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?
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To solve this problem, we can use similar triangles and the chain rule of differentiation.
Let's denote the length of the man's shadow as "s" and the distance he has traveled along the walkway as "x". We need to find the rate at which the length of his shadow is changing, or ds/dt, when he is 40 feet along the walkway.
We can set up a proportion between the similar triangles formed by the man, his shadow, and the lamp:
(height of the man) / (length of his shadow) = (height of the lamp) / (distance between the man and the base of the lamp)
Using the given information, we have:
6 / s = 126 / (x + 30)
To eliminate the fractions, let's cross-multiply:
6 * (x + 30) = s * 126
Now, we have an equation relating the length of the man's shadow to his position along the walkway.
Differentiating both sides of the equation with respect to time (t), we get:
6 * dx/dt = 126 * ds/dt
We can simplify this equation by dividing both sides by 6:
dx/dt = 21 * ds/dt
Now, we need to find dx/dt when x = 40. We know that the man is walking at a constant rate of 3 feet per second, so dx/dt = 3.
Substituting all the given values into the equation, we have:
3 = 21 * ds/dt
To find ds/dt, we rearrange the equation:
ds/dt = 3 / 21 = 1/7 feet per second.
Therefore, the length of the man's shadow is changing at a rate of 1/7 feet per second when he is 40 feet along the walkway past the closest point to the lamp.