(1 point) A street light is at the top of a 20 foot tall pole. A 6 foot tall woman walks away from the pole with a speed of 7 ft/sec along a straight path. At what rate is the distance from the base of pole to the tip of the shadow changing when she is 50 ft from the base of the pole?The tip of the shadow is moving at ft/sec.

This problem can be solved using similar triangles and the concept of related rates in calculus.

Step 1: Set up the problem We have a 20 foot tall pole and a 6 foot tall woman. The woman is walking away from the pole, creating a shadow. We can create two similar triangles: one with the pole and the tip of the shadow, and one with the woman and the tip of her shadow.

Step 2: Create equations Let's denote:

• x as the distance of the woman from the pole
• y as the distance from the pole to the tip of the shadow

Since the triangles are similar, we can write the following equation: 20/y = 6/(y-x)

Step 3: Differentiate with respect to time Differentiating both sides with respect to time (t), we get: -20/y^2 * dy/dt = -6/(y-x)^2 * dy/dt + 6/(y-x)^2 * dx/dt

Step 4: Solve for dy/dt We know that dx/dt = 7 ft/sec (the speed of the woman), and we want to find dy/dt when x = 50 ft. To find the value of y when x = 50 ft, we can use the similar triangles equation: 20/y = 6/(y-50) Solving for y, we get y = 70 ft.

Substitute y = 70 ft, x = 50 ft, and dx/dt = 7 ft/sec into the differentiated equation, we can solve for dy/dt: -20/70^2 * dy/dt = -6/(70-50)^2 * dy/dt + 6/(70-50)^2 * 7 Solving for dy/dt, we get dy/dt = 14 ft/sec.

So, the rate at which the distance from the base of the pole to the tip of the shadow is changing when the woman is 50 ft from the base of the pole is 14 ft/sec.