Related Rates

• November 14th 2006, 04:27 PM
drewms64
Related Rates
So im having some problems figuring out where to start on these 2 related rates problems.

1. For a 5000 gallon tank of water which will drain completely in 40 minutes, the volume of water remaining in the tank after t minutes is given by
V = 5000(1 - (t/40))^2
Find the rate at which water is draining from the tank after 10 minutes.

EDIT: I found an example which showed me that I should make the 5000(1-(t/40))^2 into
V(t)= 5000-250t+((25t^2)/8)
V'(t)= -250+(50/8)(t)
= -250+(50/8)(10min)
= -187.5 gal/min ?

2. Oil spills out of a tanker at a rate of 90 gallons per minute. The oil spreads in a circle with a thickness of 1/8 inch. Given that 1 gallon = 7.5 cubic feet, determine the rate at which the radius of the spill is increasing when the radius reaches 100 feet. V=(depth)(area)

edit: converted the 1/8th inch to ft, V=(.0104ft)(7.5*90) = 7.02ft^3/m , so would that be dV/dt? Not sure where I would go from here
• November 15th 2006, 12:00 AM
malaygoel
Quote:

Originally Posted by drewms64
So im having some problems figuring out where to start on these 2 related rates problems.

1. For a 5000 gallon tank of water which will drain completely in 40 minutes, the volume of water remaining in the tank after t minutes is given by
V = 5000(1 - (t/40))^2
Find the rate at which water is draining from the tank after 10 minutes.

EDIT: I found an example which showed me that I should make the 5000(1-(t/40))^2 into
V(t)= 5000-250t+((25t^2)/8)
V'(t)= -250+(50/8)(t)
= -250+(50/8)(10min)
= -187.5 gal/min ?

2. Oil spills out of a tanker at a rate of 90 gallons per minute. The oil spreads in a circle with a thickness of 1/8 inch. Given that 1 gallon = 7.5 cubic feet, determine the rate at which the radius of the spill is increasing when the radius reaches 100 feet. V=(depth)(area)

edit: converted the 1/8th inch to ft, V=(.0104ft)(7.5*90) = 7.02ft^3/m , so would that be dV/dt? Not sure where I would go from here

Your solution to the first qustion is correct.
One thing to note here:
Rate of spilling of oil is equal to rate of increase of volume of cylinder(a cicle with some thickness is cylinder, here height of cylinder is constant and radius is increasing).

So you have dV/dt for the cylinder.
As you said: $V=depth*area$
so, $\frac{ dV}{dt=(depth)*\frac{(area}{dt}=(depth)*(2*pi*r)\f rac{dr}{dt}$
$\frac{ dV}{dt=(depth)*(2*pi*r)\frac{dr}{dt}$