How much does the radius at the surface of the water increase per inch?

[caters]

Homework Statement

A truncated cone has smaller radius of 5 feet, larger radius of 10 feet, and a depth of 1 foot. Total volume is approximately 183 cubic feet. If I have water in there at x inches how much does the radius at the surface of the water increase per inch?

Homework Equations

f(x) = x^n
f'(x) = nx^(n-1)

The Attempt at a Solution

If I know this then I can just multiply it by the number of inches a particular volume takes up in a truncated cone with a particular smaller radius(in this case 5 feet). I would imagine that this requires me to do this derivative:

$$\frac{dr}{dx} \pi r^2 x |0\leq x\leq 12\, and\, 5 \leq r \leq 10$$

where x is the height from the bottom of the truncated cone to the surface of the water.

However I think the rate of change is constant since the slope of the metal surface is constant so that would require that I take the derivative twice. That would give me the derivative of 2pirx which since everything but the x is a constant relative to the derivative operator would be 1. But it can't change by 1 foot per inch or it would have to have a larger radius of 17 ft.

So what is $$\Delta{r}$$

 

[DEvens]

Radius increases 5 feet in 1 foot. So it increases how much in 1 inch?

Other comments:

Radius increases 5 feet in 1 foot of depth. Hmm... Kind of a very broad bowl. Ok.

Who says the surface is metal?

Why are you taking derivatives of volumes?

If you are going to do it using volumes and derivatives, you need the volume as a function of x, not as a function of r and x, since you want r as a function of x. So what is r as a function of x? Oh wait, that's the thing you are looking for...

 

[caters]

Well this is one of the problems related to my solar still so the surface is metal.

the reason I am taking derivatives of volumes is so that I can find $$\Delta{r}$$ since the derivative is the rate of change and that is exactly what I am looking for is the rate of change of the radius for every inch of water.

 

[SteamKing]

Homework Statement

A truncated cone has smaller radius of 5 feet, larger radius of 10 feet, and a depth of 1 foot. Total volume is approximately 183 cubic feet. If I have water in there at x inches how much does the radius at the surface of the water increase per inch?

Homework Equations

f(x) = x^n
f'(x) = nx^(n-1)

The Attempt at a Solution

If I know this then I can just multiply it by the number of inches a particular volume takes up in a truncated cone with a particular smaller radius(in this case 5 feet). I would imagine that this requires me to do this derivative:

$$\frac{dr}{dx} \pi r^2 x |0\leq x\leq 12\, and\, 5 \leq r \leq 10$$

where x is the height from the bottom of the truncated cone to the surface of the water.

However I think the rate of change is constant since the slope of the metal surface is constant so that would require that I take the derivative twice. That would give me the derivative of 2pirx which since everything but the x is a constant relative to the derivative operator would be 1. But it can't change by 1 foot per inch or it would have to have a larger radius of 17 ft.

So what is $$\Delta{r}$$

The problem is asking for the rate of change of the radius of the truncated cone w.r.t. depth. Why do you assume that this is a function of πr²?

You are given three pieces of information about the changing radius:
1. the smaller radius is 5 feet
2. the larger radius is 10 feet
3. the depth of the cone between these radii is 1 foot, or 12 inches.

Can you write a simple function which relates the two radii and the depth of the cone?

Hint: plot the relationship between the three pieces of data given above.