Question

Find the rate at which the volume of the sphere is changing when the radius of the sphere is 10cm and surface area is changing at Pi cm^2/min Find the rate at which the volume of the sphere is changing when the radius of the sphere is 10cm and surface area is changing at Pi cm^2/min @Mathematics

Answer 1

This is what i did (is it right?)
v=4/3 pi r^2
dv/dt = 4/3Pi*3r^2 dr/dt
dv/dt = 4Pir^2*dr/dt

Answer 2

i got 400Pi^2 cm^4/min

Answer 3

Nevermind, I take that back. Here:

Answer 4

\[V=\frac{4}{3} \pi r^3\]
\[SA=4 \pi r^2\]
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we want to find V' where r=10 and SA'=pi

Answer 5

\[ V= \frac{4}{3} \pi r^3 \]
\[\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \]
but since
\[ A = 4\pi r^2 \rightarrow \frac{dA}{dt} = 8 \pi r \frac{dr}{dt}\]
which means
\[\frac{dr}{dt} = \frac{1}{8\pi r}\frac{dA}{dt} \]
and so
\[\frac{dV}{dt} = 4 \pi r^2 \frac{1}{8 \pi r} \frac{dA}{dt} = \frac{r}{2} \frac{dA}{dt} = \frac{10cm}{2}\pi \space cm^2/min=5\pi\space cm^3/min\]

Answer 6

\[V'=4 \pi r^2 r'=SA r'\]
\[SA'=8 \pi r r'\]
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\[V'_{r=10}=400 \pi r'=SA r'\]
\[SA'_{r=10}=80 \pi r'=\pi => r'=\frac{1}{80} \]
\[V'_{r=10}=SAr'=400 \pi \cdot \frac{1}{80} =5 \pi\]
oh i'm slow

Answer 7

thanks a lot friends! i wish i can buy you guys a beer =)