A ladder leans against a vertical wall and the top of the ladder is sliding down the wall at a
constant rate of 1/2 ft/sec. At the moment when the top of the ladder is 16 feet above the
ground, the bottom of the ladder is sliding away from the wall (horizontally) at the rate of
2/3 ft/sec. At what rate will the bottom of the ladder be sliding away from the wall when the
top of the ladder is 12 feet above the ground?
if the ladder base is x feet from the wall and reaches y feet high and the ladder length is a,
x^2+y^2 = a^2
2x dx/dt + 2y dy/dt = 0
we are told that dy/dt = -1/2
so, when y=16,
2x (2/3) + 2(16)(-1/2) = 0
x=12
so, x^2+y^2 = 12^2+16^2 = 20^2
when y=12, x=16, and we have
2(16) dx/dt + 2(12)(-1/2) = 0
dx/dt = 3/8
To solve this problem, we can use related rates.
Let's denote the height of the ladder as "y" and the distance from the wall to the bottom of the ladder as "x".
Given: dy/dt = -1/2 ft/sec (the top of the ladder is sliding down the wall), and dx/dt = 2/3 ft/sec (the bottom of the ladder is sliding away from the wall horizontally).
We need to find: dx/dt when y = 12 ft.
To begin, we can relate the variables x, y, and the length of the ladder, L:
L^2 = x^2 + y^2 .... (Equation 1)
Differentiating both sides of Equation 1 with respect to time (t) using implicit differentiation, we get:
2L(dL/dt) = 2x(dx/dt) + 2y(dy/dt)
Simplifying the equation, we have:
L(dL/dt) = x(dx/dt) + y(dy/dt) .... (Equation 2)
Since the problem states that the top of the ladder is sliding down the wall at a constant rate of 1/2 ft/sec, we have dy/dt = -1/2 ft/sec.
Substituting the given values into Equation 2, we have:
L(dL/dt) = x(dx/dt) + y(-1/2)
Since the length of the ladder, L, remains constant, we have dL/dt = 0. Therefore, Equation 2 reduces to:
0 = x(dx/dt) - (1/2)y .... (Equation 3)
We are given that y = 16 ft when the top of the ladder is 16 ft above the ground. Substituting this value into Equation 3, we have:
0 = x(dx/dt) - (1/2)(16) .... (Equation 4)
Simplifying Equation 4, we have:
0 = x(dx/dt) - 8
Rearranging the equation to solve for dx/dt, we get:
x(dx/dt) = 8
Dividing both sides by x, we have:
dx/dt = 8/x
We need to find dx/dt when y = 12 ft. To find the value of x when y = 12 ft, we can use Equation 1:
L^2 = x^2 + y^2
Squaring both sides, we have:
L^2 = x^2 + (12^2)
Substituting the given value L = 16 ft and y = 12 ft, we have:
16^2 = x^2 + 12^2
256 = x^2 + 144
x^2 = 256 - 144
x^2 = 112
x = √112 (taking the positive square root since x represents a distance)
x ≈ 10.583 ft
Substituting the value of x into dx/dt = 8/x, we have:
dx/dt ≈ 8/10.583
dx/dt ≈ 0.757 ft/sec
Therefore, when the top of the ladder is 12 ft above the ground, the bottom of the ladder will be sliding away from the wall at a rate of approximately 0.757 ft/sec.
To solve this problem, we can use related rates. The key is to set up a right triangle and assign variables to the sides of the triangle. Let's call the distance between the bottom of the ladder and the wall x, and the distance from the bottom of the ladder to the ground y.
We are given that dy/dt = -2/3 ft/sec, as the bottom of the ladder is sliding away from the wall horizontally. We want to find dx/dt, the rate at which the bottom of the ladder is sliding away from the wall when y = 12 ft.
We also know that dy/dt = -2/3 ft/sec, so dy/dt = -2/3. By Pythagoras' theorem, we have:
x^2 + y^2 = 16^2
Differentiating both sides of this equation implicitly with respect to time t, we get:
2x dx/dt + 2y dy/dt = 0
Substituting the known values, we have:
2x dx/dt + 2(12)(-2/3) = 0
Simplifying the equation, we have:
2x dx/dt - 8 = 0
Now we can solve for dx/dt:
2x dx/dt = 8
dx/dt = 8 / (2x)
We are given that y = 12 ft, so substituting this into the equation, we have:
dx/dt = 8 / (2x)
= 4 / x
So when y = 12 ft, the rate at which the bottom of the ladder is sliding away from the wall is dx/dt = 4 / x.
To find the specific rate at that moment, we need to determine x. Using Pythagoras' theorem, we can substitute the value of y = 12 ft into the equation:
x^2 + 12^2 = 16^2
Simplifying the equation, we have:
x^2 + 144 = 256
x^2 = 112
x = sqrt(112)
Now we can substitute this value back into the equation for dx/dt:
dx/dt = 4 / x
= 4 / sqrt(112)
Therefore, when the top of the ladder is 12 feet above the ground, the bottom of the ladder will be sliding away from the wall at a rate of 4 / sqrt(112) ft/sec.