a 5m ladder rests against a vertical wall. The top of the ladder begins sliding down the wall at 2m/sec, while the foot of the ladder moves away from the base of the wall. How fast is the foot of the ladder moving away from the wall at the instant the foot of the ladder is 4m from the base of the wall?
when x=4, y=3
x^2+y^2 = 25
2x dx/dt + 2y dy/dt = 0
2(4) dx/dt + 2(3)(-2) = 0
dx/dt = 12/8 = 1.5 m/s
To solve this problem, we can use the concept of related rates.
Given:
- The ladder is 5m long
- The top of the ladder is sliding down the wall at a rate of 2m/sec
We need to find:
- How fast the foot of the ladder is moving away from the wall at the instant it is 4m from the base of the wall.
Let's assign variables to the quantities involved. Let:
- x be the distance from the foot of the ladder to the base of the wall (this is the value we need to find)
- y be the distance from the top of the ladder to the base of the wall
Now, let's differentiate both sides of the Pythagorean theorem equation that relates x, y, and the length of the ladder (5m) with respect to time (t):
d/dt (x^2 + y^2) = d/dt (5^2)
Differentiating the right side gives us 0 since it is a constant.
Now, differentiate the left side using the chain rule:
2x(dx/dt) + 2y(dy/dt) = 0
At the instant the foot of the ladder is 4m from the base of the wall, we know x = 4m and we are given dy/dt = -2m/sec (since the top of the ladder is sliding down the wall at 2m/sec). We need to find dx/dt, which represents the rate at which the foot of the ladder is moving away from the wall.
Substituting the given values into the equation, we have:
2(4)(dx/dt) + 2y(-2) = 0
Simplifying, we get:
8(dx/dt) - 4y = 0
Since we know y = 5m (the length of the ladder) and we need to find dx/dt, we can substitute these values:
8(dx/dt) - 4(5) = 0
8(dx/dt) - 20 = 0
8(dx/dt) = 20
(dx/dt) = 20/8
(dx/dt) = 2.5 m/sec
Therefore, the foot of the ladder is moving away from the wall at a rate of 2.5m/sec at the instant it is 4m from the base of the wall.