Question 1163173
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V(r)\ =\ \frac{4}{3}\pi{r^3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dV}{dt}\ =\ 120\,\frac{\text{cm}^3}{\text{min}}]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dV}{dt}\ =\ 4\pi{r}^2\,\frac{dr}{dt}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dr}{dt}\ =\ \frac{\frac{dV}{dt}}{4\pi{r}^2}\ =\ \frac{120\,\frac{\text{cm}^3}{\text{min}}}{4\pi{r}^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dr}{dt}\|_{r=15}\ =\ \frac{120\,\frac{\text{cm}^3}{\text{min}}}{4\pi(15)^2]


You can do this arithmetic (leave in terms of *[tex \Large \pi])


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dA}{dt}\ =\ 8\pi{r}\,\frac{dr}{dt}]


Plug in 15 for *[tex \Large r] and the value for *[tex \Large \frac{dr}{dt}] calculated above to find *[tex \Large \frac{dA}{dt}] at *[tex \Large r\ =\ 15]


Note:  *[tex \Large \frac{dr}{dt}] is in units of *[tex \Large \frac{\text{cm}}{\text{min}}] and *[tex \Large \frac{dA}{dt}] is in units of *[tex \Large \frac{\text{cm}^2}{\text{min}}]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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