Related Rates Problems

[jubi129]

I need help with this problem:

Water is leaking out of an inverted conical tank at a rate of 14400.000 cubic cm per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 8.000 meters and the diameter at the top is 5.000 meters. If the water level is rising at a rate of 20.000 cm per minute when the height of the water is 3.000 meters, find the rate at which water is being pumped into the tank in cubic cm per minute.

 

[stapel]

Water is leaking out of an inverted conical tank at a rate of 14400.000 cubic cm per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 8.000 meters and the diameter at the top is 5.000 meters. If the water level is rising at a rate of 20.000 cm per minute when the height of the water is 3.000 meters, find the rate at which water is being pumped into the tank in cubic cm per minute.

What is the formula for the volume V of a cone with height h and radius r? (Note that d = 2r.)

Draw the side view, being an inverted isosceles triangle. Draw a horizontal somewhere in the middle, indicating the height of the water. Draw the altitude line from the "base" (at the top) and the vertex (at the bottom) of the cone. Note that you now have nested (and thus similar) right triangles.

The height of the large right triangle is 8; the height of the nested smaller triangle is h.

The base of the large right triangle is 5/2; the base of the nested smaller triangle is r (because the base is actually the radius of the cone of water).

Form the proportion, based on these similar triangles, and solve for r when h = 3. Starting again with the proportion, solve for one of the variables in terms of the other.

Note that you are given that dh/dt = 20, h = 3, and r = (whatever value you found from the proportion).

Differentiate the "volume" equation with respect to time "t", plug in the known values, and solve for dV/dt.

Noting that part of the change in the volume is the 14,400 cm^3/min leakage, solve for the rate on input.

If you get stuck, please reply showing how far you have gotten. Thank you!

 

[BigGlenntheHeavy]

\(\displaystyle First, \ assume \ no \ leakage, \ then \ V_{cone} = \frac{\pi r^{2}h}{3}.\)

\(\displaystyle Given: \ \frac{dh}{dt} = \frac{20cm}{min} \ when \ h = 300cm. \ Height \ of \ tank = 800cm \ and \ radius \ of \ tank = 250cm.\)

\(\displaystyle \frac{800}{250} = \frac{h}{r} \ proportionality \ of \ triangles, \ h = \frac{16r}{5}, \frac{dh}{dt} = \frac{16dr}{5dt}, \ \frac{dr}{dt} = \frac{25cm}{4} \ per \ min, \ (\frac{dh}{dt} = \frac{20cm}{min}).\)

\(\displaystyle When \ h = 300cm, \ r = \frac{225cm}{4}, \ so \ \frac{dV}{dt} = \frac{\pi}{3}[2rh\frac{dr}{dt} + r^{2}\frac{dh}{dt}].\)

\(\displaystyle Plugging \ in \ all \ the \ values, \ we \ get \ \frac{dV}{dt} = 287,161\frac{cm^{3}}{min} \ (no \ leakage).\)

\(\displaystyle However, \ we \ have \ a \ leakage \ of \ 14,400 cm^{3} \ per \ min, \ so \ \frac{dV}{dt} = 287,161+14,400 = 301,561 \ \frac{cm^{3}}{min}.\)

\(\displaystyle I \ used \ this \ example \ to \ practiced \ my \ Latex.\)

 

[Deleted member 4993]

That was not help - that was spoon-feeding. Only "boxing" (may be underlining) the answer was left to imagination.