A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second.

Question

(b) Consider the triangle formed by the side of the house, ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 20 feet from the wall.

(c) Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 20 feet from the wall.

Answer 1

Let

t be time variable,

t1 be the time instant when the ladder is 20 feet from the wall,

l = 25 ft be the length of the ladder,

b(t) be the base of the ladder,

h(t) be the height of the ladder,

w(t) be the angle between the ladder and the wall of the house,

A(t) be the area of the tringle formed by the ladder,

b'(t) = 2 ft/s be the the derivative db/dt (which is given as a constant),

h' w', A' be the derivatives of h, w, A.

Assuming the normal situation of the house not leaning, and the ground

not sloping, we have a right angle triangle. That gives us the identities

h = (l² - b²)^1/2

tan(w) = b/h (1)

A = (1/2)bh

Below we will need three results concerning h:

h'(t) = (1/2)(l² - b²)^-1/2(-2bb') = -bb'(l² - b²)^-1/2

At time t1, i.e., when b(t1) = 20:

h(t1) = (25² - 20²)^1/2 = 15

h'(t1) = -20×2/15 = -2.7 (approximately)

(b)

The rate of change of A is

A' = (1/2)(b'h + bh')

To calculate A' at the time t1 we substitute actual numbers:

A'(t1) = (1/2)(2×15 + 20×(-2.7)) = -12 ft²/s (approximately)

At the time the ladder is 20 feet from the wall the area is getting reduced

at the rate of about 12 ft²/s.

(c)

From (1)

w = arctan(b/h)

w' =(b'h - bh') / (h²(1 + (b/h)²)

Substituting actual numbers

w'(t1) = (2×15 - 20×(-2.7))/(15²(1 + (20/15)²)) = 0.13 (approximately)

When the base of the ladder is 20 feet from the wall,

the angle between the ladder and the wall of the house is increasing at a rate

of about 0.13 s-1.