Air is pumped into a spherical balloon at a rate of 8cm^3 per second. At what rate is the radius of the sphere changing when its volume is 36 pie cm^3?

Question

anonymous · Answer

related rate problem right?  hold on a second i am in the middle of another problem, but the idea is to write the equation for the volume of the sphere in terms of the radius, differentiate wrt r and then substitute 8 for $V'$ and 36 for $V$

anonymous · Answer

you got the formula for the volume of the sphere?

anonymous · Answer

yes it is v=(4/3)[pi\]r^3

anonymous · Answer

$$V=\frac{2}{3}\pi r^3$$ take the derivative of both sides wrt $t$ and get 
$$V'=2\pi r^2 r'$$ now make the replacements and solve for $r'$

anonymous · Answer

ok that was wrong, it is 
$$V=\frac{4}{3}\pi r^3$$so 
$$V'=4\pi r^2 r'$$

anonymous · Answer

so i will plug in 8 and get 8=4[pi \]r^2r'

anonymous · Answer

then i would solve for r' but how would i find the radius?

anonymous · Answer

you know that 
$$V=\frac{4}{3}\pi r^3$$and you are asked for $r'$ when $V=36$ so solve $36=\frac{4}{3}\pi r^3$ for $r$

anonymous · Answer

okay thank you. the answer would be 2/9pi