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# Rate of change: conical tank [Solved!]

### My question

Please help me solve a rate of change problem about a conical tank wit vertex down. i dont know the equation i have to use

4. Related Rates

### What I've done so far

I read the examples on the page, but none of them were like my one.

X

Please help me solve a rate of change problem about a conical tank wit vertex down. i dont know the equation i have to use
Relevant page

<a href="/applications-differentiation/4-related-rates.php">4. Related Rates</a>

What I've done so far

I read the examples on the page, but none of them were like my one.

## Re: Rate of change: conical tank

Hello Ana

You will need to provide more details before I can help you.

4. The Graph of a Function
(go to the bottom of the page.)

Regards

X

Hello Ana

You will need to provide more details before I can help you.

<a href="/functions-and-graphs/4-graph-of-function.php">4. The Graph of a Function</a>
(go to the bottom of the page.)

Regards

## Re: Rate of change: conical tank

The question says the tank has diameter 3 m at the top, and is 5 m high. What is the rate of change of the height of the water at time t?

X

The question says the tank has diameter 3 m at the top, and is 5 m high. What is the rate of change of the height of the water at time t?

## Re: Rate of change: conical tank

You need to show us your attempts at the problem.

Did you look at that page I suggested?

Hint 1: What is the formula for the volume of a cone?

Hint 2: What is the rate of flow of the water?

X

You need to show us your attempts at the problem.

Did you look at that page I suggested?

<b>Hint 1: </b> What is the formula for the volume of a cone?

<b>Hint 2: </b> What is the rate of flow of the water?

## Re: Rate of change: conical tank

I did look at that example, but it says r = h and my question does not have that. So I got stuck.

Anyway, volume of a cone is pir2h/3

The question doesn't say what the water flow rate is.

X

I did look at that example, but it says r = h and my question does not have that. So I got stuck.

Anyway, volume of a cone is pir2h/3

The question doesn't say what the water flow rate is.

## Re: Rate of change: conical tank

I encourage you to use the math input system, so we can read your answer more easily.

You just need to put the following in between back ticks,

V = (pi r^2 h)/2

So it looks like this:

V=(pi r^2 h)/3

You're right - the question doesn't give us a number for the flow rate, but we can just give it a letter, say f and assume it is constant at liters per minute, say.

Hint 3: How long will it take the water to flow out?

Hint 4: How do you find the rate of change of the volume?

X

I encourage you to use the math input system, so we can read your answer more easily.

You just need to put the following in between back ticks,

<code>V = (pi r^2 h)/2</code>

So it looks like this:

V=(pi r^2 h)/3

You're right - the question doesn't give us a number for the flow rate, but we can just give it a letter, say f and assume it is constant at liters per minute, say.

<b>Hint 3: </b> How long will it take the water to flow out?

<b>Hint 4: </b> How do you find the rate of change of the volume?

## Re: Rate of change: conical tank

If it flows at f liters per minute, then it will take V/f minutes to empty out.

Do we use (dV)/(dt) to find rate of change of volume?

Then what?

X

If it flows at f liters per minute, then it will take V/f minutes to empty out.

Do we use (dV)/(dt) to find rate of change of volume?

Then what?

## Re: Rate of change: conical tank

Yes, (dV)/(dt) is correct.

On the right hand side of our volume equation, what is constant and what varies as the water empties out?

X

Yes, (dV)/(dt) is correct.

On the right hand side of our volume equation, what is constant and what varies as the water empties out?

## Re: Rate of change: conical tank

pi is constant, but r and h vary as time goes on.

I use product rule:

so (dV)/(dt) = pi( r^2 (dh)/(dt) + (h) 2r (dr)/(dt))

But I'm lost again.

X

pi is constant, but r and h vary as time goes on.

I use product rule:

so (dV)/(dt) = pi( r^2 (dh)/(dt) + (h) 2r (dr)/(dt))

But I'm lost again.

## Re: Rate of change: conical tank

What you have is correct, but it's more complicated than it needs to be.

What's the relationship between r and h?

Can you simplify V now?

X

What you have is correct, but it's more complicated than it needs to be.

What's the relationship between r and h?

Can you simplify V now?

## Re: Rate of change: conical tank

Radius is 1.5 m and height is 5 m, so

r/h = 1.5/5 = 0.3

So r = 0.3h

So now

V = (pi r^2 h)/3  = (pi (0.3h)^2 h)/3  = 0.03 pi h^3

I see it now

(dV)/(dt) = 3(0.03 pi h^2)(dh)/(dt) =0.09 pi h^2(dh)/(dt)

Where do we use f?

X

Radius is 1.5 m and height is 5 m, so

r/h = 1.5/5 = 0.3

So r = 0.3h

So now

V = (pi r^2 h)/3  = (pi (0.3h)^2 h)/3  = 0.03 pi h^3

I see it now

(dV)/(dt) = 3(0.03 pi h^2)(dh)/(dt) =0.09 pi h^2(dh)/(dt)

Where do we use f?

## Re: Rate of change: conical tank

Well, f is just the rate of flow, so it's equal to the change in volume of the water. So we have

f=0.09 pi h^2(dh)/(dt)

Can you get the expression of the rate of change of the height now?

X

Well, f is just the rate of flow, so it's equal to the change in volume of the water. So we have

f=0.09 pi h^2(dh)/(dt)

Can you get the expression of the rate of change of the height now?

## Re: Rate of change: conical tank

So is it this?

(dh)/(dt) = f/(0.09pih^2)

X

So is it this?

(dh)/(dt) = f/(0.09pih^2)

## Re: Rate of change: conical tank

Yes, you are correct. What are the units?

X

Yes, you are correct. What are the units?

## Re: Rate of change: conical tank

The units will be m/min

X

The units will be m/min

## Re: Rate of change: conical tank

Correct.

X

Correct.