Math Is Fun Forum

R.C.

Guest

Rate of change (cone problem)

Water is leaking out of a conical tank (Vertex of the cone pointing down) at a rate of 10,000 cm^3/min at the same time that water is being pumped into the tank at a constant rate. The tank has a height of 6m and the diameter at the top is 4m. If the water level is rising at a rate of 20cm/min when the height of the water is 2m, find the rate at which water is being pumped into the tank.

V=1/3(pi)r^2h

Once I replace r in terms of h and then differentiate both sides I get:

dV/dt=(pi)h^2/9*dh/dt.

Then I plug in the converted height of 200cm and 20cm for dh/dt. Is that correct?

And lastly I add the 10,000cm^3/min to the result of the previous equation?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Rate of change (cone problem)

Hi;

Then I plug in the converted height of 200cm and 20cm for dh/dt. Is that correct?

Your equation for this step is? Need to see it to check it.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

R.C.

Guest

Re: Rate of change (cone problem)

dV/dt= (pi)200^2/9*20.

The answer I got from that equation is 279252 cm^3/min. Then I added the 10,000 cm^3/min to get a rate of
289252 cm^3/min.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Rate of change (cone problem)

Hi;

That answer is correct and so is your dV / dt

Actually though when the answer is rounded it is closer to 289 253 cm^3/ min

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

gnitsuk

Member

Registered: 2006-02-09

Posts: 121

Re: Rate of change (cone problem)

*THIS LINE IS INCORRECT - SEE MY POST BELOW*

Radius a top of cone = 200cm, cone is 600cm high so radius at 200cm high is 200/3 cm. so

so we have

So my answer is

I appreciate I differ in the above expression for dv/dh by a factor of 3. Is it my mistake somewhere?

Last edited by gnitsuk (2011-08-11 01:33:51)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Rate of change (cone problem)

Hi gnitsuk;

I am getting a 9 there. Let me check my notes and see if what I have is okay.

I can not find if you made a mistake, here is what I did back then.

R is the rate that the water is coming in.

Boxed area looks like a chain rule.

Forgot units!

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

gnitsuk

Member

Registered: 2006-02-09

Posts: 121

Re: Rate of change (cone problem)

Quite right. Good fun. The problem is mine.

I write:

And that is true, at any given instant in time for the cone of water.

I next need to calculate

BUT, for the water cone, r is a function of h, in fact r = h/3.

However, I differentiated as if r were a constant wrt h.

I should have substituted in r = h/3 to give:

And THEN differentiated wrt h. Then we of course agree.

Lesson learned (again!).

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Rate of change (cone problem)

Hi gnitsuk;

Thanks for pointing that out. I did not see that either.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

TMorgan

Member

Registered: 2011-04-13

Posts: 25

Re: Rate of change (cone problem)

I got the same answer but there did not need to be any calculus involved. At 2m the level is rising at 20cm/min. So at that instant the amount of volume increase per minute is a cylinder with a radius of 2/3m and height of 0.2m. That volume is (pi)*(2/3)^2*0.2 and throw in a factor of 10^6 to turn m^3 into cm^3. Add the 10,000cc/min that is going out and that's it. Same answer.

This is an algebra/geometry problem all dressed up to look like a calculus problem. Watch out for these, especially when taking a timed test.