Optimization question - water in a conical tank

In summary, a water tank in the shape of an inverted conical cone is being filled at a rate of 0.1m^3/min and leaking at a rate of 0.1h^3/min, where h is the height of water in the tank. Using the formula for the volume of a cone, the rate of change of the depth of water when it is 5m deep is 7.1697 * 10^(-4) m/min. For part (b), the same approach can be used with the adjusted rate of leakage. For part (c), the maximum capacity of the tank can be determined by finding the height at which the rate of filling equals the rate of leakage.
  • #1
billmccai
14
0

Homework Statement



A water tank is in the shape of an inverted conical cone with top radius of 20m and

depth of 15m. Water is flowing into the tank at a rate of 0.1m^3/min.

(a) How fast is the depth of water in the tank increasing when the depth is 5m?

Water is now leaking from the tank at a rate that depends on the depth h, (h= height of

water in the tank) this rate is 0.1h^3/min.

(b) How fast is the depth of water in the tank changing when the depth is 5m?

(c) How full can the tank get?



Homework Equations





The Attempt at a Solution



Ok for part A:

Tan(angle) = 20/15 = 3/4

So i got a formula for r... r = 4/3 h

Which i put into the formula for the volume of a cone and got:

V = 16pi/27 * h^3

and then differentiated V with respect to time.

dV/dt = 16pi/27 * (3h^2)dh/dt

and i know h and dV/dt so i subbed those in and got

dh/dt = 7.1697 * 10^(-4) m/min


which I'm pretty sure is right.




Now I'm not sure how to do part b.

Do I set dV/dt as 0.1 - 0.1h^3

and then just do the same thing? Or is this wrong. Also, what about part c?

I tend to get a bit lost with these sorts of wordy questions. Please help.
 
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  • #2
(a) Correct
(b) Yes, that's the right idea.
(c) If you're still stuck after completing (b), post again.
 

1. How do you calculate the volume of water in a conical tank?

The volume of water in a conical tank can be calculated using the formula V = (1/3)πr2h, where V is the volume, r is the radius of the base of the tank, and h is the height of the water in the tank.

2. How can you optimize the amount of water in a conical tank?

To optimize the amount of water in a conical tank, you can adjust the height of the water in the tank. Keeping the radius constant, a higher water level will result in a greater volume of water in the tank.

3. What is the maximum volume of water that can be held in a conical tank?

The maximum volume of water that can be held in a conical tank is when the tank is completely filled. This can be calculated using the formula V = (1/3)πr2h, where h is the height of the cone, which is equal to the height of the tank.

4. How does the shape of the tank affect the amount of water it can hold?

The shape of the tank, specifically the ratio of the radius to the height, affects the amount of water it can hold. A taller, narrower tank with a smaller ratio of radius to height will hold less water compared to a shorter, wider tank with a larger ratio of radius to height.

5. Can you use calculus to optimize the amount of water in a conical tank?

Yes, you can use calculus to optimize the amount of water in a conical tank. By finding the derivative of the volume formula and setting it to zero, you can find the maximum volume of water that can be held in the tank.

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