A 36-cm high cone w/ base radius 12 cm is filled with water @ 3cm^3/sec. What is the rate of change...?

[Lucas456]

The question is:

A 36 cm high cone with a base of radius 12cm is filled with water at a constant rate of 3 cm^3 /sec. What is the rate of change of height in cm/sec when the water in the cone has a depth of 9cm?

Any help would be greatly appreciated

 

[stapel]

A 36 cm high cone with a base of radius 12cm is filled with water at a constant rate of 3 cm^3 /sec. What is the rate of change of height in cm/sec when the water in the cone has a depth of 9cm?

Any help would be greatly appreciated

We'll be glad to help! But in order to do that, we'll need to see what you've tried and where you're getting stuck. (Or, if you're too lost even to get started, we'll need to work with you to figure out what lessons you need to study, first.)

So please re-read the "Read Before Posting" message, and reply with the requested information. Thank you!

 

[khansaheb]

The question is:

A 36 cm high cone with a base of radius 12cm is filled with water at a constant rate of 3 cm^3 /sec. What is the rate of change of height in cm/sec when the water in the cone has a depth of 9cm?

Any help would be greatly appreciated

Is the cone being filled from the bottom or through the apex?

Please provide a sketch pertaining the problem - explaining where/how you are stuck.

 

[Lucas456]

Attached is an image of what I have done so far. I'm not sure if this is right. Can you please see if I made any mistakes

 

[Aminta_1900]

The question is:

A 36 cm high cone with a base of radius 12cm is filled with water at a constant rate of 3 cm^3 /sec. What is the rate of change of height in cm/sec when the water in the cone has a depth of 9cm?

Any help would be greatly appreciated

Hi Lucas,

I am far from an expert on this subject, since I am also currently learning rates of change and also posting questions upon it. But I'm sure I could grasp this question, but I think my answer would have to be checked and corroborated by one of the resident members.

Basically you are looking for dh/dt, which is the rate of change of height per sec. You can get this by following the formula

dh/dt = dV/dt divided by dV/dh

dV/dt is the rate of change of the volume per sec, which if you read the question carefully you'll notice that you already have this, it is 3 cm^2/sec (because it gives you information in volume cm^3 and time in sec). So

dV/dt = 3

Next you need to find dV/dh, which is the derivative of the function V = f(h). But since the formula for the volume for a cone of (1/3)pi * r^2 * h is in terms of r and h, you'll need to change that r to its equivalent in h to properly keep the function in respect of h (it's like having y = 2x + z and you don't want the z there, in this case we don't want the r) so using trig to find the r in terms of h, and using the angle theta that separates the height and the slant height of the cone...

tan theta = r/h

tan theta = 12/36

tan theta = 1/3

1/3 = r/h

h * 1/3 = r

r = h/3

So replace the r in the formula for the volume of a cone with h/3 and you now have V = (pi*h^3)/27. Since you now have your V = f(h) you can get the derivative dV/dh = (pi*h^2)/9.

So now you can use dV/dt divided by dV/dh and you get

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

This is used to find the rate of change of height over time, and since you want to know the rate of change when the height is 9cm, punch this into dh/dt

dh/dt = 27/(pi*9^2) = 27/(pi * 27) = 1/pi cm/sec, or 0.318 cm/sec

This would seem a lot easier if you write the equations on paper and thus in a better format.

 

[Dr.Peterson]

dh/dt = 27/(pi*9^2) = 27/(pi * 27) = 1/pi cm/sec, or 0.318 cm/sec

But 9^2 is not 27.

Otherwise, though your work is a little different in detail from what I am used to, it is correct.

I would just write V as a function of h, differentiate with respect to t (using the chain rule), and solve for dh/dt using the given data.

 

[Aminta_1900]

But 9^2 is not 27.

Otherwise, though your work is a little different in detail from what I am used to, it is correct.

I would just write V as a function of h, differentiate with respect to t (using the chain rule), and solve for dh/dt using the given data.

Thanks for pointing out. I often make silly mistakes from associations (3, 9, 27, 81..). Must stop rushing.

