# Determine parameters for constant area rate of change

• Feb 3rd 2010, 08:01 AM
grs
Determine parameters for constant area rate of change
A hollow cylinder's internal radius is expanding at a constant rate (b) {the external radius is fixed} and each end of the cylinder is receding at the same constant rate (b). In terms of the internal radius what should the length (height) of the cylinder be in order for the combinded internal surface area of the cylinder and the areas of each end of the cylinder to remain constant.

I have solved this problem algebrically (see below) but believe that it should be solvable using partial derivatives. Unfortunately, it has many years since I've done work in calculus and need some direction. Thank you.

A0 = Cylinder Surface Area at Time 0
At = Cylinder Surface Area at Time t
r0 = Internal Radius of Cylinder at Time 0
rt = Internal Radius of Cylinder at Time t
re = External Radius of Cylinder
l0 = Cylinder Length at Time 0
lt = Cylinder Length at Time t
Initial surface area: A0 = 2πr0l0+2π(re2-r02)
Surface area at time t: At = 2πrtlt+2π(re2-rt2)

If we want the surface area to remain constant then A0 = At

2πr0l0+2π(re2-r02)=2πrtlt+2π(re2-rt2)
r0l0-r02=rtlt-rt2

Note that: rt=r0+bt and lt=l0-2bt then substituting terms and solving for l0 results in:
l0=4r0+3bt

since the surface areas at the start and end of the process must be equal then:
bt=re-r0 where t is the total time expended.
Substituting and simplifying terms results in: l0=3re+r0 which is a confirmed solution.
• Feb 3rd 2010, 09:14 AM
Soroban
Hello, grs!

A fascinating problem . . . hope I got it right.

Quote:

A hollow cylinder's internal radius is expanding at a constant rate $\displaystyle (b)$ (the external radius is fixed),
and the cylinder is lengthening at the same constant rate $\displaystyle (b).$

In terms of the internal radius, what should the length (height) of the cylinder be
in order for the combined internal surface area of the cylinder
and the areas of each end of the cylinder to remain constant.

Code:

                :    R    :     - * - * - - * - - * - *     : |///|    :    |///|     : |///|    :    |///|     : |///|    :    |///|     h |///|    :    |///|     : |///|    :    |///|     : |///|    :    |///|     : |///|    :    |///|     - * - * - - * - - * - *                 :  r  :
The outer radius is $\displaystyle R$ (constant).
The inner radius is $\displaystyle r$.
The height is $\displaystyle h.$

The lateral area of the cylinder is: .$\displaystyle 2\pi rh$

The ends are "rings", each with area: .$\displaystyle \pi(R^2-r^2)$

The total surface area is: .$\displaystyle S \;=\;2\pi rh + 2\pi(R^2-r^2) \;=\;2\pi rh + 2\pi R^2 - 2\pi r^2$

The total area is to be a constant: .$\displaystyle 2\pi rh + 2\pi R^2 - 2\pi r^2 \;=\;K$

Differentiate with respect to time: .$\displaystyle 2\pi r\,\frac{dh}{dt} + 2\pi h\,\frac{dr}{dt} - 4\pi r\,\frac{dr}{dt} \;=\;0$

Since $\displaystyle \frac{dh}{dt} = \frac{dr}{dt} = b$, we have: .$\displaystyle 2\pi r(b) + 2\pi h(b) - 4\pi r(b) \:=\:0$

Solve for $\displaystyle h\!:\;\;2\pi h - 2\pi r \:=\:0 \quad\Rightarrow\quad\boxed{ h \:=\:r}$

• Feb 3rd 2010, 09:35 AM
In grs's original definition of the problem:
Quote:

each end of the cylinder is receding at the same constant rate (b)
I took this to mean that both ends of the cylinder were expanding at rate b.

If I'm right then instead of Soroban's:

Quote:

Originally Posted by Soroban
Since $\displaystyle \frac{dh}{dt} = \frac{dr}{dt} = b$, we have: .$\displaystyle 2\pi r(b) + 2\pi h(b) - 4\pi r(b) \:=\:0$

Solve for $\displaystyle h\!:\;\;2\pi h - 2\pi r \:=\:0 \quad\Rightarrow\quad\boxed{ h \:=\:r}$

We would need:

Since $\displaystyle \frac{dh}{dt} = 2b$ and $\displaystyle \frac{dr}{dt} = b$, we have: .$\displaystyle 2\pi r(2b) + 2\pi h(b) - 4\pi r(b) \:=\:0$

But this means $\displaystyle 2\pi \:h=0$

which is not a very interesting solution! ?
• Feb 3rd 2010, 11:41 AM
grs
Please note that the ends are receding (contracting) not expanding so
$\displaystyle \frac{dh}{dt} = -2b$ and $\displaystyle \frac{dr}{dt} = b$ implying that $\displaystyle 2\pi r(-2b) + 2\pi h(b) - 4\pi r(b) \:=\:0$

Solving for h: $\displaystyle -8\pi r(b) + 2\pi h(b) \:=\:0 \quad\Rightarrow h \:=\:4 r$

However, I am certain the answer is: $\displaystyle h \:=\:3R-r$

Thoughts?
• Feb 3rd 2010, 11:50 AM
grs
Sorry. Result should be $\displaystyle h \:=\:3R+r$