A ladder 25feet long 8s leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2feet per second. Find

Question

How fast is the top moving down the wall when the base of the ladder is 7feet, 15feet and 24feet from the wall
Consider the triangle formed by the side of the house, the ladder and the ground. Find the rate at which the area of the triangle is changing is changing when the base of the ladder is 7feet from the wall
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 7feet from the wall

asked by Sinachi

Answer 1

To solve this problem, we can use the Pythagorean theorem and trigonometric functions.

1) How fast is the top moving down the wall when the base of the ladder is 7 feet, 15 feet, and 24 feet from the wall?

Let x be the distance of the ladder's base from the wall, and let y be the height of the ladder on the wall.
Using the Pythagorean theorem, we have:

x^2 + y^2 = 25^2

Differentiating both sides with respect to time, we get:

2x(dx/dt) + 2y(dy/dt) = 0

Since dx/dt is given as 2 ft/s, we can solve for dy/dt at different values of x.

a) When x = 7 ft:

7^2 + y^2 = 25^2
49 + y^2 = 625
y^2 = 576
y = 24 ft

Plugging in x = 7 ft and dy/dt = ? into the derived equation, we have:

2(7)(2) + 2(24)(dy/dt) = 0
28 + 48(dy/dt) = 0
48(dy/dt) = -28
dy/dt = -28/48
dy/dt ≈ -0.583 ft/s

Therefore, when the base of the ladder is 7 ft from the wall, the top is moving downward at approximately 0.583 ft/s.

b) When x = 15 ft:

15^2 + y^2 = 25^2
225 + y^2 = 625
y^2 = 400
y = 20 ft

Plugging in x = 15 ft and dy/dt = ? into the derived equation, we have:

2(15)(2) + 2(20)(dy/dt) = 0
60 + 40(dy/dt) = 0
40(dy/dt) = -60
dy/dt = -60/40
dy/dt = -1.5 ft/s

Therefore, when the base of the ladder is 15 ft from the wall, the top is moving downward at 1.5 ft/s.

c) When x = 24 ft:

24^2 + y^2 = 25^2
576 + y^2 = 625
y^2 = 49
y = 7 ft

Plugging in x = 24 ft and dy/dt = ? into the derived equation, we have:

2(24)(2) + 2(7)(dy/dt) = 0
96 + 14(dy/dt) = 0
14(dy/dt) = -96
dy/dt = -96/14
dy/dt ≈ -6.857 ft/s

Therefore, when the base of the ladder is 24 ft from the wall, the top is moving downward at approximately 6.857 ft/s.

2) Find the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall.

Let A be the area of the triangle formed by the ladder, wall, and ground.
The area of a triangle can be calculated as A = (1/2)bh, where b is the base and h is the height.

Using the Pythagorean theorem, when x = 7 ft, we have:

7^2 + y^2 = 25^2
49 + y^2 = 625
y^2 = 576
y = 24 ft

Plugging in x = 7 ft, b = 7 ft, h = 24 ft, and taking the derivative with respect to time, we get:

A = (1/2)(7)(24)
A = 84 sq ft

dA/dt = (1/2)(db/dt)(h) + (1/2)(b)(dh/dt)
dA/dt = (1/2)(2)(24) + (1/2)(7)(dy/dt)
dA/dt = 48 + 7(dy/dt)

Plugging in dy/dt = -0.583 ft/s (from part 1a), we have:

dA/dt = 48 + 7(-0.583)
dA/dt = 48 - 4.081
dA/dt ≈ 43.919 sq ft/s

Therefore, when the base of the ladder is 7 ft from the wall, the area of the triangle is changing at approximately 43.919 sq ft/s.

3) 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 7 feet from the wall.

Let θ be the angle between the ladder and the wall.
Using trigonometric functions, we have:

sin(θ) = y/25
cos(θ) = x/25

Differentiating both equations with respect to time, we get:

cos(θ)(dθ/dt) = dy/dt(1/25)
-sin(θ)(dθ/dt) = dx/dt(1/25)

Using the Pythagorean theorem, when x = 7 ft, we have:

7^2 + y^2 = 25^2
49 + y^2 = 625
y^2 = 576
y = 24 ft

Plugging in x = 7 ft, y = 24 ft, dx/dt = 2 ft/s, and solving for dθ/dt, we have:

cos(θ)(dθ/dt) = dy/dt(1/25)
cos(θ)(dθ/dt) = -0.583(1/25)
cos(θ)(dθ/dt) ≈ -0.02332
dθ/dt ≈ -0.02332/cos(θ)

To find the value of cos(θ), we can use the equation cos(θ) = x/25:

cos(θ) = 7/25
cos(θ) = 0.28

Plugging in cos(θ) = 0.28, we have:

dθ/dt ≈ -0.02332/0.28
dθ/dt ≈ -0.0833 rad/s

Therefore, when the base of the ladder is 7 ft from the wall, the angle between the ladder and the wall of the house is changing at approximately -0.0833 rad/s.