A ladder 13 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 2m/sec. 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 use the Pythagorean theorem and implicit differentiation.

### 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/sec. We need to find how fast the height of the ladder on the wall is decreasing when the foot of the ladder is 5 m away from the wall.

### Step 2: Set up the variables
Let:
- \( x \) = distance from the wall to the foot of the ladder (in meters)
- \( y \) = height of the ladder on the wall (in meters)
- The length of the ladder is constant: \( L = 13 \) m

### Step 3: Apply the Pythagorean theorem
According to the Pythagorean theorem:
\[ x^2 + y^2 = L^2 \]
Substituting \( L = 13 \):
\[ x^2 + y^2 = 13^2 = 169 \]

### Step 4: Differentiate with respect to time
Differentiating both sides of the equation with respect to time \( t \):
\[ \frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(169) \]
This gives us:
\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]

### Step 5: Solve for \(\frac{dy}{dt}\)
Rearranging the equation:
\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \implies x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \]
Thus:
\[ \frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt} \]

### Step 6: Substitute known values
We know:
- \( \frac{dx}{dt} = 2 \) m/sec (the rate at which the foot of the ladder is moving away from the wall)
- When \( x = 5 \) m, we need to find \( y \).

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

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

### Step 7: Interpret the result
The negative sign indicates that the height is decreasing. Therefore, the height on the wall is decreasing at a rate of \(\frac{5}{6}\) m/sec.

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

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