Question 1163170
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V\ =\ \frac{4}{3}\pi{r^3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dV}{dt}] is given.


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


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


*[tex \Large r_i] for given *[tex \Large V_i]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r_i\ =\ \(\frac{3V_i}{4\pi}\)^{\small{\frac{1}{3}}]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dr}{dt}\|_{r=r_i}\ =\ \frac{dV/dt}{4\pi\(\frac{3V_i}{4\pi}\)^{\small{\frac{2}{3}}}}\ =\ \frac{dV/dt}{\(4\pi\)^{\small{\frac{1}{3}}}\(3V_i\)^{\small{\frac{2}{3}}}}]


You are given *[tex \Large \frac{dV}{dt}] and *[tex \Large V_i]. Plug in the numbers and do the arithmetic.


Note: Volume is measured in cubic centimeters and the radius is measured in centimeters.  The time unit is minutes.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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