# Related rates: water level in a cone

rxh140630
Homework Statement:
Given a water tank in the shape of a right circular cone with radius of the base 4 feet and altitude of 10 feet. Water is added to the tank at a constant rate of 5 cubic feet per minute. Find the rate at which the water level is rising when the depth of the water is 5 feet when:
Relevant Equations:
V=(1/3)pir^2h
Firstly I'm having trouble understanding what water level means.

I tried a quick google search and got the following: " water level(Noun) The level of a body of water, especially when measured above a datum line. "

That doesn't help me. Is water level the distance from the base to where the water ends from the line that's perpendicular to the center of the base, or is it the line from the outer part of the base following the container from the outside to where the water ends?

Homework Helper
It's the vertical distance from the base to the surface of the water.

• rxh140630
rxh140630
It's the vertical distance from the base to the surface of the water.

I figured. This answer makes sense: it would be hard(impossible?) to find a relation if it was the other example (from outer part of base following the container to the water level,) because, we are given the change in volume.

I'm working my calculations right now, I'll be back with an answer or another question shortly.

rxh140630
So the point of this problem is to find r when the water level is 5 ft. that's what I am trying to do now

Mentor
Draw a picture of the tank. I'm assuming that the conical tank is oriented with the narrow part downward. With this orientation, the vertex (bottom) is at the point (0, 0), and the right-most point of the rim will be at the point (4, 10). Find a relationship between x and y for the sloped side of the tank.

Your textbook is likely to have a similar problem that you can draw on to help you with the rest.

rxh140630
So a cone is the object of a right triangle that is revolved once?

rxh140630
At this point I don't care if it takes me a year to do this problem, I will do it (by myself.)

Mayhem
A 10 second google search revealed this: It's not your problem exactly, but the same type. The water is the height ##h##.

So yes, it's a right triangle that has been revolved once. Do you know how to solve the rest?

rxh140630
r=2 @ h=5

Mayhem
r=2 @ h=5

rxh140630
let the pointed end of the cone be situated at (0,0) so one point of its cross section is located at (4,10)

then slope = 5/2

y=5x/2

when y = 5, x=2

x is the radius of the cone when the water level is at height 5.

Mayhem
let the pointed end of the cone be situated at (0,0) so one point of its cross section is located at (4,10)

then slope = 5/2

y=5x/2

when y = 5, x=2

x is the radius of the cone when the water level is at height 5.
Is that what you were asked to find, though?

rxh140630
Is that what you were asked to find, though?
no, I was just giving an update that I did find the radius

• Mayhem
Mentor
So a cone is the object of a right triangle that is revolved once?
no, I was just giving an update that I did find the radius
That's not what the problem is asking for.
Find the rate at which the water level is rising when the depth of the water is 5 feet
IOW, you are asked to find dh/dt at the moment when the water is 5 ft deep in the cone. You'll need to use some calculus to find dh/dt.

Based on your replies, above, you really should take a look at some worked examples in your textbook.

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rxh140630
I understand that's not what the problem is asking for, I just thought that I might need r=2 at h=5 for the problem.

I understand I'm looking for dH/dT but there has to be some trick. I came to the same formula using two different methods but it's the wrong formula.

$V_{cone}=\frac13{\pi}r^2h$

we're given dV/dT= 5 ft^3/min

$dV=5dT$
$v=5t$

$5t=\frac13{\pi}r^2h$
...
$dH/dT=15{\pi}^{-1}r^{-2}$

but this is the wrong formula.

My logic is wrong somewhere but I got the same result another way:

$\frac{dH}{dT}=\frac{dH}{dV}\frac{dV}{dT}$

dv/dt is given, = 5

by solving:$V_{cone}=\frac13{\pi}r^2h$

for h..

$dH/dV = 3{\pi^1}r^{-2}$

I don't see where my logic is wrong

rxh140630
I think it has to do since there are 3 things and they all depend on each other (radius, height, and volume) I'm missing one of them in my equation.

rxh140630
Okay, I got the same answer as in my book. I'll be editing this post and doing a write up as to how I got it. Making this post in case someone happens to stop by this thread, so they don't waste their time.

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Firstly, I'm writing this post to further my own learning and to help me cement these comments, and maybe the users here will also be able to write their own comments in regards to my thought process or anything else. I'm also writing this for anyone

Where do I start? Well, my book didn't really give me a proper explanation on related rates (Apostol's Calculus volume. The author lumped related rates into the section on the Chain Rule, and only gave one example that was not similar to this one.

Originally what I did was:

$$\frac{dH}{dT} = \frac{dH}{dV} \frac{dV}{dT}$$

where $\frac{dH}{dV} = 3v{\pi^{-1}}r^{-2}$

and I merely plugged in the value you would get for r at a height of 5, r=2, then multiply
$\frac{dH}{dV}$ and $\frac{dV}{dT}$

This is wrong. r is not some fixed value, it is related to the volume and height. I think the author wanted me to come to this realization through problems without explaining (though he would have saved me a lot of time.) Essentially what I did in the above example in $\frac{dH}{dV}$ is say that h changes with respect to v, with a fixed value of r. R is NOT fixed!

So I realized that we have to take into account everything that is related, and the following equation gave me the solution:

$\frac{dH}{dt}=\frac{dH}{dR}\frac{dR}{dV}\frac{dV}{dT}$

using h=(5/2)r

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• Mayhem
Mentor
I think it has to do since there are 3 things and they all depend on each other (radius, height, and volume) I'm missing one of them in my equation.
The radius and height are related by the equation you found in post #11. In modified form, to suit this problem, h = (5/2)r, or equivalently, r = (2/5)h
In your volume formula, you can replace r by (2/5)h, which will give you V in terms of h alone.

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• rxh140630
rxh140630
The radius and height are related by the equation you found in post #11. In modified form, to suit this problem, h = (5/2)r, or equivalently, r = (2/5)h
In your volume formula, you can replace r by (2/5)h, which will give you V in terms of h alone.

Yes I understand now. We have to take into account everything that changes in the problem.

Mayhem
Yes I understand now. We have to take into account everything that changes in the problem.
When I was learning related rates, explicitly writing out which were functions of time at the beginning of the problem helped me maintain focus.

• rxh140630