Calculus Please help!!!!
Consider a spherical cow (!) that is consuming a GREAT deal of hay. She is fattening herself up at a rate of 200 liters/day. How quickly is her radius growing (in cm/day) when she can't fit through the circular stall door (which is 2 meters wide)?

Question

Calculus Please help!!!!
Consider a spherical cow (!) that is consuming a GREAT deal of hay. She is fattening herself up at a rate of 200 liters/day. How quickly is her radius growing (in cm/day) when she can't fit through the circular stall door (which is 2 meters wide)?

kainui · Answer

Lol ok so what have you figured out so far?

anonymous · Answer

I'm so frustrated! I've been stuck on this question for a while.  
V = 4/3 Pi r^3

kainui · Answer

Ok awesome, so you have the volume of the cow related to the radius of the cow which is great, now the strategy I find that works best for these is to identify what you're given and list them out. So here they say:

"She is fattening herself up at a rate of 200 liters/day."

What this really means to me is that the volume is changing over time, in symbols I'll write this:

$$\frac{dV}{dt} = 200 \frac{Liters}{day}$$

(I keep the units cause I notice they're using cm somewhere else in the problem so I'll have to convert... ugh... lol)

So ok, if you're unfamiliar with this dV/dt or something else I've said, go ahead and ask, and while you're at it try to see if you can write anything else in terms of symbols.

Also you have $V=\tfrac{4}{3} \pi r^3$, and the problem gives us how the cow's volume changes over time, $\tfrac{dV}{dt}$, so see if you can get the derivative of this with respect to time.

anonymous · Answer

wait what do i find the derivative with the respect of time of?  which equation? Thank you for doing this!

kainui · Answer

Yeah specifically this equation here:

$$V=\tfrac{4}{3} \pi r^3$$

This might better be written as:


$$V(t)=\tfrac{4}{3} \pi [r(t)]^3$$

Since the Volume of the cow depends on time V(t) and the radius of the cow depends on time r(t).

This is what we'll need to take the derivative of, since this will tell us how the change of the volume in time relates to the change of radius in time, since ultimately what we wanna get is: "How quickly is her radius growing (in cm/day)" which is in terms of symbols, $\tfrac{dr}{dt}$ or we could just as well write $r'(t)$. Whichever notation you like better, they're both the same.

anonymous · Answer

so then it would be $$4πr^3t^2$$ right?