Question

A ladder 13 m long is leaning against the 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 the height on the wall decreasing when the foot of the ladder is 12 m away from the wall?

Answer 1

Ladder Sliding Down the Wall

Let the length of the ladder be constant:

\( l = 13 \, \text{m} \)

Let \( x \) be the distance of the foot of the ladder from the wall (along the ground), and \( y \) be the height of the top of the ladder from the ground (along the wall).

Step 1: Use Pythagoras Theorem

\[ x^2 + y^2 = 13^2 = 169 \tag{1} \]

Step 2: Differentiate with Respect to Time \( t \)

\[ \frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(169) \] \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \] \[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \tag{2} \]

Step 3: Use Given Values

• \( x = 12 \, \text{m} \)
• \( \frac{dx}{dt} = 2 \, \text{m/s} \)

Step 4: Find \( y \) Using Equation (1)

\[ x^2 + y^2 = 169 \Rightarrow 12^2 + y^2 = 169 \Rightarrow 144 + y^2 = 169 \Rightarrow y^2 = 25 \Rightarrow y = 5 \, \text{m} \]

Step 5: Substitute into Equation (2)

\[ 12 \cdot 2 + 5 \cdot \frac{dy}{dt} = 0 \Rightarrow 24 + 5 \cdot \frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = -\frac{24}{5} = -4.8 \, \text{m/s} \]

Final Answer:

The height on the wall is decreasing at the rate of \( \boxed{-4.8 \, \text{m/s}} \).