A 13 m long ladder is leaning against a wall, touching the wall at a certain height from the ground level. The bottom of the ladder is pulled away from the wall, along the ground, at the rate of 2 m/s. How fast is the height on the wall decreasing when the foot of the ladder is 5 m away from the wall ?

Answer

To solve the problem step-by-step, we will follow the reasoning laid out in the video transcript.

### Step 1: Understand the Problem
We have a ladder of length 13 m leaning against a wall. The bottom of the ladder is being pulled away from the wall at a rate of 2 m/s. We need to find out how fast the height at which the ladder touches the wall is decreasing when the foot of the ladder is 5 m away from the wall.

### Step 2: Draw a Diagram
Draw a right triangle where:
- The ladder is the hypotenuse (length = 13 m).
- The distance from the wall to the foot of the ladder is the base (let's denote it as \( x \)).
- The height at which the ladder touches the wall is the vertical side (let's denote it as \( y \)).

### Step 3: Apply the Pythagorean Theorem
According to the Pythagorean theorem:
\[ x^2 + y^2 = 13^2 \]
This simplifies to:
\[ x^2 + y^2 = 169 \]

### Step 4: Differentiate with Respect to Time
Since both \( x \) and \( y \) are changing with respect to time, we differentiate the equation with respect to \( t \):
\[ \frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(169) \]
Using the chain rule:
\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]
Dividing through by 2 gives:
\[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \]

### Step 5: Rearrange the Equation
Rearranging the equation gives us:
\[ y \frac{dy}{dt} = -x \frac{dx}{dt} \]
Thus,
\[ \frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt} \]

### Step 6: Substitute Known Values
We know:
- \( x = 5 \) m (the distance from the wall when we want to find the rate).
- \( \frac{dx}{dt} = 2 \) m/s (the rate at which the foot of the ladder is moving away from the wall).

Now we need to find \( y \) when \( x = 5 \):
Using the Pythagorean theorem:
\[ 5^2 + y^2 = 13^2 \]
\[ 25 + y^2 = 169 \]
\[ y^2 = 169 - 25 = 144 \]
\[ y = 12 \text{ m} \]

### Step 7: Calculate \( \frac{dy}{dt} \)
Now substitute \( x \), \( y \), and \( \frac{dx}{dt} \) into the equation:
\[ \frac{dy}{dt} = -\frac{5}{12} \cdot 2 \]
\[ \frac{dy}{dt} = -\frac{10}{12} = -\frac{5}{6} \text{ m/s} \]

### Conclusion
The height on the wall is decreasing at a rate of \( \frac{5}{6} \) m/s when the foot of the ladder is 5 m away from the wall.