A 5.8-ft-tall person walks away from a 9-ft lamppost at a constant rate of 3.4 ft/sec. What is the rate that the tip of the person's shadow moves away from the lamppost when the person is 11 ft away from the lampost

9.56 ft/sec

Step-by-step explanation:

We are told that a 5.8-ft-tall person walks away from a 9-ft lamppost at a constant rate of 3.4 ft/sec.

I've attached an image showing  triangle that depicts this;

Thus; dx/dt = 3.4 ft/sec

From the attached image and using principle of similar triangles, we can say that;

9/y = 5.8/(y - x)

9(y - x) = 5.8y

9y - 9x = 5.8y

9y - 5.8y = 9x

3.2y = 9x

y = 9x/3.2

dy/dx = 9/3.2

Now, to find how fast the tip of the shadow is moving away from the lamp post, it is;

dy/dt = dy/dx × dx/dt

dy/dt = (9/3.2) × 3.4

dy/dt = 9.5625 ft/s ≈ 9.56 ft/sec

a = 11 : 4

b = 38

c = 1 for every 6