[tex \Large r(t)].  We are given the rate of change of *[tex \Large r], namely 2 cm/s.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ dr\ =\ 2\,dt]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int\,dr\ =\ \int\,2\,dt]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r(t)\ =\ 2t\ +\ c]


where the constant of integration, *[tex \Large c], represents the initial the initial radius, so *[tex \Large c\ =\ r_o]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r(t)\ =\ 2t\ +\ r_o]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  A(t)\ =\ 4\pi(2t\,+\,r_o)^2\,+\,c\ =\ 4\pi\(4t^2\,+\,4tr_o\,+\,r_o^2\)]


																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
</font>