Question

Gravel is being dumped from a conveyor belt at a rate of 30ft^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 10ft high?

Answer 1

Given:\[\frac{ dv }{ dt }=30ft ^{3}/\min\]
Reequired:\[\frac{ dh }{ dt }|_{h = 10}\]First write a related equation. That is an equation which relates the variable quantities of the problem.

Answer 2

@KC33 what is your related equation here?

Answer 3

I'm not sure what the related equation would be. I know that we have to find dV/dt but I'm not sure how to set it up.

Answer 4

Well, we have a base, a radius, and volume of a cone. Thus the related equation for this problem would be the equation for the volume of a cone.

Answer 5

which is V=1/3pi r^2h

Answer 6

so at this point I'm not sure if we go substitute the values of the cone to get the Volume and take the derivative

Answer 7

Right\[V =\frac{ 1 }{ 3 }\pi r ^{2}h\]Now the first rule (and this is important!!) in solving related rate problems is to remember that the number of variables in your related equation must equal to the total number of rates given and required in the problem.

@KC33 how many variables do we have in our related equation and how many total rates were given and required.

Answer 8

we have 2 variables in the related equation which is the r and h but only one rate of 30ft^3/min.

Answer 9

NO. We have three variables V, r, and h in our related equation and a total of two rates. One given (dV/dt) and one required (dh/dt). Understand?

Answer 10

ok but even so, the rates that's listed doesn't match the variables that you said was very important.

Answer 11

Yes so that means that we need to replace one of the variables. Which one. Well let's see. We were given dV/dt so we have to keep V. Similarly we were required to find dh/dt so we need to keed h. That means that r is the nuisance variable. So what do we do with it? Well remember in the question we are told that the base diameter is equal to the height? OK so then this implies that 2r = h or r = h/2. So now we replace r with h/2 in the equaton. So now rewrite the related equation in terms of V and h only. Understand @KC33 ??

Answer 12

so the equation would be dv/dt = pi/12(3h^2 dh/dt)

Answer 13

when you pull out the constant, it'll be 3(pih^2/12 dh/dt) = pih^2/4 dh/dt

Answer 14

Yes\[\frac{ dV }{ dt }=\frac{ 3\pi }{ 12 }h ^{2}\frac{ dh }{ dt }\]OR\[\frac{ dV }{ dt }=\frac{ \pi }{ 4 }h ^{2}\frac{ dh }{ dt }\]Understand?

Answer 15

Now find dh/dt when dv/dt = 30 and h = 10.

Answer 16

@KC33 are you there?? Do you understand??

Answer 17

yep.. I'm here and I believe I understand.

Answer 18

OK so then can you give me the exact answer for dh/dt in terms of pi?

Answer 19

I'm getting 6/5pi ft^3/min

Answer 20

Very good!! From now on remember that before you differentiate the related equation, the number of variables in your equation must equal to the number of rates given and sought.

Answer 21

:) thanks so much for your help!

Answer 22

Anytime! Anything else??

Answer 23

nope that's it. Do you tutor here a lot? Can you be requested?

Answer 24

Oh just one thing. I just noticed that in your final answer you wrote ft cube but it is just ft.

Answer 25

ok

Answer 26

Yes. I'm a teacher who loves math plus teaching so whenever I have some free time, I sign in.

Answer 27

ok.. thank you very much.

Answer 28

Welcome. Bye for now.