A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

Answer

To solve the problem step-by-step, we will use the Pythagorean theorem and implicit differentiation. Here’s how we can approach the problem:

### Step 1: Understand the problem and set up the variables
Let:
- \( x \) = the distance from the wall to the bottom of the ladder (in meters)
- \( y \) = the height of the ladder on the wall (in meters)
- The length of the ladder is constant at 5 m.

### Step 2: Use the Pythagorean theorem
According to the Pythagorean theorem, we have:
\[ x^2 + y^2 = 5^2 \]
This simplifies to:
\[ x^2 + y^2 = 25 \]

### Step 3: Differentiate with respect to time
Differentiating both sides with respect to time \( t \):
\[ \frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(25) \]
Using the chain rule, we get:
\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]
Dividing the entire equation by 2 gives:
\[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \]

### Step 4: Substitute known values
We know:
- \( \frac{dx}{dt} = 2 \) cm/s (which is \( 0.02 \) m/s since we need to keep units consistent)
- When the foot of the ladder is 4 m away from the wall, \( x = 4 \) m.

Now, we need to find \( y \) when \( x = 4 \):
\[ 4^2 + y^2 = 25 \implies 16 + y^2 = 25 \implies y^2 = 9 \implies y = 3 \text{ m} \]

### Step 5: Substitute values into the differentiated equation
Now substituting \( x = 4 \), \( y = 3 \), and \( \frac{dx}{dt} = 0.02 \) m/s into the differentiated equation:
\[ 4(0.02) + 3 \frac{dy}{dt} = 0 \]
\[ 0.08 + 3 \frac{dy}{dt} = 0 \]
\[ 3 \frac{dy}{dt} = -0.08 \]
\[ \frac{dy}{dt} = -\frac{0.08}{3} = -\frac{8}{300} = -\frac{8}{3} \text{ cm/s} \]

### Conclusion
The height of the ladder on the wall is decreasing at a rate of \( \frac{8}{3} \) cm/s when the foot of the ladder is 4 m away from the wall.

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