Posted by Joel on October 30, 2002 at 01:46:40:
In Reply to: Related Rates Problems posted by C. Slayden on October 29, 2002 at 18:06:30:
: 5.A conical tank (with vertex down) is 10 feet acros the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rete of change of the depth of the water when the water is 8 feet deep.
: For some reason couldn't figure this one at in any shape or form. Again, thank you in advance
: C. Slayden
Very similar to previous ones but in the opposite direction, and here you have to set up the volume as a function of h instead of r. But you know the relationship between h and r, so that shouldn't be too difficult. Same kind of formula, dv/dt = dv/dh * dh/dt, except now you're given dv/dt - the rate of flow into the tank is the same as the rate of increase in volume, right?
The tank is 10 feet across the top, so the radius is 5 feet. And the height is 12 feet, so you know that AT ANY LEVEL the radius is 5/12 of the depth. That ratio is constant.
r = (5/12)*h
v = (1/3)*pi*r^2*h
v = (1/3)*pi*[(5/12)*h]^2*h
v = (25*pi/432)*h^3
dv/dh = (25*pi/144)*h^2
Now, if dv/dt = (dv/dh)*(dh/dt)
then dh/dt = (dv/dt)/(dv/dh)
dh/dt = 10/((25*pi*h^2)/144)
and when h = 8 ft that simplifies to
dh/dt = 9/(10*pi) or about 0.2865 ft/min
You should check that by figuring out the volume at 8 feet & again at 8.3 feet & you'll see that the result is reasonable.
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