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. How quickly is her radius growing when she can't fit through the circular stall door (which is 2 m wide)?

Answer 1

@sourwing

Answer 2

lol spherical cow

Answer 3

teacher's crazy

Answer 4

V = (4pi/3)r^3

Answer 5

So, )4pi/3)(200)^3 right>?

Answer 6

i got 3,200,000pi

Answer 7

@SithsAndGiggles sourwing disappeared on me

Answer 8

\[V=\frac{4}{3}\pi r^3~~\Rightarrow~~\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\]
Given: \(\dfrac{dV}{dt}=200\)

The cow won't be able to fit through the door when its diameter matches the width of the door: \(d=2r=2\text{m}~~\Rightarrow~~r=1\text{m}\).

Keep in mind that \(1\text{L}=0.001\text{m}^3\).
\[200\text{L}=0.2\text{m}^3=4\pi (1\text{m})^2\frac{dr}{dt}~~\Rightarrow~~\frac{dr}{dt}=\cdots\]

Answer 9

so, the derivative of 4pi(1m)^2 right?

Answer 10

No, we already found the derivatives we need. All you need to do is solve for \(\dfrac{dr}{dt}\).

Answer 11

k i sec

Answer 12

4pi?

Answer 13

Not quite, check your algebra:
\[\frac{dr}{dt}=\frac{0.2}{4\pi}\frac{\text{m}}{\text{s}}=\cdots\]

Answer 14

It's a big number. i can't get fraction form. it's like 0.0159...

Answer 15

You mean the answer is supposed to be a big number?

Answer 16

0.015915494309?

Answer 17

Yeah, that's what I got.

Answer 18

so, there's no fraction form right?

Answer 19

You could leave it as \(\dfrac{0.2}{4\pi}\), which is equal to \(\dfrac{1/5}{4\pi}=\dfrac{1}{20\pi}\)

Answer 20

Yes! I see :D