Question

Please help, gravel is being dumped from a conveyor belt at a rate of 30 ft^3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 10 feet high?

Answer 1

Volume of cone = 1/12 x pi x d^2 x h
when the height & diameter is 10
=>V = 1/12 x pi x 10 x 10 x 10 = 11000/42 = 261.90 ft^3
In the next minute V = 261.90 + 30 = 291.90 ft^3,let the hight now is h(=d)=>291.90 = 1/12 x 22/7 x h^3
=>h^3 = 1114.55
=>h = 10.37 ft
Increase = 0.37 ft/min
6/(5 x 3.14) = 6/15.7 = 0.38

Answer 2

1/12?

Answer 3

is it multiple choice?

Answer 4

no that is the right answer, but it has to be from calculus standpoint

Answer 5

ooohhh

V = (1/3)πr²h

h = 2r

dV/dt = 30 cubic feet per minute

V = (1/3)πh³/4

V = (1/12)πh³

dV/dt = (3/12)πh²(dh/dt)

dV/dt = (1/4)πh²(dh/dt)

(dV/dt)(4)/πh² = dh/dt

30(4)/π(10)² = dh/dt

120/100π = dh/dt

6/5π = dh/dt

Answer 6

Sorry I'm trying to follow this

Answer 7

Do r would be 5 correct?

Answer 8

yes

Answer 9

And I thought you would have to differentiate h because it's asking for dh/dt so that would be all that's left like h would be gone? Or can't you do that because there asking for the amount at a given time?

Answer 10

@dogzcatz

Answer 11

because of given time I can't really do that

Answer 12

Oh okay

Answer 13

Thanks for help

Answer 14

@zepdrix

Answer 15

I'm just confused where you got the h^3?

Answer 16

@freckles

Answer 17

I'm confused

Answer 18

sup? :O

Answer 19

Confused I need an explanation I'm so bad at Related Rate problems :(

Answer 20

I think your trying to find dh/dt

Answer 21

Are you ok with this diagram?

Answer 22

Yes

Answer 23

Volume of a cone is given by,\[\large\rm V=\frac13 \pi r^2h\]In this problem, since the diameter is the height h, the radial length is always half of that,\[\large\rm V=\frac13 \pi \left(\frac{h}{2}\right)^2h\]

Answer 24

I plugged in 5 not h/2 that might help

Answer 25

Maybe we simplify before we take our derivative,\[\large\rm V=\frac{\pi}{12}h^3\]Ok with that step? :o
Lemme know what you think.

Answer 26

Ughhhhh

Answer 27

No no no, the height is `changing`.
It is varying, a variable.

We can't plug anything in for it before differentiating.
Our derivative measures change, we can't hold h constant.

Answer 28

pi/12? Am I missing something And I'm so bad at that

Answer 29

\[\large\rm \frac{1}{3}\pi \color{orangered}{\left(\frac{h}{2}\right)^2}h\quad=\quad \frac{\pi}{3}\color{orangered}{\frac{h^2}{2^2}}h\]From that point, since multiplication is commutative, we can sort of move things around how we want.
(Keeping denominators as denominators of course).

Answer 30

Ohhhhhhhhh duhhhhhh

Answer 31

We can take a derivative from this point,\[\large\rm V=\frac{\pi}{12}h^3\]Are you ok with doing that step? :)
Power rule and chain rule business ya?

Answer 32

Yay :)

Answer 33

I prefer using the prime notation,
but whatever works for you,\[\large\rm V'=\frac{3\pi}{12}h^2h'\]And this is the information we started with:
\(\rm h=10\)
\(\rm V'=30\)
Solve for \(\rm h'\)

Answer 34

Just a second

Answer 35

Okay

Answer 36

Ya, so plug the stuff in, solve... DO IT! :)

Related rates are awesome sauce,
you gotta keep studying,
they're so great XD

Answer 37

6/5pi

Answer 38

6/(5pi)
good good good :)

Answer 39

Can we do one more? I really want to do well on my text and youn explain really well

Answer 40

Open it in another thread please?
I'll try to come look at it, but I'm really busy right now :D

Answer 41

Sure! It's okay if you don't respond right away, I'll be up a while

Answer 42

k cool c: