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
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
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.
To solve these problems, we can apply trigonometry and calculus concepts. Let's go step-by-step:
1. How fast is the top moving down the wall when the base of the ladder is 7 feet from the wall?
Let's assume the distance between the top of the ladder and the ground is 'h' feet. To solve this problem, we need to find the rate at which 'h' is changing with respect to time. We can use the Pythagorean theorem to relate the variables:
h^2 + 7^2 = 25^2
Differentiating both sides of the equation with respect to time, we get:
2h(dh/dt) = 0 + 2(7)(7(d7/dt))
Simplifying, we have:
2h(dh/dt) = 2(7)(7(d7/dt))
Now we need to substitute the known values:
h = sqrt(25^2 - 7^2) = sqrt(576) = 24 feet
d7/dt = -2 feet/second (since the base is being pulled away from the wall)
Plugging in these values, we can calculate the rate of change:
2(24)(dh/dt) = 2(7)(7(-2))
48(dh/dt) = -196
dh/dt = -196/48
dh/dt ≈ -4.08 feet/second
So, when the base of the ladder is 7 feet from the wall, the top is moving down the wall at a rate of approximately 4.08 feet/second.
2. How fast is the top moving down the wall when the base of the ladder is 15 feet from the wall?
Using the same method as above, we can calculate the height 'h' when the base is 15 feet from the wall:
h = sqrt(25^2 - 15^2) = sqrt(400) = 20 feet
Since the rate of change of the base distance remains the same (-2 feet/second), we can use the formula:
dh/dt = (7)(7)(-2)/h = -14h/20
Plugging in the value of 'h', we get:
dh/dt = -14(20)/20
dh/dt = -14 feet/second
So, when the base of the ladder is 15 feet from the wall, the top is moving down the wall at a rate of 14 feet/second.
3. How fast is the top moving down the wall when the base of the ladder is 24 feet from the wall?
Using the same method as above, we can calculate the height 'h' when the base is 24 feet from the wall:
h = sqrt(25^2 - 24^2) = sqrt(7) = 7 feet
Since the rate of change of the base distance remains the same (-2 feet/second), we can use the formula:
dh/dt = (7)(7)(-2)/h = -14h/7
Plugging in the value of 'h', we get:
dh/dt = -14(7)/7
dh/dt = -14 feet/second
So, when the base of the ladder is 24 feet from the wall, the top is moving down the wall at a rate of 14 feet/second.
4. Rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall:
The area of the triangle formed by the ladder, the ground, and the side of the house can be calculated as:
A = (1/2)(base)(height)
Here, the base is given as 7 feet. We can substitute this value into the area formula:
A = (1/2)(7)(h)
To find the rate of change of the area (dA/dt), we need to differentiate the area formula with respect to time (t):
dA/dt = (1/2)(7)(dh/dt)
Substituting the known values:
dh/dt = -2 feet/second (rate of change of the base)
dA/dt = (1/2)(7)(-2)
dA/dt = -7 square feet/second
So, when the base of the ladder is 7 feet from the wall, the area of the triangle is changing at a rate of -7 square feet/second.
5. 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:
To find the rate at which the angle between the ladder and the wall is changing, we need to differentiate the equation:
tan(theta) = h/7
Differentiating both sides with respect to time, we get:
sec^2(theta)(d(theta)/dt) = (dh/dt)/7
Substituting the known values:
theta = tan^(-1)(h/7)
dh/dt = -2 feet/second (rate of change of the base)
Now, we need to determine the value of sec^2(theta) to calculate d(theta)/dt. Using the identity sec^2(theta) = 1 + tan^2(theta):
sec^2(theta) = 1 + (h/7)^2
sec^2(theta) = 1 + (h^2)/49
Substituting the sec^2(theta) expression, we can solve for d(theta)/dt:
(1 + (h^2)/49)(d(theta)/dt) = (-2)/7
d(theta)/dt = (-2)/(7 + (h^2)/49)
Plugging in the value of 'h' (computed in the first question) into the above equation, we get:
d(theta)/dt = (-2)/(7 + (24^2)/49)
d(theta)/dt = -2/23 radians/second
So, when the base of the ladder is 7 feet from the wall, the angle between the ladder and the wall is changing at a rate of -2/23 radians/second.