Question

a 25 foot ladder leaning against a wall slips so that its base moves away from the wall at a rate of 2 ft/sec. how fast will the top of the ladder be moving down the wall when its base is 9 feet away from the wall?

Answer 1

When the base of the ladder is 9 feet away from the wall, the top of the ladder is moving down the wall at a rate of approximately 0.77 ft/sec.

Step-by-step explanation:

To find the rate at which the top of the ladder is moving down the wall, we need to use similar triangles and apply the concept of related rates. Let's denote the distance of the base of the ladder from the wall as x (in feet) and the distance of the top of the ladder from the ground as y (in feet).

According to the problem, dx/dt (the rate at which x is changing) is given as 2 ft/sec. We need to find dy/dt (the rate at which y is changing) when x = 9 ft.

Using the Pythagorean theorem, we have x^2 + y^2 = 25^2. Differentiating both sides with respect to t (time), we get:

2x(dx/dt) + 2y(dy/dt) = 0

Substituting the given values, we have:

2(9)(2) + 2y(dy/dt) = 0

36 + 2y(dy/dt) = 0

2y(dy/dt) = -36

dy/dt = -36/(2y)

Since we want to find dy/dt when x = 9 ft, we substitute x = 9 into our original equation:

9^2 + y^2 = 25^2

y^2 = 25^2 - 9^2

y = √(625 - 81)

y = √544

y ≈ 23.32 ft

Now, substituting the values of y and dx/dt into our earlier equation:

dy/dt = -36/(2y)

dy/dt = -36/(2(23.32))

dy/dt ≈ -0.77 ft/sec

Therefore, when the base of the ladder is 9 feet away from the wall, the top of the ladder is moving down the wall at a rate of approximately 0.77 ft/sec.