A ladder 5 m in length is resting against vertical wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of 1.5m/sec. The length of the highest point of the ladder when the foot of the ladder 4.0 m away from the wall decreases at the rate of

Options

A. 2m/sec
B. 3m/sec
C. 2.5m/sec
D. 1.5m/sec

Explanation

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

### Step 1: Understand the Problem
We have a ladder of length 5 m leaning against a wall. The bottom of the ladder is being pulled away from the wall at a rate of 1.5 m/s. We need to find the rate at which the height of the ladder (the point where it touches the wall) is decreasing when the bottom of the ladder is 4 m away from the wall.

### Step 2: Set Up the Variables
Let:
- \( x \) = the distance from the wall to the bottom of the ladder (horizontal distance)
- \( y \) = the height of the ladder on the wall (vertical distance)
- The length of the ladder is constant at 5 m.

According to the Pythagorean theorem:
\[ x^2 + y^2 = 5^2 \]
\[ x^2 + y^2 = 25 \]

### Step 3: Differentiate with Respect to Time
We differentiate both sides of the equation with respect to time \( t \):
\[ \frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(25) \]
Using the chain rule, we have:
\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]

### Step 4: Solve for \(\frac{dy}{dt}\)
Rearranging the equation gives:
\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]
Dividing through by 2:
\[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \]
Now, solving for \(\frac{dy}{dt}\):
\[ y \frac{dy}{dt} = -x \frac{dx}{dt} \]
\[ \frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt} \]

### Step 5: Substitute Known Values
We know:
- \( x = 4 \) m (the distance from the wall)
- We need to find \( y \) when \( x = 4 \):
\[ x^2 + y^2 = 25 \implies 4^2 + y^2 = 25 \implies 16 + y^2 = 25 \implies y^2 = 9 \implies y = 3 \text{ m} \]
- The rate at which the bottom of the ladder is pulled away from the wall is \(\frac{dx}{dt} = 1.5\) m/s.

Now substituting these values into the equation for \(\frac{dy}{dt}\):
\[ \frac{dy}{dt} = -\frac{4}{3} \cdot 1.5 \]
Calculating this gives:
\[ \frac{dy}{dt} = -\frac{6}{3} = -2 \text{ m/s} \]

### Conclusion
The height of the highest point of the ladder is decreasing at the rate of 2 m/s when the foot of the ladder is 4 m away from the wall.

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