A street light is at the top of a 15 ft pole. A 5 ft tall girl walks along a straight path away from the pole with a speed of 5 ft/sec.

(A)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 28 ft away from the pole? (B)How fast is her shadow lengthening?

To solve this problem, we need to use similar triangles and related rates.

Let's denote the distance between the girl and the pole as x and the length of her shadow as y at any given moment.

(A) To find the rate at which the tip of her shadow is moving away from the light, we need to find dy/dt when x = 28 ft.

We can consider two similar triangles: one formed by the pole, the girl, and her shadow, and another formed by the pole, the length of the shadow, and the total length of the shadow.

Let's set up the proportion between these two triangles:

(15 ft / y) = ((15 + x) ft / (y + 5) ft)

To isolate y in terms of x, we can cross-multiply:

15(y + 5) = (15 + x)y

Expanding and simplifying:

15y + 75 = 15y + xy
75 = xy

Now, we differentiate both sides with respect to time t:

d(75)/dt = d(xy)/dt

Since the length of the shadow is constant over time (we're interested in how fast the tip of the shadow is moving, not its length), the derivative of xy with respect to t is 0. Therefore:

0 = x(dy/dt) + y(dx/dt)

Substituting the given values:

0 = (28 ft)(dy/dt) + (y)(5 ft/sec)

Simplifying:

0 = 28(dy/dt) + 5y

Now, we can solve for dy/dt:

28(dy/dt) = -5y
(dy/dt) = -5y/28

Substituting the value of x as 28:

(dy/dt) = -5y/28 = -5(75)/28 = -375/28 ft/sec

Therefore, the rate at which the tip of her shadow is moving away from the light when the girl is 28 ft away from the pole is -375/28 ft/sec.

(B) To find how fast her shadow is lengthening, we need to find dy/dt when x = 28 ft.

We can differentiate both sides of the equation 75 = xy with respect to t:

d(75)/dt = d(xy)/dt

Again, as the length of the shadow is constant:

0 = x(dy/dt) + y(dx/dt)

Substituting the given values:

0 = (28 ft)(dy/dt) + (y)(5 ft/sec)

Simplifying:

0 = 28(dy/dt) + 5y

Now, we can solve for dy/dt:

28(dy/dt) = -5y
(dy/dt) = -5y/28

Substituting the value of x as 28:

(dy/dt) = -5y/28 = -5(75)/28 = -375/28 ft/sec

Therefore, the rate at which her shadow is lengthening when the girl is 28 ft away from the pole is -375/28 ft/sec.