A man 6 feet tall walks along a walkway which is 30 feet from a the base of a
lamp which is 126 feet tall. The man walks at a constant rate of 3 feet per
second. How fast is the length of his shadow
changing when he is 40 feet along the walkway past
the closest point to the lamp?
To solve this problem, we can use similar triangles and related rates.
Let's denote the length of the man's shadow as "s" and the distance he has walked along the walkway as "x". We can also denote the height of the man as "h" and the height of the lamp as "H".
According to similar triangles, we have the following ratio:
h/s = H/(s + x)
We are given that the man's height is 6 feet and the height of the lamp is 126 feet. Therefore:
6/s = 126/(s + x)
To find the rate of change of the length of his shadow, we can take the derivative of both sides of the equation with respect to time (t):
d/dt (6/s) = d/dt (126/(s + x))
To solve for ds/dt, the rate at which the length of his shadow is changing with respect to time, we need to find ds/dt on the left side of the equation and dx/dt on the right side of the equation.
First, let's differentiate the left side with respect to time using the quotient rule:
d/dt (6/s) = (s * d/dt (6) - 6 * d/dt (s)) / s^2
= (0 - 6 * ds/dt) / s^2
= -6 * ds/dt / s^2
Now, let's differentiate the right side with respect to time:
d/dt (126/(s + x)) = d/dt (126) / (s + x) + 126 * d/dt (s + x)^(-1)
= 0 / (s + x) + 126 * (-1) * d/dt (s + x) / (s + x)^2
= -126 * ds/dt / (s + x)^2
Now we can substitute these values back into the equation:
-6 * ds/dt / s^2 = -126 * ds/dt / (s + x)^2
Notice that ds/dt appears on both sides of the equation. We can now solve for ds/dt by canceling out the common factors:
-6 / s^2 = -126 / (s + x)^2
Cross-multiplying gives:
-6 * (s + x)^2 = -126 * s^2
Expanding and rearranging the equation, we get:
6s^2 + 12sx + 6x^2 = 126s^2
Combine like terms:
120s^2 - 12sx -6x^2 = 0
Divide both sides by 6:
20s^2 - 2sx - x^2 = 0
We now have a quadratic equation in terms of "s" and "x". To find the value of "s" when the man is 40 feet along the walkway (x = 40), we can substitute x = 40 into the equation and solve for s using the quadratic formula.
Once we have the value of "s", we can substitute it into the equation -6 / s^2 = -126 / (s + x)^2 to find the rate of change of the length of his shadow, ds/dt, when he is 40 feet along the walkway.