A ladder 25 feet is leaning against the wall of the house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second.
how fast is the top of the ladder moving down the wall when the base is the ladder is 7 feet away 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 is changing when the base of the ladder is 7 feet 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 7 feet from the wall.
To find the rate at which the top of the ladder is moving down the wall when the base is 7 feet away, we can use the concept of related rates.
Let's denote the distance between the wall and the top of the ladder as y, and the distance between the wall and the base of the ladder as x. We know that y is decreasing as the base of the ladder is pulled away from the wall. We also know that the ladder has a fixed length of 25 feet.
Using the Pythagorean theorem, we have the equation: x^2 + y^2 = 25^2.
Taking the derivative with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0.
We are given that dx/dt = 2 ft/s (the rate at which the base of the ladder is pulled away), and we want to find dy/dt when x = 7.
Plugging in these values, we get:
2(7)(2) + 2y(dy/dt) = 0.
Simplifying the equation, we have:
28 + 2y(dy/dt) = 0.
Now, we need to solve for dy/dt. Rearranging the equation, we get:
2y(dy/dt) = -28.
Dividing both sides by 2y, we find:
dy/dt = -28 / (2y).
To find the value of y when x = 7, we can substitute this value into the equation x^2 + y^2 = 25^2:
7^2 + y^2 = 25^2.
49 + y^2 = 625.
y^2 = 625 - 49.
y^2 = 576.
y = sqrt(576).
y = 24 feet.
Substituting y = 24 into the equation for dy/dt, we get:
dy/dt = -28 / (2*24) = -7/12 ft/s.
Therefore, the top of the ladder is moving down the wall at a rate of -7/12 ft/s when the base is 7 feet away.
Now, let's move on to the second part of the question, i.e., 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 the triangle is given by A = (1/2) * x * y.
Differentiating both sides with respect to time, we get:
dA/dt = (1/2) * (x * dy/dt + y * dx/dt).
Substituting the given values x = 7, y = 24, dx/dt = 2, and dy/dt = -7/12, we have:
dA/dt = (1/2) * (7 * (-7/12) + 24 * 2).
Simplifying the expression, we find:
dA/dt = (1/2) * (-49/12 + 48) = (1/2) * (-49/12 + 576/12) = (1/2) * 527/12.
Therefore, the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall is 527/24 square feet per second.
Finally, let's determine 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.
Since the triangle formed by the ladder, the side of the house, and the ground is a right triangle, we can use the tangent function to relate the angle θ to the sides of the triangle.
We know that tan(θ) = y/x.
Differentiating both sides of the equation with respect to time, we get:
(sec^2(θ)) * (dθ/dt) = (dy/dt * x - y * dx/dt) / (x^2).
Substituting the given values x = 7, y = 24, dx/dt = 2, and dy/dt = -7/12, we have:
(sec^2(θ)) * (dθ/dt) = ((-7/12) * 7 - 24 * 2) / (7^2).
Simplifying the expression, we find:
(sec^2(θ)) * (dθ/dt) = (-49/12 - 48) / 49 = -145/588.
To find (dθ/dt), we need to divide both sides of the equation by (sec^2(θ)):
dθ/dt = (-145/588) / (sec^2(θ)).
Now, we need to determine the value of (sec^2(θ)) when x = 7 and y = 24:
sec^2(θ) = 1 + tan^2(θ) = 1 + (y/x)^2 = 1 + (24/7)^2.
Substituting this value along with the rate (dθ/dt), we have:
dθ/dt = (-145/588) / (1 + (24/7)^2).
Evaluate this expression to find the rate at which the angle between the ladder and the wall of the house is changing.
Note: If a decimal approximation is needed, substitute the known values into the expressions and calculate the result.
given:
x^2 + y^2 = 25^2
dx/dt = 2
when x=7, y=25
x dx/dt + y dy/dt = 0
plug in your numbers to find dy/dt
a = 1/2 xy
da/dt = 1/2 (y dx/dt + x dy/dt)
tanθ = y/x
sec^2θ dθ/dt = (x dy/dt - y dx/dt)/y^2