# Water flowing out of a conical tank

member 392791

## The Attempt at a Solution

For this problem, it reminds me of related rates problems, except I'm not given a constraint such as dr/dt = something at r = something.

I tried using partial derivatives to solve it, but I'm not seeing a way to get rid of the r terms, or its derivative

#### Attachments

• 2.9 attempt 1.pdf
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• problem 2.9.pdf
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SteamKing
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Since you know the dimensions of the conical tank, you can eliminate 'h' as a separate variable by making it a function of 'r'. This will simplify your calculations.

haruspex
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Gold Member
Use geometry to find the relationship between r and h.

SteamKing
Staff Emeritus
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I'm going to amend my earlier post and say you should eliminate 'r' instead of 'h'.

member 392791
Do you mean to say that

r=√3V/(hπ)?? Because then V is reintroduced into the equation. Or is there another geometrical relationship that I am missing that will remove both V and r?

or rather plugging in the initial values given

V = (1/3)πr^2h

r = 3 and h =5

(1/3)π(3)^2(5) = 47.12

47.12 = (1/3)πr^2h
45 = (r^2)(h)
r = √(45/h)

If this is the case, what is dr/dt? Is it just √(45/h) dh/dt?

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SteamKing
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Nope. Look at the tank itself. Imagine that the cone is closed at the bottom, for argument's sake. At the bottom, h = 0, and r = 0 (essentially). At the top of the tank, h = 5 m, and r = 3 m. Don't you think you can find a relationship for 'r' in terms of 'h' using these two pieces of information?

verty
Homework Helper
Woopydalan, it looks like you are heading towards having a differential equation to solve. There is a better way, although it takes some ingenuity to see. If you know your Fundamental Theorem, you should be able to find a more direct method.

member 392791
Nope. Look at the tank itself. Imagine that the cone is closed at the bottom, for argument's sake. At the bottom, h = 0, and r = 0 (essentially). At the top of the tank, h = 5 m, and r = 3 m. Don't you think you can find a relationship for 'r' in terms of 'h' using these two pieces of information?

r = 3/5h?

would that mean dr/dt = 3/5h dh/dt or is it 3/5 dh/dt? I think its the latter, but not sure. I'm rusty with the math, summer came and went =(

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member 392791
Woopydalan, it looks like you are heading towards having a differential equation to solve. There is a better way, although it takes some ingenuity to see. If you know your Fundamental Theorem, you should be able to find a more direct method.

Which Fundamental Theorem?

member 392791
Here is my 2nd attempt, I'm interested in the method that doesn't rely on a differential equation, but is this correct?

#### Attachments

• 2.9 attempt 2.pdf
260 KB · Views: 195
SteamKing
Staff Emeritus
Homework Helper
Here is my 2nd attempt, I'm interested in the method that doesn't rely on a differential equation, but is this correct?

It seems OK. The original rate of flow out of the tank was given in units of m^3/min. Where did you convert this to m^3/s? Otherwise, the time for the outflow will be in minutes rather than seconds.

• 1 person
member 392791
Upon further inspection, in my calculator I had it in degree mode, and in actuality the integral evaluates as -1.64 if in radian mode, which means time is negative, which shouldn't be the case.

Did something go wrong in my solving for the differential equation?

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member 392791
bumping, this problem is still incomplete as far as I know. Anyone have some clues for other avenues I should look or where in my analysis I have gone wrong?

verty
Homework Helper
I think, because ##dh## was negative, the left hand side should be negative.

member 392791
I mean shouldn't doing the integral from 5 -> 3.15 m take care of that though?