A 17 foot ladder is placed against a wall. if the foot of the ladder is pushed toward the wall at 0.5 feet per second, how fast is the top of the ladder rising when the foot is 8 feet from the wall?
at the moment in question, the height is 15. If the base of the ladder is x from the wall, and the height is y,
x^2+y^2 = 17^2
2x dx/dt + 2y dy/dt = 0
2(8)(-.5) + 2(15) dy/dt = 0
dy/dt = 4/15
To solve this problem, we will use the concept of related rates.
Let's assume that the distance between the foot of the ladder and the wall is represented by "x," and the height of the ladder is represented by "y." We are given the rate at which the foot of the ladder is moving, which is dx/dt = 0.5 ft/s.
By drawing a diagram, we can see that the ladder, the wall, and the ground form a right triangle. The ladder is the hypotenuse, and the foot of the ladder and the wall are the legs of the triangle.
Using the Pythagorean theorem, we know that x^2 + y^2 = 17^2 (since the ladder is 17 feet long).
Differentiating both sides of the equation with respect to time t, we get:
2x(dx/dt) + 2y(dy/dt) = 0
We are given dx/dt = 0.5 ft/s, so the equation becomes:
2x(0.5) + 2y(dy/dt) = 0
Simplifying, we have:
x + y(dy/dt) = 0
We need to find dy/dt when x = 8 feet. Plugging in the values, we have:
8 + y(dy/dt) = 0
To find y, we can use the Pythagorean theorem:
y = √(17^2 - x^2)
Substituting this value into the equation, we have:
8 + √(17^2 - 8^2)(dy/dt) = 0
Solving for dy/dt, we get:
dy/dt = -8 / √(17^2 - 8^2)
Now, we can calculate the value of dy/dt using the equation above:
dy/dt = -8 / √(289 - 64)
dy/dt = -8 / √(225)
dy/dt = -8 / 15
Therefore, when the foot is 8 feet from the wall, the top of the ladder is rising at a rate of -8/15 ft/s. The negative sign indicates that the top of the ladder is actually descending.