Question

Consider a spherical cow that is consuming a great deal of hay. He is fattening himself up at a rate of 200 liters/day, so how quickly is her radius growing (in cm/da) when he can't fit through the circular stall door (that's 2 meters wide)?

Answer 1

I can help you with the first part.
\[V(sphere) = (4/3)\pi r^{3}\]

Answer 2

We're given dV/dt = 200 L/day; This problem is basically saying at what rate is the radius, r, increasing.

Answer 3

Wait...
200 = 4(pI)(r)^2(dv/dt) right?

Answer 4

I think (dr/dt) will be on the right side instead, because we're deriving the function with respect to t. Man, it's a little hazy to me. I hope someone else comes around to help out.

Answer 5

(dV/dt) = 4pi*r^2*(dr/dt)

Answer 6

200/40,000(pi) = 4(pi)(100)^2(dr/dt)?

Answer 7

Ohhh, I know what I'm doing now. Let's forget about the derivative right now. We do need both parts of the problem to figure this out.

Answer 8

Okay.

Answer 9

So, the circular door is 2m wide. Therefore, when the cow's radius is over 1m, she can't fit through the door! We'll have to find the volume of the cow at 1m.

Answer 10

We do need to do some conversions in the units though. The problem gives dV/dt in L/day, and they want dr/dt in cm/day. It's alright though.
1m * (100cm/1m) = 100 cm is the maximum radius of the cow.
When we find the volume of the cow at 100 cm, units will be in cm^3.

Answer 11

For dV/dt = 200 L/day = (200 L/day)(1000 mL/L)(1 cm^3 / mL)

Answer 12

So, dV/dt = 200,000 cm^3/day

Answer 13

NOW, we can find dr/dt using our derivative.
dV/dt = 4*pi*r^2*(dr/dt)

Answer 14

Solve for dr/dt, plug in the numbers and boom. The units will be in cm/day

Answer 15

Is it 1/200(pi)?

Answer 16

Sounds like I'm ranting, it takes alot of words to write out related rates problems.
No, I got 5/pi

Answer 17

Use dV/dt value that has units cm^3/day & use the radius value that is in cm.

Answer 18

Remember it's 0.02 m right?

Answer 19

No, that's the diameter.

Answer 20

Okay, so it should be 5/pi?

Answer 21

Should be...

Answer 22

Okay, thanks.