More Related Rates
While on vacation in Mexico, a tanker accident has spilled oil in the Gulf of Mexico. Scuba Steve volunteered to help by spreading oil-eating bacteria which is gobbling 7 ft^3 per hour. the oil slick has the form of a circular cylinder. When the radius of the cylinder is 250 feet, the thickness of the slick is 0.02 feet and decreasing at a rate of 0.002 ft/hr.
1. At what rate is the area of the circular top of the slick changing at this time?
2. Is the area of the slick increasing or decreasing?
To solve this problem you have to identify what the problem is giving you in terms of information and find out what it is asking of you. The problem talks about a cylinder and the thickness of it, so one must realize that the problem will require the use of the volume formula for a cylinder.
V = πr^2 h
dV/dt = -7 ft^3/hr dh/dt = -0.002 ft/hr r = 250 ft h = 0.02 dA/dt = ?
Because of the two variables r and h, you have to use the Product Rule.
dV/dt = π(r^2)(dh/dt) + 2πr(dr/dt)h
You solve for dr/dt, by using substitution.
-7 = π(250^2)(-0.002) + 2π(250)(dr/dt)(0.02)
-7 = -125π + 10π(dr/dt)
dr/dt = (-7 + 125π)/(10π)
dr/dt = 12.2772
Now to solve the area question, we use the formula for the area of a circle.
A = πr^2
dA/dt = 2πr(dr/dt)
Now use substitution to solve for dA/dt.
dA/dt = 2π(250)(12.2772)
dA/dt = 19284.9807 ft^2/hr
If you see the word rate in a question, you have to use the derivative function.
For the Prince of Calculand project, the questions are at http://calculandquest.blogspot.com and the solutions are at http://calculandans.blogspot.com
The next scribe is Prince.