# rate of change problem

• Nov 16th 2008, 01:03 PM
billbarber
rate of change problem
Ok. I have a problem, I thought I was doing this problem correctly but my answers aren't making since logically.

A tank of water in the shape of a cone is leaking water at a constant rate of 2 ft^3/hour. The base of the radius of the tank is 5 ft. and the height of the tank is 14 ft.
a) At what rate is the depth of the water in the tank changing when the depth of the water is 6 ft?
b) At what rate is the radius of the top of the water in the tank changing when the depth of the water is 6 ft?

Now I set up the first problem using the Volume formula V=pi(r^2)(h)/3. And then multiplied each side by d/dt. Since the problem told us dV/dt=2 ft^3/hr and the radius is 5 I could solve for dh/dt and I got .0764 ft/hr for a). I then took the original height of the cone 14 ft minus the new depth of the water 6 ft which gives me a change of 8 ft and divided that by .0764 to get the number of hours that it took for the depth to change that much. Which gave me 104.7120 hrs I multiplied that times 2 ft^3/hr to find the amount of volume that had drained out in this time and got 209.4241 ft^3. I subtracted this from the original volume 366.5191 and got 157.0950 which is the volume when the height is 6. I then plugged that volume in to find the radius of what is left of the water and the answer was slightly larger than the original radius which shouldn't happen since it gets smaller as you get closer to the tip of the cone. Where did I go wrong?

Thanks. Any help would be greatly appreciated.
• Nov 16th 2008, 01:56 PM
skeeter
$\frac{dV}{dt} = -2 \frac{ft^3}{hr}$

$\frac{r}{h} = \frac{5}{14}$

$r = \frac{5h}{14}$

$V = \frac{\pi}{3}\left(\frac{5h}{14}\right)^2 h$

simplify the formula for V in terms of h

take the derivative of V w/r to time

calculate the value of $\frac{dh}{dt}$ when $h = 6$

$\frac{dr}{dt} = \frac{5}{14} \cdot \frac{dh}{dt}$
• Nov 16th 2008, 04:17 PM
billbarber
Ok. So let's see if I'm following this correctly.

I started with:

V=pi/3*(5h/14)^2*h

then I differentiated both sides with respect to time and got:

dV/dt=pi/3*(5h/14)^2*dh/dt

then I squared the 5h/14 and I substituted the DV/dt so now it looks like:

2=pi/3*(25h^2/196)*(dh/dt)

I multiplied the pi/3 * 25h^2/196 and multiplied each side by the denominator so now I'm at:

1176=pi*(25h^2)*dh/dt

I then substituted the 6 in for h and multiplied the pi*25*36 and got:

1176=2827.433388*(dh/dt)

then dividing the 1176 by the 2827.4333...... I got

dh/dt=.415924918

Did I understand this correctly?

Oh, by the way.....thanks for the help.
• Nov 16th 2008, 05:52 PM
skeeter
you did not do this correctly.

$V = \frac{\pi}{3}\left(\frac{5h}{14}\right)^2 h$

simplify the formula for V in terms of h ...

$V = \frac{25\pi}{588} h^3$

take the derivative of V w/r to time ...

$\frac{dV}{dt} = \frac{25\pi}{196} h^2 \cdot \frac{dh}{dt}$

substitute in the given values ...

$-2 = \frac{25\pi}{196} \cdot 36 \cdot \frac{dh}{dt}$

now, solve for $\frac{dh}{dt}$.
• Nov 16th 2008, 06:23 PM
billbarber
ok. so let me try again here.

-2=900pi/196*dh/dt

-392=900pi*dh/dt

-.1386416393=dh/dt

then:

dr/dt=5/14*-.1386416393

dr/dt=-.0495148712

my question then is shouldn't I be able to solve for dr/dt using the same method you used to solve for dh/dt and get the same answer .0495148712?

But when I solve that way I get a different answer.

V=pi/3*(r^2)(6) the 6 is the height now.

so V=6pi/3*(r^2) and.....

dV/dt=6pi/3*(2r)(dr/dt) r=2.142857143 according to r=5h/14 so...

-2=6pi/3*(2*2.142857143)(dr/dt)

-2=80.7838111/3*(dr/dt)

-6=80.7838111*(dr/dt)

dr/dt=-.0742723068

Shouldn't I be able to find the same answer solving both ways? Is my logic flawed here?
• Nov 16th 2008, 07:18 PM
skeeter
your logic is flawed.

you cannot plug in 6 for h before you take the derivative ... h is a changing quantity.

$V = \frac{\pi}{3} r^2 h$

product rule ...

$\frac{dV}{dt} = \frac{\pi}{3}(r^2 \frac{dh}{dt} + 2rh \frac{dr}{dt})$
• Nov 16th 2008, 07:30 PM
billbarber
Ahh......yes. got it now.

Thanks again man.