1a. 15 ft ladder is placed against a vertical wall. the bottom of the ladder is sliding away from the wall at a rate of 2 ft/sec. How fast is the top of the ladder sliding down the wall when the bottom of the ladder is 9 ft from the wall?
1b. Using the data in the first part, Find the rate of change of the area of the triangle formed by the ladder, the ground, and the wall when the bottom of the ladder is 9ft from the wall.
-1.5ft/sec
To solve both parts of this problem, we can use related rates, which involve finding the rate at which one variable changes with respect to the rate of change of another related variable.
1a. To find how fast the top of the ladder is sliding down the wall, we need to relate the variables involved. Let's call the distance between the bottom of the ladder and the wall x, and let's call the distance between the top of the ladder and the ground y. We are given that dx/dt (the rate at which x changes) is 2 ft/sec. We need to find dy/dt (the rate at which y changes) when x = 9 ft.
Using the Pythagorean theorem, we know that x^2 + y^2 = 15^2, since the ladder forms a right triangle with the wall and the ground. Differentiating both sides of this equation with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0
Now, we can substitute the given values: x = 9 ft and dx/dt = 2 ft/sec:
2(9)(2) + 2y(dy/dt) = 0
Simplifying this equation, we get:
36 + 2y(dy/dt) = 0
Now, solve for dy/dt:
2y(dy/dt) = -36
dy/dt = -36/(2y)
Since we are interested in finding dy/dt when x = 9 ft, we can substitute x = 9 into the Pythagorean theorem equation and solve for y:
9^2 + y^2 = 15^2
81 + y^2 = 225
y^2 = 144
y = 12 ft
Now substitute y = 12 into dy/dt = -36/(2y) to find the rate at which the top of the ladder is sliding down the wall when x = 9 ft:
dy/dt = -36/(2(12))
dy/dt = -36/24
dy/dt = -3/2 ft/sec
Therefore, when the bottom of the ladder is 9 ft from the wall, the top of the ladder is sliding down the wall at a rate of -3/2 ft/sec.
1b. To find the rate of change of the area of the triangle formed by the ladder, the ground, and the wall, we need to use the formula for the area of a triangle: A = (1/2)bh, where b is the base (the distance between the bottom of the ladder and the wall) and h is the height (the distance between the top of the ladder and the ground).
We are given that db/dt = 2 ft/sec (the rate at which the base is changing). We need to find dA/dt (the rate at which the area changes) when b = 9 ft.
Using the formula A = (1/2)bh, we can differentiate both sides of this equation with respect to time (t):
dA/dt = (1/2)(dh/dt)b + (1/2)(db/dt)h
Again, using the Pythagorean theorem, we know that x^2 + y^2 = 15^2. Differentiating both sides of this equation with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since we're interested in finding dh/dt when b = 9 ft, we know that h = 12 ft (as we found in 1a).
Substituting the given values into dA/dt = (1/2)(dh/dt)b + (1/2)(db/dt)h, we get:
dA/dt = (1/2)(dh/dt)(9) + (1/2)(2)(12)
We already found that dh/dt = -3/2 ft/sec in 1a, so we can substitute that value into the equation:
dA/dt = (1/2)(-3/2)(9) + (1/2)(2)(12)
dA/dt = -27/4 + 12
dA/dt = -27/4 + 48/4
dA/dt = 21/4
Therefore, when the bottom of the ladder is 9 ft from the wall, the area of the triangle is changing at a rate of 21/4 square feet per second.