- #1
KingNothing
- 880
- 4
Hi. I am getting absolutely embarassed by these related rates problems. Here is one that I simply keep getting wrong:
The volume of an expanding sphere is increasing at a rate of 12 cubed cm per second. When the volume is 36\pi, how fast is the surface area increasing?
V=\frac {4*pi*r^3}{3}S=4*pi*r^2
(how the heck do you use pi in latex? I know it's \pi, but that doesn't work right when Iput it in!)
\frac {dV}{dt}=4 \cdot \pi \cdot r^2 \cdot \frac {dr}{dt}
Since volume is 36 \cdot \pi, r=3. Correct?
The volume of an expanding sphere is increasing at a rate of 12 cubed cm per second. When the volume is 36\pi, how fast is the surface area increasing?
V=\frac {4*pi*r^3}{3}S=4*pi*r^2
(how the heck do you use pi in latex? I know it's \pi, but that doesn't work right when Iput it in!)
\frac {dV}{dt}=4 \cdot \pi \cdot r^2 \cdot \frac {dr}{dt}
Since volume is 36 \cdot \pi, r=3. Correct?
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