illuminated by the rays of the setting Sun, Kelly rides along on a merry-go-round casting a shadow on a wall. The merry-go-round is turning 36 deg per sec. Kelly is 25 ft from its center, and the Sun's rays are perpendicular to the wall.

At what rate is the shadow moving when it is 15 ft from the point on the wall that is closest to the merry-go-round?

Draw a figure with a circle of radius 25 and a flat sceen behind it. Draw a line through the center of the circle that extends back to the screen, which will be perpendicular to the line. Let the angle of Kelly's location from the sun direction line be A.

A = sin^-1 15/25 = 36.8 degrees

The location of the shadow relative to the closest position (where the line meets the wall) is

X = R sin A
The rate the shadow moves is
dX/dt = R cos A* dA/dt
where dA/dt is in radians/s
dA/dt = (2 pi/10 rad)/1 s
= 0.6283 rad/s

dX/dt = 25 ft*0.8*0.6283 rad/s= 12.6 ft/s

To find the rate at which the shadow is moving when it is 15 ft from the point on the wall closest to the merry-go-round, we can use related rates. Let's break down the problem into smaller steps:

Step 1: Find the rate at which the angle between the Sun's rays and the merry-go-round is changing.
Given that the merry-go-round is turning at 36 degrees per second, the angle between the rays and the ground is also changing at the same rate. Therefore, the rate at which the angle is changing is 36 degrees per second.

Step 2: Determine the initial angle between the Sun's rays and the line connecting Kelly to the merry-go-round.
Since the Sun's rays are perpendicular to the wall, they are also perpendicular to the line connecting Kelly to the merry-go-round. Therefore, the initial angle between the Sun's rays and that line is 90 degrees.

Step 3: Calculate the rate at which the angle between the Sun's rays and the line connecting Kelly to the merry-go-round is changing.
Since the initial angle is 90 degrees and it is changing at a rate of 36 degrees per second, we can use the formula dθ/dt = w for the rate of change of the angle, where w is the angular velocity of the merry-go-round. Therefore, dθ/dt = 36 degrees per second.

Step 4: Determine the length of the line connecting Kelly to the point on the wall closest to the merry-go-round.
The distance from Kelly to the merry-go-round is given as 25 ft, and we can use the Pythagorean theorem to calculate the length of the line connecting Kelly to the wall. Let's call this line L.
L = sqrt((25^2) - (15^2)) = sqrt(400) = 20 ft

Step 5: Determine the rate at which the shadow is moving (i.e., the rate at which the distance from the point on the wall closest to the merry-go-round to that point's shadow is changing). Let's call this rate dx/dt.
We know that dx/dt (rate of change of x) is the rate we're looking for, so we need to calculate it.

Step 6: Relate the angles and lengths to solve for dx/dt.
Using the properties of similar triangles, we can relate the angle θ between the line connecting Kelly to the merry-go-round and the Sun's rays to the rate dx/dt. Since tan(θ) = (distance from the point on the wall closest to the merry-go-round to the shadow) / (distance from Kelly to the merry-go-round), we can differentiate both sides with respect to time (t) to get:

sec^2(θ)*(dθ/dt) = (dx/dt) / (L - x),

where x is the distance from the point on the wall closest to the merry-go-round to the shadow.

Since we know dθ/dt = 36 degrees per second, L = 20 ft, and we need to find dx/dt when x = 15 ft, we can plug in these values to solve for dx/dt:

sec^2(θ) * (36 degrees per second) = (dx/dt) / (20 ft - 15 ft).
sec^2(θ) = (dx/dt) / 5.

Step 7: Calculate sec^2(θ).
From the properties of right triangles, we can determine that sec^2(θ) = (distance from Kelly to the merry-go-round) / (distance from the point on the wall closest to the merry-go-round to the merry-go-round), which is 25 ft / 15 ft = 5/3.

Step 8: Substitute sec^2(θ) and solve for dx/dt.
Now, we substitute sec^2(θ) = 5/3 into the equation from step 6:

(5/3) * (36 degrees per second) = (dx/dt) / 5.
Multiply both sides by 5:

(5/3) * (36 degrees per second) * 5 = (dx/dt).
Simplify:

(5/3) * (36 degrees per second) * 5 = dx/dt.
(5/3) * 36 * 5 = dx/dt.
600/3 = dx/dt.
200 = dx/dt.

Therefore, the rate at which the shadow is moving when it is 15 ft from the point on the wall closest to the merry-go-round is 200 ft per second.