Related Rates Problems

[suicoted]

I understand related rates, but I don't know how to tackle a related rates problem. I can solve it once I know how to set up the formulas and stuff. Thx.

Q: Water is leaking out of an inverted conical tank at a rate of 10 000 cm^3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank.

Q2: A water trough is 10 m long and a cross-section has the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 80 cm wide at the top and has height 50 cm. If it is being filled with water at the rate of 0.2 m^3/min, how fast is the water level rising when the water is 30 cm deep?

Q3: Why is it called 'related rates' anyways?

 

[tkhunny]

Q: Water is leaking out of an inverted conical tank at a rate of 10 000 cm^3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank.

Tank volume - Right, circular cone (1/3)*pi*r²*h.

There's a secret in most such problems. It's the constant ratio of the height of the water to the radius of the top of the water, no matter the depth of the water. That's why the cone is inverted. It doesn't work the other way.

"The tank has height 6 m and the diameter at the top is 4 m."

There it is! Height/Radius = 6/2, making Radius = (1/3)*Height.

I could go one of two ways at this point. If we look at the problem statement, we see which way to go.

"If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m"

We don't know anything about the radius. We had better work the problem in terms of the height.

V(r,h) = (1/3)*pi*r²*h

V(h) = (1/3)*pi*((1/3)*h)²*h = (1/27)*pi*h³

We're talking about change, so we'll need the derivative. Change is over time, so we'll assume that the volume and the height are functions of time.

V(t) = (1/27)*pi*h(t)³
dV = (1/9)*pi*h²*dh

OK, now what?

We need dV or dh or we're stuck.

"If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m"

OK, then h = 2 m and dh = 20 cm/min

dV = (1/9)*pi*(2 m)²*(20 cm/min)

"Water is leaking out of an inverted conical tank at a rate of 10 000 cm^3/min at the same time that water is being pumped into the tank at a constant rate."

Well, that's evil. We are given only half of dV. dV = rate in - rate out = dIN - 10000 cm²/min

dIN - 10000 cm²/min = (1/9)*pi*(2 m)²*(20 cm/min)

That's all the information we have. It is hoped that it can be solved from there. In this case, there are some small unit inconsistencies that will have to be repaired, but it looks pretty straight-forward after that. Change "2 m" to centimeters and you are on your way. (or, it may be simpler to change all the cm to m).

Just one step at a time. Be VERY deliberate. Do NOT try to solve the whole problem in your head without writing anything. Proper notation will help you organize your thoughts and relieve you from having to remember everything all at once. Do NOT be concerned that you do not have a clear solution in mind when you start. Just write down what you know and see where the information takes you.

 

[suicoted]

Thanks!

Thanks! I have a daunting test on rates tmr . I'm sure whatever you posted helped, now I need to work through it. This site is great. Once I learn my calculus, maybe I can help, hehe .