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- Mimi_x3

The volume of a sphere radius r increases at 12pi cm^3/minute by inflation. What is the rate of growth of its surface area?

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- Mimi_x3

- chestercat

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- Mimi_x3

Noo..? This is a maths question.

- anonymous

This is what's called related rate problem

- Mimi_x3

rate of change.

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- anonymous

volume of sphere
= 4/3 Pi r^3

- anonymous

|dw:1324353965977:dw|

- Mimi_x3

I know, i got stuck, there's not enough info :/

- anonymous

first find the derivative of V first

- Mimi_x3

I already know.
I got stuck in dA/dt = 8pi3*dr/dt
There's not enough info.

- anonymous

your missing the rate that the radius is increasing.

- anonymous

youre*

- Mimi_x3

I found that the radius is sqrt(3)

- anonymous

how did you come about that? there is no volume given

- anonymous

i mean the derivative of volume

- anonymous

write equation relating volume to surface area
v= 4/3 Pi r^3
S= 4 Pi r^2
v= s/3 r
dv/ds= 1/3r
dv= 1/3 r ds

- anonymous

We would still need r though right? or do you think they want the answer in terms of r?

- anonymous

boo i already gave you a medal lolol

- Hero

Mimi skipped basic math and went straight to calculus.

- Mimi_x3

Shut up Hero, people don't need to know that i dont know basic maths.
v= s/3r? where did that come from?

- anonymous

*minor correction
v= 4/3 Pi r^3
S= 4 Pi r^2
v= (s r)/3
dv/ds= r/3
dv= r/3 ds
(12*3 Pi)/r= ds

- Mimi_x3

ty.

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