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
To solve these problems, we can use the concepts of related rates and the Pythagorean theorem.
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's denote the distance between the base of the ladder and the wall as x (in feet). We are given that dx/dt = -2 ft/s (negative because the base is moving away from the wall).
Using the Pythagorean theorem, we know that the height h (in feet) of the ladder on the wall satisfies the equation:
x^2 + h^2 = 25^2
Differentiating both sides with respect to time t, we get:
2x(dx/dt) + 2h(dh/dt) = 0
Plugging in the given values, we get:
2(7)(-2) + 2h(dh/dt) = 0
-28 + 2h(dh/dt) = 0
Solving for dh/dt, we have:
2h(dh/dt) = 28
dh/dt = 28/(2h)
dh/dt = 14/h
When x = 7 feet, we can find the height h by substituting into the equation:
7^2 + h^2 = 25^2
49 + h^2 = 625
h^2 = 576
h = 24 feet
Therefore, when the base of the ladder is 7 feet from the wall, the top is moving down the wall at a rate of (14/24) feet per second.
Similarly, when the base is 15 feet from the wall, we find h = 20 feet. Thus, the top is moving down the wall at a rate of (14/20) feet per second.
When the base is 24 feet from the wall, we find h = 7 feet. Thus, the top is moving down the wall at a rate of (14/7) feet per second, or 2 feet per second.
2. To find the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall, we use the formula for the area of a triangle:
Area = (1/2) * base * height
Let's denote the area as A and the height as h. We already know that the height when the base is 7 feet from the wall is 24 feet.
Differentiating both sides with respect to time t, we get:
dA/dt = (1/2) * (db/dt) * h + (1/2) * b * (dh/dt)
But db/dt is given as -2 ft/s, and dh/dt is calculated earlier as 14/24 ft/s.
Substituting the values, we have:
dA/dt = (1/2)(-2)(24) + (1/2)(7)(14/24)
Simplifying, we get:
dA/dt = -48/2 + 98/48
dA/dt = -24 + 49/24
dA/dt = (49 - 48)/24
dA/dt = 1/24 square feet per second
Therefore, when the base of the ladder is 7 feet from the wall, the area of the triangle is changing at a rate of 1/24 square feet per second.
3. To find the rate at which the angle between the ladder and the wall is changing when the base of the ladder is 7 feet from the wall, we use the trigonometric relationship between the angle and the sides of the triangle.
Let's denote the angle as θ (in radians). We know that the tangent of θ is equal to the height h divided by the base x:
tan(θ) = h/x
Differentiating both sides with respect to time t using the chain rule, we get:
sec^2(θ) * (dθ/dt) = (dh/dt)/x - h(dx/dt)/x^2
Plugging in the given values:
sec^2(θ) * (dθ/dt) = (14/24)/(7) - (24)(-2)/(7^2)
Simplifying, we get:
sec^2(θ) * (dθ/dt) = 7/24 + 48/49
Taking the square root of both sides and simplifying further, we have:
(dθ/dt) = sqrt((7/24 + 48/49)/(sec^2(θ)))
To find the value of sec^2(θ), we use the Pythagorean theorem:
x^2 + h^2 = 25^2
7^2 + 24^2 = 25^2
49 + 576 = 625
625 = 625
Therefore, sec^2(θ) = 1, which means that (dθ/dt) = sqrt((7/24 + 48/49)/1)
Simplifying further, we get:
(dθ/dt) = sqrt(49/24)
Taking the square root, we have:
(dθ/dt) = 7/2 radians per second
Therefore, 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 7/2 radians per second.
To solve these questions, let's denote the distance of the base from the wall as x (in feet) and the distance of the top of the ladder from the ground as y (in feet).
1. To find how fast the top is moving down the wall, we can use the Pythagorean theorem:
x^2 + y^2 = 25^2
Differentiating both sides of the equation with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0 [since 25^2 is a constant]
Rearranging the equation, we have:
dy/dt = -(x/y)(dx/dt)
Now, we can substitute the given values of x to find the corresponding values of dy/dt:
i. When x = 7 feet,
dy/dt = -(7/y)(2 feet/second)
ii. When x = 15 feet,
dy/dt = -(15/y)(2 feet/second)
iii. When x = 24 feet,
dy/dt = -(24/y)(2 feet/second)
2. To find the rate at which the area of the triangle is changing, we need to use the formula for the area of a triangle:
Area = (1/2)xy
Differentiating both sides of the equation with respect to time (t), we get:
d(Area)/dt = (1/2)((dx/dt)y + x(dy/dt))
We know that dx/dt = 2 feet/second (the base of the ladder being pulled away from the wall at this rate).
To find the value of dy/dt when x = 7 feet, we can use the previously derived formula for dy/dt and substitute the given values.
Substituting these values, we can calculate d(Area)/dt.
3. To find the rate at which the angle between the ladder and the wall is changing, we can use the trigonometric relationship for a right-angled triangle:
tan(theta) = y/x
Differentiating both sides of the equation with respect to time (t), we get:
sec^2(theta)(d(theta)/dt) = (x(dy/dt) - y(dx/dt)) / x^2
We know that dx/dt = 2 feet/second and the value of dy/dt when x = 7 feet.
Substituting these values, we can calculate d(theta)/dt.