Derivative of a cylinder's volume

[NP04]

Homework Statement

why does the derivative of a cylinder's volume (its surface area) = 2πrh+2πr^2

Relevant Equations

Power rule: [d/dx]x^n = nx^(n-1)
Product Rule: y' = ab' + ba'

Volume of a cylinder = (πr^2)h
Surface Area of a cylinder = 2πrh+2πr^2

Using product rule, we have:
[d/dx] (πr^2)(h)
= (πr^2)(1 ) + (2πr)(h)

Why is the two there? V = 2 πrh+2πr^2
The derivative of h is 1, not 2.
Please help!

 

[PeroK]

Homework Statement: why does the derivative of a cylinder's volume (its surface area) = 2πrh+2πr^2
Homework Equations: Power rule: [d/dx]x^n = nx^(n-1)
Product Rule: y' = ab' + ba'

Volume of a cylinder = (πr^2)h
Surface Area of a cylinder = 2πrh+2πr^2

Using product rule, we have:
[d/dx] (πr^2)(h)
= (πr^2)(1 ) + (2πr)(h)

Why is the two there? V = 2 πrh+2πr^2
The derivative of h is 1, not 2.
Please help!

What is $x$ here?

 

[NP04]

What is $x$ here?

If you are referencing the "a" term, that is πr^2.

 

[PeroK]

If you are referencing the "a" term, that is πr^2.

No. I mean here:

Using product rule, we have:
[d/dx] (πr^2)(h)
= (πr^2)(1 ) + (2πr)(h)

 

[NP04]

No. I mean here:

I think I understand what you mean. The x should be a v, right?

 

[PeroK]

I think I understand what you mean. The x should be a v, right?

I'm not sure what it should be. $r$ or $h$ are the obvious candidates, but that doesn't give you what you're looking for.

Where did you get this problem?

 

[NP04]

Deriving the surface area is only the first step.

Here is the full problem (related rates):
Water flows at 8 cubic feet per minute into a cylinder with radius 4 feet. How fast is the water level rising when the water is 2 feet high? Answer should be expressed in terms of feet per minute.

 

[PeroK]

Deriving the surface area is only the first step.

Here is the full problem (related rates):
Water flows at 8 cubic feet per minute into a cylinder with radius 4 feet. How fast is the water level rising when the water is 2 feet high. Answer should be expressed in terms of feet per minute.

That's what you should have posted. What you were doing is not relevant.

Here's what I would do first, to get a grip on the problem:

Assume the water is 2 feet high at some time. A minute later 8 cubic feet of water have been added. How high is the water then?

 

[LCKurtz]

Your problem has nothing to do with surface area. I would start with the formula for the volume of the cylinder: $V = \pi r^2 h = 16\pi h$. Remember $r=4$ is a constant. Also note that there is an unwritten variable, time $t$ which is frequently not explicitly written in the equation, as in this case. Both $V$ and $h$ depend on $t$. I would start by differentiating this equation with respect to $t$ and go from there.

 

[Mark44]

Using product rule, we have:
[d/dx] (πr^2)(h)
= (πr^2)(1 ) + (2πr)(h)

This makes no sense. You are differentiating with respect to x, an expression that doesn't involve x. The total derivative (a term you might not have seen yet) of a function of two or more variables involves partial derivatives.

In any case, as has already been mentioned, surface area is irrelevant in this problem. Although that it's true that the derivative of the volume of a sphere is the surface area, this relationship does not hold for a cylinder.

 

[HallsofIvy]

One basic problem is that the volume of a cylinder, $V= \pi r^2h$, depends upon two variables, r and h. Which one do you want to differentiate with respect to? (You tried to sweep that under the rug by writing "d/dx"!). You could be talking about the two partial derivatives, or the "total differential", $dv= 2\pi rh dr+ \pi r^2 dh$, or perhaps, h and r are both functions of some single variable, t, that you didn't mention. As stated, your problem makes no sense!