 

[Aminta_1900]

Hi Lucas,

I am far from an expert on this subject, since I am also currently learning rates of change and also posting questions upon it. But I'm sure I could grasp this question, but I think my answer would have to be checked and corroborated by one of the resident members.

Basically you are looking for dh/dt, which is the rate of change of height per sec. You can get this by following the formula

dh/dt = dV/dt divided by dV/dh

dV/dt is the rate of change of the volume per sec, which if you read the question carefully you'll notice that you already have this, it is 3 cm^2/sec (because it gives you information in volume cm^3 and time in sec). So

dV/dt = 3

Next you need to find dV/dh, which is the derivative of the function V = f(h). But since the formula for the volume for a cone of (1/3)pi * r^2 * h is in terms of r and h, you'll need to change that r to its equivalent in h to properly keep the function in respect of h (it's like having y = 2x + z and you don't want the z there, in this case we don't want the r) so using trig to find the r in terms of h, and using the angle theta that separates the height and the slant height of the cone...

tan theta = r/h

tan theta = 12/36

tan theta = 1/3

1/3 = r/h

h * 1/3 = r

r = h/3

So replace the r in the formula for the volume of a cone with h/3 and you now have V = (pi*h^3)/27. Since you now have your V = f(h) you can get the derivative dV/dh = (pi*h^2)/9.

So now you can use dV/dt divided by dV/dh and you get

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

This is used to find the rate of change of height over time, and since you want to know the rate of change when the height is 9cm, punch this into dh/dt

dh/dt = 27/(pi*9^2) = 27/(pi * 27) = 1/pi cm/sec, or 0.318 cm/sec

This would seem a lot easier if you write the equations on paper and thus in a better format.

Correction: dh/dt = 27/(pi*9^2) = 27/(pi * 81) = 1/(3pi) = 0.106 cm/sec

 

[Lucas456]

Correction: dh/dt = 27/(pi*9^2) = 27/(pi * 81) = 1/(3pi) = 0.106 cm/sec

1/3pi approximately 0.106
so we both got the same answer!

 

[Lucas456]

But 9^2 is not 27.

Otherwise, though your work is a little different in detail from what I am used to, it is correct.

I would just write V as a function of h, differentiate with respect to t (using the chain rule), and solve for dh/dt using the given data.

I have another question. If the cone were the other way around, with the apex at the top, would it still be the same answer?

 

[blamocur]

I have another question. If the cone were the other way around, with the apex at the top, would it still be the same answer?

Good question! My answer is "no", but I am curious to hear what you think and why.
Also, you can modify your question: at which height will the rate be the same in the upside-down cone as it is at 9cm in the upright one?

 

[khansaheb]

I have another question. If the cone were the other way around, with the apex at the top, would it still be the same answer?

Like the previous problem, sketch the situation!

 

[ashley2]

The question is:

A 36 cm high cone with a base of radius 12cm is filled with water at a constant rate of 3 cm^3 /sec. What is the rate of change of height in cm/sec when the water in the cone has a depth of 9cm?

Any help would be greatly appreciated

1. First, calculate the volume of water in the cone when the depth is 9 cm. You can use the formula for the volume of a cone:
   Volume = (1/3) * π * r^2 * h
   Where:
   • r is the radius of the base (12 cm).
   • h is the height of the water (9 cm).
2. Calculate the volume when h = 9 cm:
   Volume = (1/3) * π * (12 cm)^2 * (9 cm)
3. Now, differentiate the volume with respect to time to find the rate of change of volume:
   dV/dt = (1/3) * π * 2 * 12 * 3 * (dh/dt)
   Where:
   • dV/dt is the rate of change of volume (3 cm^3/sec, given in the problem).
   • dh/dt is the rate of change of height (which we want to find).
4. Plug in the values and solve for dh/dt:
   3 cm^3/sec = (1/3) * π * 2 * 12 * 3 * (dh/dt)
5. Simplify and solve for dh/dt:
   dh/dt = (3 cm^3/sec) / (2 * π * 12 * 3)
   dh/dt ≈ 0.0131 cm/sec

So, the rate of change of height when the water in the cone has a depth of 9 cm is approximately 0.0131 cm/sec.