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.