The Solution
3.25 feet
12 feet
We know that dr⁄dt = 1 inch⁄second.
We also know that the width of the rainbow is a constant 12 ft.
We want dA⁄dt when r = 3.25 feet.
However, notice that we have two different units. We can convert all of our feet to inches by multiplying by 12.
dr⁄dt = 1 in⁄sec
r = 39 in
width = 144 in
Now we need to come up with a formula to determine A, the area of the rainbow.
We can define A as the area of the outer semicircle minus the area of the inner semicircle.
If diameter of the rainbow is a constant 144 inches, the radius must be 72 in.
The area of a circle is defined as
A = πr2
Therefore, the area of the outer semicircle is equal to
Aouter = 1⁄2π(72)2
or
Aouter = 2592π
since a semicircle is half of a circle.
Now we have to determine the equation for the inner semicircle.
Since the total radius of the outer semicircle is 72 inches, the radius of the inner must be 72 minus the thickness of the rainbow.
Therefore,
Ainner = 1⁄2π(72-r)2
This means that our full equation for the area of our rainbow is
A = 2592π - 1⁄2π(72-r)2
If we take the derivative of this equation, we get
dA⁄dt = 0 - 1⁄2π*-2(72-r)dr⁄dt
= π(72-r)dr⁄dt
Once we plug in our known values, we get
dA⁄dt = π(72-(39 in))(1 in⁄sec)
= π(33)
= 33π in2⁄sec
When the thickness of the rainbow is 3.25 feet, the area of the rainbow is decreasing at a rate of 33π in2⁄sec