Question

Gravel is being dumped from a conveyor belt at a rate of 20 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 14 feet high?

Recall that the volume of a right circular cone with height h and radius of the base r is given by
V=1/3πr^2h


Rate of change of height = feet per minute

Answer 1

@zepdrix

Answer 2

iv finished most of it, but now im not sure where to go with it

Answer 3

this is what i got so far (1/3)pi(2r((1/2)dh/dt)h+r^2dh/dt)

Answer 4

No you don't want to take the derivative while you have 2 variable in there. That's no good.

Answer 5

well thats what we did in class, so we can start over if thats easier for you

Answer 6

Either we're missing some information, or I'm just misunderstanding something perhaps.

If it's a `right circular` cone... does that ... oh oh oh they tell us that the diameter and height are equal, ok that's what I was missing.

Answer 7

brb

Answer 8

\[\large V=\frac{1}{3}\pi r^2 h\]

\[\large r=\frac{1}{2}d=\frac{1}{2}h\]We can determine this since the height and diameter are equal.
Substituting in,\[\large V=\frac{1}{3}\pi\left(\frac{1}{2}h\right)^2h\]

Answer 9

Unlike the last problem, this time, let's make sure we simplify it down before taking a derivative.

Answer 10

We get something like this,
\[\large V=\frac{1}{12}\pi h^3\]

Take a look at those steps, make sure they make sense to you.

Answer 11

yup that makes sense

Answer 12

The givens are:\[\large V'=20\]\[\large h=14\]

Take a derivative, then solve for \(\large h'\)

Answer 13

It will be just a tad different than the previous problem since you'll need to isolate the h' after differentiating.

Answer 14

V=1/12πh^3 so dV/dt=1/4pih^2

Answer 15

You are missing that important term again. :O

Answer 16

sorry which was?

Answer 17

dV/dt=1/4pih^2 dh/dt

Answer 18

oh ya lol my bad

Answer 19

so dh/dt=20/(1/4pi(14)^2)

Answer 20

ya looks good.

Answer 21

thank you so much

Answer 22

no probs c:

Answer 23

now hopefully i can pass my test tomorrow for once lol

Answer 24

heh

Answer 25

have a great night