A ladder 25feet long is 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 related rates. Let's set up some variables:
- Let x represent the distance between the base of the ladder and the wall.
- Let y represent the distance between the top of the ladder and the ground.
- Let A represent the area of the triangle formed by the ladder, the wall, and the ground.
- Let θ represent the angle between the ladder and the wall.
1) How fast is the top moving down the wall when the base of the ladder is 7 feet from the wall?
The ladder, the wall, and the ground form a right triangle. Using the Pythagorean theorem, we can write:
x^2 + y^2 = 25^2
Differentiating both sides with respect to time t, we get:
2x(dx/dt) + 2y(dy/dt) = 0
We know that dx/dt = 2 ft/s (given), and we need to find dy/dt when x = 7 ft.
Plugging in the given values:
2(7)(2) + 2y(dy/dt) = 0
14 + 2y(dy/dt) = 0
2y(dy/dt) = -14
dy/dt = -14/(2y)
Using the Pythagorean theorem, we can solve for y when x = 7 ft:
7^2 + y^2 = 25^2
49 + y^2 = 625
y^2 = 576
y = 24 ft
Now we can calculate dy/dt:
dy/dt = -14/(2 * 24)
dy/dt = -7/24 ft/s
Therefore, the top is moving down the wall at a rate of -7/24 ft/s when the base of the ladder is 7 feet from the wall.
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 approach as before, we can calculate y when x = 15 ft:
15^2 + y^2 = 25^2
225 + y^2 = 625
y^2 = 400
y = 20 ft
Using the same related rates equation:
2x(dx/dt) + 2y(dy/dt) = 0
2(15)(2) + 2(20)(dy/dt) = 0
60 + 40(dy/dt) = 0
40(dy/dt) = -60
dy/dt = -60/40
dy/dt = -3/2 ft/s
Therefore, the top is moving down the wall at a rate of -3/2 ft/s when the base of the ladder is 15 feet from the wall.
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 approach as before, we can calculate y when x = 24 ft:
24^2 + y^2 = 25^2
576 + y^2 = 625
y^2 = 49
y = 7 ft
Using the same related rates equation:
2x(dx/dt) + 2y(dy/dt) = 0
2(24)(2) + 2(7)(dy/dt) = 0
96 + 14(dy/dt) = 0
14(dy/dt) = -96
dy/dt = -96/14
Therefore, the top is moving down the wall at a rate of -96/14 ft/s when the base of the ladder is 24 feet from the wall.
4) Finding the 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 a triangle can be calculated using the formula A = (1/2) * base * height.
In this case, the base is x = 7 ft and the height is y = 24 ft (from previous calculation).
A = (1/2) * 7 * 24
A = 84 sq.ft
Differentiating both sides with respect to time t, we get:
dA/dt = (1/2)(7)(dy/dt) + (1/2)(24)(dx/dt)
We know dy/dt = -7/24 ft/s (from previous calculation) and dx/dt = 2 ft/s (given).
Plugging in the values:
dA/dt = (1/2)(7)(-7/24) + (1/2)(24)(2)
dA/dt = -49/48 + 24
dA/dt = 24 - 49/48
dA/dt = 24 - 1.02
dA/dt ≈ 22.98 sq.ft/s
Therefore, the rate at which the area of the triangle is changing is approximately 22.98 sq.ft/s when the base of the ladder is 7 feet from the wall.
5) Finding 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.
To find the angle, we can use the trigonometric function cosine.
Cos(θ) = x/25
θ = arccos(x/25)
Differentiating both sides with respect to time t, we get:
d(θ)/dt = (1/sqrt(1 - (x/25)^2))(dx/dt)
We know dx/dt = 2 ft/s (given) and x = 7 ft (given).
Plugging in the values:
d(θ)/dt = (1/sqrt(1 - (7/25)^2))(2)
d(θ)/dt = (1/sqrt(1 - 49/625))(2)
d(θ)/dt = (1/sqrt(576/625))(2)
d(θ)/dt = (1/(24/25))(2)
d(θ)/dt = (25/24)(2)
d(θ)/dt = 25/12 rad/s
Therefore, the rate at which the angle between the ladder and the wall of the house is changing is 25/12 rad/s when the base of the ladder is 7 feet from the wall.
To solve these problems, we can use the concept of related rates. Here's how we can approach each question step-by-step:
1. Determine the relationship between the variables:
Let x represent the distance between the base of the ladder and the wall (in feet), and y represent the distance between the top of the ladder and the ground (in feet). We know that the ladder is fixed at 25 feet, so we have the equation: x^2 + y^2 = 25^2.
2. Determine the rates:
We are given that dx/dt = 2 feet per second, which represents the rate at which x is changing. We need to find dy/dt, which represents the rate at which y is changing.
3. Find dy/dt when x = 7 feet:
Using the equation x^2 + y^2 = 25^2, we can differentiate both sides of the equation with respect to time (t) and solve for dy/dt.
2x(dx/dt) + 2y(dy/dt) = 0
2(7)(2) + 2y(dy/dt) = 0
28 + 2y(dy/dt) = 0
2y(dy/dt) = -28
dy/dt = -28/(2y)
We can substitute the value of x = 7 into the original equation to find y:
7^2 + y^2 = 25^2
49 + y^2 = 625
y^2 = 576
y = 24 (taking the positive value)
Now, substitute y = 24 into dy/dt = -28/(2y):
dy/dt = -28/(2*24)
dy/dt = -7/12 feet per second
Therefore, when the base of the ladder is 7 feet from the wall, the top is moving downwards at a rate of -7/12 feet per second.
4. Find dy/dt when x = 15 feet:
Using the same process, we can find y when x = 15 and calculate dy/dt.
5. Find dy/dt when x = 24 feet:
Again, using the same process, we can find y when x = 24 and calculate dy/dt.
Now let's move on to the next question:
6. Find the 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 is given by A = (1/2) * x * y.
To find the rate at which the area is changing, we need to differentiate A with respect to time (t) and solve for dA/dt.
dA/dt = (1/2) * (x * dy/dt + y * dx/dt)
Given the values of x = 7, y = 24, dx/dt = 2, and dy/dt = -7/12 (from previous calculations), we can substitute them into the equation to find dA/dt.
Finally, let's move on to the last question:
7. 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:
To find the rate at which the angle is changing, we need to differentiate the equation x^2 + y^2 = 25^2 with respect to time (t) and solve for dt/dθ (the rate of change of t with respect to θ).
2x(dx/dt) + 2y(dy/dt) = 0
2x(dx/dt) = -2y(dy/dt)
(dx/dt) = -y(dy/dt)/x
Again, substitute the values of y = 24, dy/dt = -7/12, and x = 7 into the equation to find dx/dt. Additionally, note that dx/dt is equal to -dy/dt.
I hope this helps you solve the problem step-by-step!