Rate of change in surfacearea on a cone with constant volume

[Autious]

Homework Statement

A cone has the base radius of 8 cm and height of 10 cm. The height of the cone is changing at at a rate of 1cm/hour whilst radius of the base is changing with it keeping the volume constant.

At what rate is the surface area of the entire cone changing at this exact moment?

Homework Equations

None were given in the assignment. Such info is assumed to have been memorized.

The Attempt at a Solution

We know the following.

height
$$h=10cm$$
radius
$$r=8cm$$
change in height
$$dh/dt = 1cm/hour$$

Starting out here, I figured that the formula for the volume of the cone was relevant as it gives a relationship between the height, radius and the volume.

$$V= (\pi r^2 * h )/3$$

Given that the volume is constant, I can calculate it...

$$V =(\pi 640)/3$$

...and replace it in the equation

$$(\pi 640)/3 = (\pi r^2 * h )/3$$

...using some algebra I can get the following equation.

$$r = \sqrt{640/h}$$

Which means I have a way to express r as a function of h.

Here comes the part that makes me think that I'm way off.

The formula for the full surface of the cone now is

$$A = \pi r s + \pi r^2$$

Where s is

$$s = \sqrt{ r^2 + h^2 }$$

Finally gives me a function of the cones total surface as a function of the cones height that is

$$A = \pi sqrt{640/h} \sqrt{640/h + h^2} + \pi 640/h$$

Which I can then derivate to calculate the rate of change in the surface of the area.

At this point I start getting the feeling that I've taken the wrong turn somewhere as getting to the answer seems harder than it's "supposed to be".

Can anyone see an easier solution to the problem? Am I even on the path of a solution that would be correct?

Any insight into the issue is appriciated.

 

[Mark44]

Welcome to Physics Forums!

Homework Statement

A cone has the base radius of 8 cm and height of 10 cm. The height of the cone is changing at at a rate of 1cm/hour whilst radius of the base is changing with it keeping the volume constant.

At what rate is the surface area of the entire cone changing at this exact moment?

Homework Equations

None were given in the assignment. Such info is assumed to have been memorized.

The Attempt at a Solution

We know the following.

height
$$h=10cm$$
radius
$$r=8cm$$
change in height
$$dh/dt = 1cm/hour$$

Starting out here, I figured that the formula for the volume of the cone was relevant as it gives a relationship between the height, radius and the volume.

$$V= (\pi r^2 * h )/3$$

Given that the volume is constant, I can calculate it...

$$V =(\pi 640)/3$$

...and replace it in the equation

$$(\pi 640)/3 = (\pi r^2 * h )/3$$

...using some algebra I can get the following equation.

$$r = \sqrt{640/h}$$

Which means I have a way to express r as a function of h.

Here comes the part that makes me think that I'm way off.

The formula for the full surface of the cone now is

$$A = \pi r s + \pi r^2$$

Where s is

$$s = \sqrt{ r^2 + h^2 }$$

Finally gives me a function of the cones total surface as a function of the cones height that is

$$A = \pi sqrt{640/h} \sqrt{640/h + h^2} + \pi 640/h$$

Which I can then derivate to calculate the rate of change in the surface of the area.

At this point I start getting the feeling that I've taken the wrong turn somewhere as getting to the answer seems harder than it's "supposed to be".

Can anyone see an easier solution to the problem? Am I even on the path of a solution that would be correct?

Any insight into the issue is appriciated.

The key here is that the volume of the cone remains constant, which you have noticed, but your mistake is setting the height and radius values. These values are not constant, but are instead functions of time. What you are given is that r(0) = 8 (cm) and h(0) = 10 (cm), but otherwise you don't know what they are at other times.

Use the fact that volume is constant to get a relationship between r and h.

Since V = (1/3) pi r²h = constant, then r²h is also constant, so you can say that r²h = K. From this equation you can solve for h as a function of r, and you can use this to solve for the slant height s as another function of r.

To answer this question you want to evaluate dA/dt at the time when t = 0.

BTW, you don't "derivate" a function - you differentiate it to get the derivative. Derivate might be a word in English, but it is not a verb.

 

[Autious]

Welcome to Physics Forums!
The key here is that the volume of the cone remains constant, which you have noticed, but your mistake is setting the height and radius values. These values are not constant, but are instead functions of time. What you are given is that r(0) = 8 (cm) and h(0) = 10 (cm), but otherwise you don't know what they are at other times.

Use the fact that volume is constant to get a relationship between r and h.

Since V = (1/3) pi r²h = constant, then r²h is also constant, so you can say that r²h = K. From this equation you can solve for h as a function of r, and you can use this to solve for the slant height s as another function of r.

To answer this question you want to evaluate dA/dt at the time when t = 0.

BTW, you don't "derivate" a function - you differentiate it to get the derivative. Derivate might be a word in English, but it is not a verb.

Thanks for your help so far Mark.

I'm not quite sure i see what you mean, but I'm going to try to clarify my thought in the hopes of reaching a solution.

What i did in my attempt show above is that i only used the initial values of h and r to solve for the volume and get a slightly simpler expression. Beyond that, (i think) i treated them as being variable.

Then i ended up with a function dependant soley of h.

f(h) = A = \pi sqrt{640/h} \sqrt{640/h + h^2} + \pi 640/h

Which i can differentiate and get [tex] dA/dh [tex]

and then solve for for $$dA/dt$$ because i know $$dh/dt$$

and already being given $$dh/dt$$ i can thanks to the the chain rule.

$$dA/dh * dh/dt = dA/dt$$

Can i not?

The problem is that the derivative of A is an gargantuan of an expression and i can't help to think that I'm doing something wrong. I can't figure out what though.

p.s

Thanks for correcting me on using the word "derivate".

 

[Mark44]

Thanks for your help so far Mark.

I'm not quite sure i see what you mean, but I'm going to try to clarify my thought in the hopes of reaching a solution.

What i did in my attempt show above is that i only used the initial values of h and r to solve for the volume and get a slightly simpler expression. Beyond that, (i think) i treated them as being variable.

Then i ended up with a function dependant soley of h.

$$f(h) = A = \pi sqrt{640/h} \sqrt{640/h + h^2} + \pi 640/h$$

My thinking was flawed. Since the volume is constant, you get the same volume no matter what the radius and height are, as long as you use values for these two variables that occur at the same time. So yes, V = 640pi/3.

And yes, your formula for surface area as a function of h is correct.

Which i can differentiate and get $$dA/dh$$

and then solve for for $$dA/dt$$ because i know $$dh/dt$$



and already being given $$dh/dt$$ i can thanks to the the chain rule.

$$dA/dh * dh/dt = dA/dt$$

Can i not?

Yes, this makes sense.

The problem is that the derivative of A is an gargantuan of an expression and i can't help to think that I'm doing something wrong. I can't figure out what though.

p.s

Thanks for correcting me on using the word "derivate".

Combine the two radicals in the first part to get
$$A = \pi \left( \frac{640^2 + 640h^3}{h^2}\right)^{1/2} + \text{the other term}$$

You can simplify this first term a bit by bringing out a factor of 1/h. That will make differentiation a little easier.

 

[Autious]

Thanks for your help Mark.

I guess this thread was more about basic algebra and a lack of understanding when it comes to multiplying radicals more than anything else.