Related Rates/ cone question

[tmt1]

A water tank has the shape of an inverted circular cone with base radius 2 m
and height 4 m. If water is being pumped into the tank at a rate of 2 m^3/ min, find the rate at which the water level is rising when the water is 3 m deep.

The answer to this question is $$\frac{8}{9\pi}$$But I got a different answer, here is what I did:

We have

$$\d{V}{t} = 2 m^3 /min$$

and we need

$$\d{h}{t} = ?$$

The volume of a cone is

$$V = \frac{1}{3}\pi r^2 h$$

Since $$2/4 = r/h$$, $$r= h/2$$ and I can change the volume formula to:

$$V = \frac{1}{12}\pi h^3$$

we derive this:

$$\d{V}{t} = \frac{\pi}{4}h^2 \d{h}{t}$$

We already know dV/dt:

$$2 m^3 /min= \frac{\pi}{4}h^2 \d{h}{t}$$

We isolate dh/dt:

$$\frac{8}{\pi 16} = \d{h}{t}$$

h is equal to 4:

$$\frac{8}{\pi h^2} = \d{h}{t}$$

Thus the answer is:

$$\d{h}{t} = \frac{1}{\pi 2} $$

I've done this question multiple times and I keep getting the same answer which is wrong.

EDIT: I solved it, h is equal to 3 not 4.

 

[vacuum]

So the answer is $$\frac{3}{\pi}$$

Hi there! Thank you for sharing your approach to solving this problem. It looks like you made a small mistake in your calculations. The volume formula for a cone is actually V = (1/3)πr^2h, not V = (1/12)πh^3. This is why your answer is different from the correct one.

To solve the problem, we can use the given information to set up an equation:

V = (1/3)π(2)^2h

We know that dV/dt = 2 m^3/min, so we can substitute this in and solve for dh/dt:

2 = (1/3)π(2)^2(dh/dt)

dh/dt = (3/4π) m/min

Substituting in h = 3 (since the water level is 3 m deep), we get:

dh/dt = (3/4π) m/min = 0.239 m/min

Therefore, the rate at which the water level is rising when the water is 3 m deep is approximately 0.239 m/min. I hope this helps clarify the solution. Keep up the good work!