How to Solve Related Rates Problems Involving Water Leaks and Pumping Rates?

[catch.yossarian]

Hi guys, I'm completely stuck here when doing related rates questions. Here is the question, and following the setup I have so far.
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Water is leaking out of an inverted conical tank
at a rate of 10,00cm^3/min at the same time that water is
being pumped into the tank at a constant rate. The tank has
height 6m and diameter at top is 4m. IF the water level is
rising at a rate of 20cm/min when the height of water is
2m, find the rate at which water is being pumped into the
tank.



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d = 4m
h = 6m
2m = 20cm/min.
v = (1/3) (pi) r^2 h
dl/dt = 10,000cm^3/min (rate at which water leaves cone)
dp/dt = ? (rate at which water is being pumped into cone)

thats really all I have. I don't know where to go after that. Any help is appreciated.

 

[HallsofIvy]

this belongs under "Calculus" not "Differential Equations".k

The volume of a cone of radius r and height h is (1/3)πr²h. In this case entire tank has r= 3 and h= 4. Whatever the level of the water, the ratio of height to radius must be the same: r= (3/4)h so the volume is given by
V= (1/3)π(9/16)h³= (3/16)πh³ (I've written r in terms of h because we are given the rate of change of the height, h).
Now differentiate with respect to t: dV/dt= (9/16)πh² (dh/dt).

You are told that h= 2 and dh/dt= 20 cm/min= .2 m/min so dV/dt= (9/16)π(.04)(20)= 0.4 &pi cubic m /min= 400000 cubic cm/min. Since we are also told that "Water is leaking out of an inverted conical tank at a rate of 10,00cm^3/min", in order to account for the water leaking out AND the increase in volume, the water must be coming in at 400010 cubic cm./min.

 

[M98Ranger]

Hi there, it looks like you have set up the problem correctly so far. To find the rate at which water is being pumped into the tank (dp/dt), we can use the related rates formula:

dp/dt = dl/dt - dh/dt

Where dl/dt is the rate at which water is leaving the cone (which is given as 10,000cm^3/min) and dh/dt is the rate at which the height of water is changing (which is given as 20cm/min).

Now, to find the value of dh/dt, we can use the formula for the volume of a cone (V = (1/3)πr^2h) and differentiate it with respect to time:

dV/dt = (1/3)πr^2dh/dt

Since we know the volume (V) and the radius (r), we can solve for dh/dt:

dh/dt = (3dV/dt)/(πr^2)

Plugging in the values, we get:

dh/dt = (3(10,000cm^3/min))/(π(2m)^2) = 477.46 cm/min

Now, we can plug this value back into the related rates formula to find dp/dt:

dp/dt = 10,000cm^3/min - 477.46cm/min = 9522.54 cm/min

Therefore, the rate at which water is being pumped into the tank is approximately 9522.54 cm/min. I hope this helps! Remember, when solving related rates problems, it's important to carefully consider all the given information and use the appropriate formulas to find the unknown rate. Keep practicing and you'll get the hang of it!