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
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Noo..? This is a maths question.

imranmeah91
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This is what's called related rate problem

Mimi_x3
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rate of change.

imranmeah91
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volume of sphere
= 4/3 Pi r^3

moneybird
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dw:1324353965977:dw

Mimi_x3
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I know, i got stuck, there's not enough info :/

moneybird
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first find the derivative of V first

Mimi_x3
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I already know.
I got stuck in dA/dt = 8pi3*dr/dt
There's not enough info.

joemath314159
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your missing the rate that the radius is increasing.

joemath314159
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youre*

Mimi_x3
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I found that the radius is sqrt(3)

joemath314159
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how did you come about that? there is no volume given

moneybird
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i mean the derivative of volume

imranmeah91
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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

joemath314159
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We would still need r though right? or do you think they want the answer in terms of r?

joemath314159
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boo i already gave you a medal lolol

Hero
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Mimi skipped basic math and went straight to calculus.

Mimi_x3
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Shut up Hero, people don't need to know that i dont know basic maths.
v= s/3r? where did that come from?

imranmeah91
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*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
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ty.