1. ## Another Related Rates

A 100ft. long cable of diameter 4 inches is submerged in seawater. Due to corrosion, the surface area of the cable decreases at a rate of 750 in^2 / year. Ignoring the corrosion at the ends of the cable, find the rate at which the diameter is decreasing.

2. Surface area (S) of a cylinder: $S = 2\pi r^{2} + 2\pi r h$

Since you are dealing with the diameter, it would be nice to express the radius in terms of its diameter (D):

$S = 2\pi \left(\frac{1}{2}D\right)^{2} + 2 \pi \left(\frac{1}{2}D\right)h$
$S = \frac{\pi}{2}D^{2} + \pi Dh$

Differentiate with respect to time:
$\frac{dS}{dt} = \frac{\pi}{2} \underbrace{\left(2D \frac{dD}{dt}\right)}_{\mbox{Chain Rule}} + \pi\underbrace{\left(h \frac{dD}{dt} + D\frac{dh}{dt}\right)}_{\mbox{Product Rule}}$

And solve for $\frac{dD}{dt}$

3. Originally Posted by smschaefer
A 100ft. long cable of diameter 4 inches is submerged in seawater. Due to corrosion, the surface area of the cable decreases at a rate of 750 in^2 / year. Ignoring the corrosion at the ends of the cable, find the rate at which the diameter is decreasing.
Always nice to see a fellow Schaeffer (or any other spelling) here!

4. Originally Posted by o_O
Surface area (S) of a cylinder: $S = 2\pi r^{2} + 2\pi r h$

Since you are dealing with the diameter, it would be nice to express the radius in terms of its diameter (D):

$S = 2\pi \left(\frac{1}{2}D\right)^{2} + 2 \pi \left(\frac{1}{2}D\right)h$
$S = \frac{\pi}{2}D^{2} + \pi Dh$

Differentiate with respect to time:
$\frac{dS}{dt} = \frac{\pi}{2} \underbrace{\left(2D \frac{dD}{dt}\right)}_{\mbox{Chain Rule}} + \pi\underbrace{\left(h \frac{dD}{dt} + D\frac{dh}{dt}\right)}_{\mbox{Product Rule}}$

And solve for $\frac{dD}{dt}$
but would you use the lateral surface area of a cylinder (2 pi r h) or the regular surface area of a cylinder, because corrosion doesnt happen at the ends