A man 6ft. tall walks at the rate of 5 ft/sec toward a streelight that is 16 ft. above the ground. At what rate is the length of his shadow changing when he is 10ft. from the base of the light?

x is shadow tip from streetlight

y is man from streetlight
from similar triangles
x/16 = (x-y)/6
16 x - 16 y = 6 x
16 y = 10 x
x = 1.6 y
dx/dt = 1.6 dy/dt
= 1.6*5

To find the rate at which the length of the man's shadow is changing when he is 10ft. from the base of the light, we can apply the concept of similar triangles and the chain rule from calculus.

Let's consider a right triangle formed by the man, his shadow, and the light pole. The height of the light pole is 16ft, and the length of the man's shadow is x ft.

The man's height of 6ft and the height of the light pole form a similar triangle. The corresponding sides of similar triangles are proportional. So, we can set up the following proportion:

(Length of the man's shadow) / x = 6 / 16

Simplifying this proportion, we get:

x / x = 6 / 16
1 = 6 / 16
x = (6 / 16) * x
x = 0.375x

Now, we can differentiate both sides of the equation with respect to time (t) using the chain rule:

d(x) / dt = d(0.375x) / dt

The left side represents the rate of change of the length of the shadow with respect to time (the quantity we're trying to find). The right side represents the rate of change of 0.375x with respect to time. We know the man's vertical rate of change (walking towards the lamp) is 5ft/sec. We need to find the rate at which the length of his shadow is changing, so we need to find d(x) / dt.

Let's continue differentiating:

d(x) / dt = 0.375 * d(x) / dt

Now, we substitute the given rate (5ft/sec) into the equation:

d(x) / dt = 0.375 * 5

Simplifying further:

d(x) / dt = 1.875 ft/sec

Therefore, the rate at which the length of the man's shadow is changing when he is 10ft. from the base of the light is 1.875 ft/sec.