Final answer:
To find the rate at which the base of the ladder is sliding away from the wall, we can use related rates of change. Given that the rate at which the top of the ladder is sliding down the wall is 6 ft/sec, we can solve for dx/dt when x = 15 ft. By forming a right triangle with the ladder, the wall, and the floor, and using the Pythagorean theorem and implicit differentiation, we can determine that dx/dt is approximately 3.2 ft/sec.
Step-by-step explanation:
To find the rate at which the base of the ladder is sliding away from the wall, we can use related rates of change. Let's denote the distance of the base from the wall as x and the height of the ladder as y. We are given that dy/dt (the rate at which the top of the ladder is sliding down the wall) is 6 ft/sec and we want to find dx/dt (the rate at which the base of the ladder is sliding away from the wall) when x = 15 ft.
We can form a right triangle with the ladder, the wall, and the floor. Using the Pythagorean theorem, we have x^2 + y^2 = 17^2 (the ladder length is 17 ft). Differentiating both sides of this equation with respect to time, we get 2x(dx/dt) + 2y(dy/dt) = 0. Plugging in the given values, we can solve for dx/dt when x = 15 ft.
Using the equation, we have 2(15)(dx/dt) + 2y(6) = 0. Since the ladder is sliding down the wall, the height is decreasing. Therefore, dy/dt = -6 ft/sec. Plugging this into the equation, we have 30(dx/dt) - 12y = 0. To solve for dx/dt, we need to find y. Using the Pythagorean theorem, we can find y when x = 15 ft.
Using the equation x^2 + y^2 = 17^2, we have (15)^2 + y^2 = (17)^2. Solving for y, we get y = 8 ft. Plugging this value into the previous equation, we have 30(dx/dt) - 12(8) = 0. Solving for dx/dt, we find that dx/dt ≈ 3.2 ft/sec.