Question

4) A light is on the top of a 12 ft tall pole and a 5 ft 6 in tall person is walking away from the pole at a rate of 2 ft/sec. a. At what rate is the tip of the shadow moving away from the pole when the person is 25 ft from the pole? b. At what rate is the tip of the shadow moving away from the person when the person is 25 ft from the pole?

Answer 1

The rate at which the tip of the shadow is moving away from the pole when the person is 25 ft from the pole is 0 ft/sec. The rate at which the tip of the shadow is moving away from the person when the person is 25 ft from the pole is 0 ft/sec.

Step-by-step explanation:

To solve for the rates at which the tip of the shadow is moving away from the pole and from the person, we can use similar triangles and related rates. Let's solve each part step by step:

a. Rate the tip of the shadow is moving away from the pole:

We have a right triangle formed by the pole, the person, and the tip of the shadow. Let x be the distance between the person and the tip of the shadow.

Since the person is moving away from the pole at a rate of 2 ft/sec, dx/dt = 2 ft/sec.

Using the similar triangles, we have:

(12 ft + 5.5 ft) / x = 12 ft / (x + 25 ft)

Simplifying the equation gives:

17.5x + 437.5 = 12x + 300

5.5x = 137.5

x = 25 ft

Now, let's differentiate both sides of the equation with respect to time t:

17.5 (dx/dt) = 12 (dx/dt) + 0

17.5 (dx/dt) - 12 (dx/dt) = 0

5.5 (dx/dt) = 0

dx/dt = 0 ft/sec

So, the rate at which the tip of the shadow is moving away from the pole when the person is 25 ft from the pole is 0 ft/sec.

b. Rate the tip of the shadow is moving away from the person:

Using the similar triangles, we have:

(12 ft + 5.5 ft) / x = 5.5 ft / 25 ft

Simplifying the equation gives:

17.5x = 137.5

x = 7.857 ft

Now, let's differentiate both sides of the equation with respect to time t:

17.5 (dx/dt) = 5.5 (dx/dt) + d(7.857)/dt

12 (dx/dt) = d(7.857)/dt

dx/dt = (d(7.857)/dt) / 12

dx/dt = (0 ft/sec) / 12

So, the rate at which the tip of the shadow is moving away from the person when the person is 25 ft from the pole is 0 ft/sec.

Answer 2

A) 48/13 ft/sec

B) 22/13 ft/sec

Step-by-step explanation: Given that:

A light is on the top of a 12 ft tall pole and a 5ft 6in tall person is walking away from the pole at a rate of 2 ft/sec.

A) At what rate is the tip of the shadow moving away from the pole when the person is 25 ft from the pole?

Using similar triangles, we can say:

12/L = 55/(L - x)

Cross multiply to get:

12(l - x) = 5.5l

12l - 12x = 5.5l

6.5l = 12x

x = (6.5/12)l

Taking the derivative with respect to time we get:

dx/dt = (6.5/12)dl/dt or

dl/dt = (12/6.5)(dx/dt)

Since dx/dt = 2 so

dl/dt = (12/6.5)(2)

= 24/6.5

= 48/13 ft/sec

B) At what rate is the tip of the shadow moving away from the person when the person is 25 ft from the pole ?

Subtract the rate the shadow is going from the rate the man is going. Therefore

48/13 - 2 = 22/13 ft/sec.