View Full Version : Relatively easy math problem...

10-17-2005, 05:16 PM
We're learning related rates, and I'm still trying to completely understand them, so I'm not sure how correct my answer to this question is, and I just wanted some help on it.

A container has the shape of an open right circular cone. The height of the cone is 10 cm and the diameter of the opening is 10 cm. Water in the container is evaporating so that its depth h is changing at the constant rate of -3/10 cm/hr.

Find the rate of change of the volume of water in the container with respect to time, when h = 5 cm.

How I think it should be done:

let h = depth of water in container
let r = radius of surface of water
V(cone) = 1/3(pi)(r^2)(h)

h/r = 10/5 = 2, so r = h/2

Since I'm trying to relate dh/dt and dV/dt, I solved for r and will plug that into the formula for volume of a cone.

V = 1/3(pi)[(h^2)/4](h)
V = [(pi)(h^3)]/12

I'll differentiate that:

dV/dt = [2h^2(pi)/3](dh/dt)

Substitute 5 cm for h and -3/10 cm/hr for dh/dt.

dV/dt = [10(pi)/3](-3/10)

Thus, dV/dt = -(pi)

Did I do this about right?

Rashakil Fol
10-17-2005, 05:53 PM
Did I do this about right?

No. Had you differentiated correctly, your answer would be correct. I think you're taking a roundabout solution to the problem, too, but it works.

Don't forget to tack units onto your answer, or else it's lost all meaning.

10-17-2005, 06:18 PM
Oh, wow. I suck at basic math.

I just wasn't paying attention when differentiating.

Well, I just wanted to make sure I've got the basic idea of it down.

10-17-2005, 06:32 PM
Ok did it a little hasty but I used the Chain Rule for One Independent Variable

dv/dt = ∂v/∂r * dr/dt + ∂v/∂h * dh/dt

So I started with:

v(r,h) = π[r(t)]^2 [h(t)] / 3
h(t) = (-3/10)t + 10 = (-3t+100)/10
r(t) = (h/2) = (-3t+100)/20 = (-3/20)t + 5

Then its a matter of finding the derivatives

∂v/∂r = 2πrh/3
∂v/∂h = (πr^2)/3
dr/dt = -3/20
dh/dt = -3/10

Plug into the above equation and combine turns / reduce / etc

dv/dt = (2πrh/3)*(-3/20) + (πr^2)/3 * (-3/10)
= (-2πrh)/20 - (πr^2)/10
= (-πr/10)(h+r)
Here I went back to the shape and used the fact that r=h/2

dv/dt = (πh/20)(h+h/2)
=(πh/20)( [3/2]h )
= (-3πh^2)/40
Here I went back and plugged in h(t) for h to get the function in terms of t instead of h

dv/dt = (-3π[3t+100]^2)/(40 * 10^2)
= (-3π)(3t+100)^2 / 4000

here I went back to h(t) and solved for t when the result was 5

5 = (-3t+100)/10
50 = -3t + 100
-50 = -3t
t = 50/3

Now we plug that t value into the above equation:

dv/dt = (-3π)(3[50/3]+100)^2 / 4000
= (-3π)(150)^2 / 4000
= (-3π)(45/8)
= (-135/8)π cm^3/h

Rashakil Fol
10-17-2005, 11:56 PM
Um.... Here's a solution.

As Δh converges towards zero, (ΔV - A*Δh) converges towards zero, where A is the surface area, since the change in volume is approximable by a cylinder. So ΔV/Δt converges towards 6.25pi Δh/Δt. Hence dV/dt = 6.25pi dh/dt = 6.25pi * -3/10 = -15pi/8 cm^3/hr.

dv/dt = (πh/20)(h+h/2)
=(πh/20)( [3/2]h )
= (-3πh^2)/40

Where'd that negative sign come from?

10-18-2005, 07:29 AM
sorry that was caused by a slight difference in my work vs what I typed. The neg came from:
= (-πr/10)(h+r) then sub in r=h/s

Edit: Hmmm after reworking it with a less tired mind I found an easier way the came up with same answer as Mr. Nazi. As much as I'd like to not hand wave I really don't want to type it up.

10-18-2005, 08:48 PM
Okay, I understand now. Thanks for the help, guys!