Depth in Calculus

Saitenji · 2005-10-14 14:23:44

A conical tank (with vertext down) is 10 ft across the top and 12 feet deep. If water is flowing into the tank at a rate of 10ft³/min, Find the rate of change of the depth of the water when the water is 8ft deep.

...I'm pretty sure it's asking for dh/dt (or dD/dt, OR Rate of change of the depth.) Problem is...I can't seem to figure out what formula to use...
Any help would be very much appreciated. Thanks in advance!

Jai Ganesh · 2005-10-14 16:28:29

Volume of the concial tank is 1/3*pi*r²*h
We know the diameter is 10 feet, therefore, radius is 5 feet.
The height is 12 feet, therefore radius is 5/12*h
V = 1/3*pi*(5/12 *h)²*h
V = 1/3*pi*5/12*h³
V = 5/36*pi*h³
dV/dt = 5/36*pi*3h² dh/dt
We know dV/dt=10 ft³/min
Therefore, when h=8,
10 = 5/12*pi*64*dh/dt
dh/dt = (10*12)/(5*pi*64)
dh/dt= 120/320*pi
dh/dt = 0.1194 ft
I haven't checked for calculation errors!

ryos · 2005-10-14 17:43:10

The formula for the volume of a cone is V = (1/3) * (pi * r²) * h.

The volume of water in the tank is a function of time:
V(t) = 10 (ft³ / min) * t (min).

As the water flows into the tank, it fills a smaller cone in the tip of the larger one, the volume of which is the volume of water that has entered the tank so far. The radius also varies with time, and this will give us trouble unless we do something about it now.

You can think of a cone as a right triangle that has been spun about its height. This triangle will let us find r as a function of h, so we can eliminate r altogether in our original formula.

For the cone given in this problem, r = h * tan[sin-¹(5/12)]. You should probably draw the triangle to convince yourself that this is true.

Ok, now plug that in for radius in our original formula:
V = (1/3) pi * h * {h * tan[sin-¹(5/12)] }²

We need h as a function of time, so start by solving for h:
h = (3V / (pi * tan[sin-¹(5/12)]² ))^(1/3)
Oof.

Remember that V is a function of time, so we can substitute that to get h as a function of time:
h(t) = (30t / (pi * tan[sin-¹(5/12)]² ))^(1/3)

Differentiate this function you must. I don't want to .

Too bad they didn't just give you a time. Instead, you have to find the rate of change of h when h = 8. You can find the time at which this happens by setting h = 8 and solving for t. Plug that t into h'(t), and you should be set.

Last edited by ryos (2005-10-14 17:46:35)

ryos · 2005-10-14 17:55:53

Heh, I took too long, ganesh beat me to it.

I also didn't check for calculation errors. And ganesh's expression for r is a lot simpler than mine. Did I mention that I drag at trig? It's pretty much my least favorite part of math.

He's right, of course. tan(some angle) = O / A, so tan[sin-¹(5/12)]h = (5/12)h. Neat!

ryos · 2005-10-14 18:02:20

Hey ganesh, I was trying to learn from you , and I found a calculation error:

V = 1/3*pi*(5/12 *h)²*h
V = 1/3*pi*5/12*h³

You forgot to square the 5/12.

Also, dh/dt should have units of ft/min. I have no idea what, if anything, went wrong.

Jai Ganesh · 2005-10-14 18:28:15

OMG...
I forgot to square 5/12.
Thats right, I hadn't checked for calculation erros.
Heisenberg's Uncertainity Principle:-
You cannot work at your 100% speed with 100% accuracy

MathsIsFun · 2005-10-14 18:41:58

You'll never catch up with my record of mistakes!

mathsyperson · 2005-10-14 21:28:30

Such as not giving insomnia any boards to mod?
Or was that intentional, for some odd reason?

justlookingforthemoment · 2005-10-14 23:32:34

Well...insomnia said he wanted to be a 'mod'. He never said he wanted to mod.

Math Is Fun Forum

#1 2005-10-14 14:23:44

Depth in Calculus

#2 2005-10-14 16:28:29

Re: Depth in Calculus

#3 2005-10-14 17:43:10

Re: Depth in Calculus

#4 2005-10-14 17:55:53

Re: Depth in Calculus

#5 2005-10-14 18:02:20

Re: Depth in Calculus

#6 2005-10-14 18:28:15

Re: Depth in Calculus

#7 2005-10-14 18:41:58

Re: Depth in Calculus

#8 2005-10-14 21:28:30

Re: Depth in Calculus

#9 2005-10-14 23:32:34

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