Answer:
Dx/dt = 4,8 f/s
Explanation:
The ladder placed against a wall, and the ground formed a right triangle
with x and h the legs and L the hypothenuse
Then
L² = x² + h² (1)
L = 26 f
Taking differentials on both sides of the equation we get
0 = 2x Dx/dt + 2h Dh/dt (1)
In this equation
x = 10 distance between the bottom of the ladder and the when we need to find, the rate of the ladder moving away from the wall
Dx/dt is the rate we are looking for
h = ? The height of the ladder when x = 10
As L² = x² + h²
h² = L² - x²
h² = (26)² - (10)²
h² = 676 - 100
h² = 576
h = 24 f
Then equation (1)
0 = 2x Dx/dt + 2h Dh/dt
2xDx/dt = - 2h Dh/dt
10 Dx/dt = - 24 ( -2 ) ( Note the movement of the ladder is downwards)
Dx/dt = 48/10
Dx/dt = 4,8 f/s
