A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground at rate 2 cm/s . How fast is its height on wall decreasing when foot of ladder I'd 4 m
To solve this problem, we can apply the concept of related rates.
Let's assign variables to the given information:
- Let x represent the distance between the foot of the ladder and the wall (in meters).
- Let y represent the height of the ladder on the wall (in meters).
We are given that dx/dt (the rate at which x is changing) is 2 cm/s (centimeters per second) and the value of x is 4 m (meters).
We need to find dy/dt (the rate at which y is changing) when x = 4 m.
To determine the relationship between x and y, we can use the Pythagorean theorem:
x^2 + y^2 = 5^2 (since the ladder is 5 m long)
Differentiating both sides of the equation with respect to time (t), we get:
2x * dx/dt + 2y * dy/dt = 0
Simplifying the equation, we have:
x * dx/dt + y * dy/dt = 0
Since we want to find dy/dt when x = 4 m, we substitute the given values into the equation:
4 * 2 cm/s + y * dy/dt = 0
8 cm/s + y * dy/dt = 0
Now, we need to solve for dy/dt:
y * dy/dt = -8 cm/s
dy/dt = -8 cm/s / y
To convert cm/s to m/s, we divide by 100:
dy/dt = -0.08 m/s / y
Substituting the value of x = 4 m into the Pythagorean theorem equation, we can solve for y:
4^2 + y^2 = 5^2
16 + y^2 = 25
y^2 = 25 - 16
y^2 = 9
y = 3 m
Substituting the value of y = 3 m into the expression for dy/dt, we can find the rate at which the height on the wall is decreasing:
dy/dt = -0.08 m/s / (3 m)
dy/dt = -0.0267 m/s
Therefore, the height on the wall is decreasing at a rate of approximately 0.0267 m/s.