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?
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
8 m/s
Valie of d0/dt will be 4/a
Final Answer is 8 m/s
To determine the rate at which the shadow from Engr. Ramos to the center of the courtyard is changing, we can use the concept of similar triangles.
Let's denote the length of the shadow as S and the radius of the courtyard as r.
We know that the length of the shadow is 1/2 r. Let's call the distance of Engr. Ramos from the center of the courtyard x. Therefore, we have the following similar triangles:
1. The larger triangle formed by the distance from Engr. Ramos to the center of the courtyard (x), the radius of the courtyard (r), and the length of the shadow (S).
2. The smaller triangle formed by the distance that Engr. Ramos travels (y), the radius of the courtyard (r), and the change in the length of the shadow (dS).
These two triangles are similar since they share the same angles.
Using the properties of similar triangles, we can set up the following proportion:
(S + dS) / dS = x / y
Since Engr. Ramos is walking at a constant speed of 5 m/sec, we can express y in terms of x:
y = 5 * t, where t is the time in seconds.
Now, let's solve the proportion for dS:
(S + dS) / dS = x / (5 * t)
Cross multiply to get:
S + dS = (x * dS) / (5 * t)
Rearrange the equation to isolate dS:
5 * t * (S + dS) = x * dS
Now, we are given that S = 1/2 r, so substitute that in:
5 * t * (1/2 r + dS) = x * dS
Simplify:
5 * t * (r/2 + dS) = x * dS
Now, we need to express dS in terms of dx/dt, the rate at which x is changing. Since x is the distance from Engr. Ramos to the center of the courtyard, dx/dt is given as 5 m/sec (as Engr. Ramos is walking at a speed of 5 m/sec). Therefore, dS/dt can be expressed as 5 * dx/dt.
Substitute dS/dt with 5 * dx/dt:
5 * t * (r/2 + 5 * dx/dt) = x * dx/dt
Now, solve for dx/dt:
5 * t * r/2 + 25 * t * dx/dt = x * dx/dt
Rearrange the equation:
25 * t * dx/dt - x * dx/dt = -5 * t * r/2
Factor out dx/dt:
dx/dt * (25 * t - x) = -5 * t * r/2
Now, divide both sides of the equation by (25t - x):
dx/dt = (-5 * t * r/2) / (25 * t - x)
Therefore, the rate at which the shadow from Engr. Ramos to the center of the courtyard is changing is given by:
dx/dt = (-5 * t * r/2) / (25 * t - x)