A 13 ft ladder is leaning against a house when its base begins to slide away. By the time the base is 3 ft from the house, the base is moving at the rate of 5 ft/sec. Answer the following:
At what speed is the top of the ladder sliding down the wall at that time?
Answer = ft/sec.
At what rate is the area of the triangle formed by the ladder, wall, and ground changing then?
Answer = ft/sec.
At what rate is the angle (theta) changing then?
Answer = ft/sec.
To find the speed at which the top of the ladder is sliding down the wall, we can use similar triangles. Let's call the distance from the top of the ladder to the ground y at a given time t.
We have a right triangle formed by the ladder, the wall, and the ground. The ladder is the hypotenuse, so its length is always fixed at 13 ft. The distance from the base of the ladder to the wall is changing over time, and we'll call it x.
Using the Pythagorean theorem, we can write the relationship between x, y, and the length of the ladder as:
x^2 + y^2 = 13^2
Now, let's take the derivative of both sides with respect to time t:
2x(dx/dt) + 2y(dy/dt) = 0
Since we know dx/dt (the rate at which the base is moving away from the house) is 5 ft/sec and x = 3 ft, we can solve for dy/dt:
2(3)(5) + 2y(dy/dt) = 0
30 + 2y(dy/dt) = 0
2y(dy/dt) = -30
dy/dt = -15/y
At this point, we need to determine y, the distance from the top of the ladder to the ground, when the base is 3 ft from the house. Again, using the Pythagorean theorem:
x^2 + y^2 = 13^2
3^2 + y^2 = 13^2
9 + y^2 = 169
y^2 = 169 - 9
y^2 = 160
y = sqrt(160)
y ≈ 12.65 ft
Now, substitute y = 12.65 ft into the equation for dy/dt:
dy/dt = -15/12.65
dy/dt ≈ -1.186 ft/sec
Therefore, at that time, the top of the ladder is sliding down the wall at a speed of approximately 1.186 ft/sec.
To find the rate at which the area of the triangle formed by the ladder, wall, and ground is changing, we can use the formula for the area of a triangle: A = (1/2)bh, where b represents the distance from the base of the ladder to the wall and h represents the distance from the top of the ladder to the ground.
Again, differentiating the area formula with respect to time t, we get:
dA/dt = (1/2)(bh)' = (1/2)(b'(h) + b(h)')
To find the values of b and h, we can use the given values:
b = x = 3 ft (the distance from the base of the ladder to the wall)
h = y ≈ 12.65 ft (the distance from the top of the ladder to the ground)
Taking the derivative of b with respect to t, we get:
db/dt = dx/dt = 5 ft/sec
Taking the derivative of h with respect to t, we get:
dh/dt = dy/dt = -1.186 ft/sec (as we calculated earlier)
Now, substitute these values into the equation for rate of change of area:
dA/dt = (1/2)(b'(h) + b(h)')
dA/dt = (1/2)(5(12.65) + 3(-1.186))
dA/dt = (1/2)(63.25 - 3.558)
dA/dt = (1/2)(59.692)
dA/dt ≈ 29.846 ft^2/sec
Therefore, at that time, the rate at which the area of the triangle formed by the ladder, wall, and ground is changing is approximately 29.846 ft^2/sec.
Finally, to find the rate at which the angle between the ladder and the ground is changing, we can use trigonometry. Let's call the angle between the ladder and the ground θ.
Since we know the values of x and y, we can use the trigonometric identity:
tan(θ) = y/x
Differentiating both sides with respect to time t:
sec^2(θ)(dθ/dt) = (dy/dt)(1/x) - (y/x^2)(dx/dt)
We can substitute the known values:
x = 3 ft
y ≈ 12.65 ft
dy/dt = -1.186 ft/sec
dx/dt = 5 ft/sec
Now, substitute these values into the equation and solve for dθ/dt:
sec^2(θ)(dθ/dt) = (-1.186/3) - (12.65/9)(5)
sec^2(θ)(dθ/dt) = (-0.3953) - (7.06)
sec^2(θ)(dθ/dt) = -7.4553
dθ/dt ≈ -1.062 rad/sec
Therefore, at that time, the angle θ is changing at a rate of approximately -1.062 rad/sec.