# More Related Rates

• Nov 24th 2006, 02:36 PM
Jacobpm64
More Related Rates
A ruptured oil tanker causes a circular oil slick on the surface of the ocean. When its radius is 150 meters, the radius of the slick is expanding by 0.1 meter/minute and its thickness is 0.02 meter. At that moment:

(a) How fast is the area of the slick expanding?
(b) The circular slick has the same thickness everywhere, and the volume of oil spilled remains fixed. How fast is the thickness of the slick decreasing?

(a) For this one, I'm not sure whether by "area" they mean surface area (since I think we'd be dealing with a cylinder), or just an area of the top circle (the top of the cylinder)?

(b) I'll come back and work this one if you guys help me with the first one.
• Nov 24th 2006, 03:34 PM
galactus
I think, for part a, they're asking for change in area

$A={\pi}r^{2}$

$\frac{dA}{dt}=2{\pi}r\frac{dr}{dt}$

You can use the volume of a cylinder formula to find part b.

Remember, volume is fixed and the answer should be negative because it's decreasing.
• Nov 24th 2006, 03:48 PM
Jacobpm64
all right, let's see..

See, for part A, i had worked it out as if they were asking for the entire surface area (the area of the cylinder part and the top and bottom of the cylinder all added together)... but, if they're asking for the area of just the top circle, I'd do it like this:

(a)
$A = \pi r^2$
$\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}$
$\frac{dA}{dt} = 2* \pi * 150 * 0.1$
$\frac{dA}{dt} = 30 \pi \approx 94.2478$ m^2/min

(b)
$V = \pi r^2 h$
$h = \frac{V}{ \pi r^2 }$
$\frac{dh}{dt} = \frac{-2V}{ \pi r^3} \frac{dr}{dt}$
$\frac{dh}{dt} = \frac{-2*( \pi 150^2 0.02 )}{ \pi 150^3 } * 0.1$
$\frac{dh}{dt} \approx -0.000026667$
decreasing at 0.000026667 m/min.

Correct?
• Nov 24th 2006, 03:58 PM
galactus
Good. That's what I got. Except maybe write your answer as $\frac{dh}{dt}=\frac{-1}{37500}$. Looks better.