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

Answer

To solve the problem, we need to use the Pythagorean theorem and related rates. Let's break down the steps:

### 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)
- \( L \) = the length of the ladder = 13 m

### Step 2: Apply the Pythagorean theorem
According to the Pythagorean theorem, we have:
\[ x^2 + y^2 = L^2 \]
Substituting the length of the ladder:
\[ x^2 + y^2 = 13^2 \]
\[ x^2 + y^2 = 169 \]

### Step 3: Differentiate with respect to time
To find the rates of change, we differentiate both sides of the equation with respect to time \( t \):
\[ \frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(169) \]
Using the chain rule, we get:
\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]
Dividing through by 2 gives:
\[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \]

### Step 4: Substitute known values
We know:
- \( \frac{dx}{dt} = 2 \) m/s (the rate at which the bottom of the ladder is moving away from the wall)
- We need to find \( \frac{dy}{dt} \) when \( x = 5 \) m.

First, we need to find \( y \) when \( x = 5 \):
\[ 5^2 + y^2 = 169 \]
\[ 25 + y^2 = 169 \]
\[ y^2 = 144 \]
\[ y = 12 \text{ m} \]

Now substitute \( x = 5 \), \( y = 12 \), and \( \frac{dx}{dt} = 2 \) into the differentiated equation:
\[ 5(2) + 12 \frac{dy}{dt} = 0 \]
\[ 10 + 12 \frac{dy}{dt} = 0 \]

### Step 5: Solve for \( \frac{dy}{dt} \)
\[ 12 \frac{dy}{dt} = -10 \]
\[ \frac{dy}{dt} = -\frac{10}{12} = -\frac{5}{6} \text{ m/s} \]

### Conclusion
The height of the ladder 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.

---