A ladder 15 feet long leans against a vertical wall. Supppose that when the bottom of the ladder is x feet from the wall, the bottom is being pushed towards the wall at the rate of 1/2x feet per second. How fast is the top of the ladder rising at the moment the top is 5 feet from the floor?

let the ladder be y ft above the ground

x^2 + y^2 = 15^2

when foot is x ft from wall
y^2 = 225-x^2
y = (225-x^2)^(1/2)

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

= -2x(1/2)x/√(225 - x^2)
= -x^2/√(225-x^2)

07

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To solve this problem, we can use related rates. Let's define the following variables:

- Let x represent the distance from the bottom of the ladder to the wall.
- Let y represent the distance from the floor to the top of the ladder.
- Let t represent time.

We are given that the bottom of the ladder is being pushed towards the wall at the rate of 1/2x feet per second. This means dx/dt = 1/2x.

We want to find dy/dt, the rate at which the top of the ladder is rising. We can start by using the Pythagorean Theorem to relate x, y, and the length of the ladder:

x^2 + y^2 = 15^2

Differentiating both sides of this equation with respect to t (time), we get:

2x(dx/dt) + 2y(dy/dt) = 0 [using the chain rule]

Now, we can substitute dx/dt = 1/2x into the equation and solve for dy/dt:

2x(1/2x) + 2y(dy/dt) = 0

1 + 2y(dy/dt) = 0

2y(dy/dt) = -1

dy/dt = -1 / (2y)

We want to find the rate at the moment the top is 5 feet from the floor, so we substitute y = 5 into the equation:

dy/dt = -1 / (2(5))
= -1 / 10
= -1/10 ft/s

Therefore, the top of the ladder is rising at a rate of -1/10 feet per second when it is 5 feet from the floor.

Note: The negative sign indicates that the top of the ladder is actually descending.