Question

(5 points) A street light is at the top of a 18 ft pole. A 5 ft tall girl walks along a straight path

away from the pole with a speed of 6 ft/sec.
At what rate is the tip of her shadow moving away from the light (ie, away from the top of the
pole) when the girl is 22 ft away from the pole?​

Answer 1

By applying related rates and the concepts of similar triangles, the rate at which the tip of the girl's shadow is moving away from the light when she is 22 ft away from the pole is computed to be approximately 7.2 ft/sec.

Step-by-step explanation:

To solve for the rate at which the tip of the girl's shadow is moving away from the light when the girl is 22 ft away from the pole, we need to apply the principles of similar triangles and the concepts of related rates in calculus. Let's denote the height of the pole by P (18 ft), the height of the girl by G (5 ft), the distance between the girl and the pole by x (which is changing over time), and the length of the shadow by s (also changing over time). Since the girl is moving away from the pole at a speed of 6 ft/sec, this is the rate at which x is increasing, so dx/dt = 6 ft/sec.

Using the similarity of triangles, we have the following proportion:

(P / (x + s)) = (G / s)

18 / (22 + s) = 5 / s

By solving the proportion for s, we can find the rate at which s is changing when x = 22 ft.

Cross-multiplying and rearranging the terms we get:

18s = 5(22 + s)

18s = 110 + 5s

13s = 110

s = 110/13

Now we can differentiate the proportion with respect to time t:

d/dt (P / (x + s)) = d/dt (G / s)

0 = (G * ds/dt) - ((G / s) * dx/dt)

Substitute values for G, s and dx/dt:

0 = (5 * ds/dt) - ((5 / ((110/13))) * 6)

ds/dt = ((5 / ((110/13))) * 6) / 5

ds/dt is the rate at which the tip of the girl's shadow is moving away from the light.

After calculating, we find that ds/dt is approximately 7.2 ft/sec.

Answer 2

2.31 ft/sec

Step-by-step explanation:

In the picture attached, a representation of the problem is shown, where w(t) is the walking distance (as a function of time) and s(t) is the position of the tip of her shadow (as a function of time).

We want to find the derivative of s(t).

From triangles similarity:

s(t)/5 = [w(t) + s(t)]/18

18s(t) = 5w(t) + 5s(t)

13s(t) = 5w(t)

w(t) = 13/5*s(t)

We know that her speed is 6 ft/sec, that is:

d(w(t))/dt = 6 ft/sec

From the previous relationship:

d(w(t))/dt = 13/5*d(s(t))/t

Replacing:

6 = 13/5*d(s(t))/t

d(s(t))/t = 5*6/13

d(s(t))/t ≈ 2.31 ft/sec