Question

Engr. Ramos is walking at the rate of 5m/sec along the diameter of of a circular courtyard. A light at one end of a diameter perpendicular to his path casts a shadow on the circular wall. How fast the shadow from the man to the center of the courtyard is 1/2 r ,where r is the number of meters in the radius of the courtyard?

Answer 1

I assume you meant to ask:

How fast is the shadow moving on the wall, when the distance from the man to the center of the courtyard is 1/2 r

Let's set this up in polar coordinates. Let's call the radius a, so we can use r as a function of θ.

Set the light at the origin, and let the equation of the wall be

r = 2acosθ

Let's also introduce an x-y system for the man's location. He is walking along the line x=a such that dy/dt = 5

tanθ = y/a
sec^2θ dθ/dt = (dy/dt)/a = 5/a
dθ/dt = 5/a sec^2θ

Now, when y=a/2, tanθ = 1/2, so sec^2θ = 5/4

So, dθ/dt = 25/4a

The speed of the shadow on the wall is the rate at with the arc length is increasing. In polar coordinates, the element of arc length is

ds = √(r^2 + r'^2) dθ
that means
ds/dt = √(r^2 + r'^2) dθ/dt
= √((2a cosθ)^2 + (-2a sinθ)^2) dθ/dt
= 2a dθ/dt
= 2a * 25/4a = 25/2 m/s

Answer 2

8 m/s

Answer 3

Valie of d0/dt will be 4/a
Final Answer is 8 m/s