A ladder is resting with a vertical wall at an angle of `30^(@)`. If a man is ascending the ladder of at the rate of 6 feet/sec, then the rate at which the man is approaching the wall, is

Options

A. 9 feet/sec
B. 2 feet/sec
C. 3 feet/sec
D. 6 feet/sec

Answer

To solve the problem step by step, we will use the relationship between the ladder's position, the angle it makes with the wall, and the rates at which the man ascends the ladder and approaches the wall.

### Step 1: Understand the Geometry
We have a right triangle formed by the wall, the ground, and the ladder. The ladder makes an angle of \(30^\circ\) with the ground. Let:
- \(L\) = length of the ladder (hypotenuse)
- \(x\) = distance from the wall to the base of the ladder (adjacent side)
- \(y\) = height of the ladder on the wall (opposite side)

### Step 2: Set Up the Relationships
From the triangle, we can use trigonometric functions:
- \(\sin(30^\circ) = \frac{y}{L}\)
- \(\cos(30^\circ) = \frac{x}{L}\)

Given that \(\sin(30^\circ) = \frac{1}{2}\) and \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\), we can express \(y\) and \(x\) in terms of \(L\):
- \(y = L \cdot \sin(30^\circ) = \frac{L}{2}\)
- \(x = L \cdot \cos(30^\circ) = L \cdot \frac{\sqrt{3}}{2}\)

### Step 3: Differentiate with Respect to Time
We know the man is ascending the ladder at a rate of \( \frac{dy}{dt} = 6 \) feet/sec. We want to find the rate at which he is approaching the wall, which is \( \frac{dx}{dt} \).

Using the relationship \(y = \frac{L}{2}\), we can differentiate both sides with respect to time \(t\):
\[ \frac{dy}{dt} = \frac{1}{2} \frac{dL}{dt} \]

### Step 4: Relate the Rates
Since the man is climbing the ladder, the length of the ladder \(L\) is constant (not changing), so \(\frac{dL}{dt} = 0\). Thus:
\[ \frac{dy}{dt} = 6 = \frac{1}{2} \cdot 0 \]
This means we need to relate \(y\) and \(x\) directly to find \(\frac{dx}{dt}\).

From the relationship:
\[ x = L \cdot \frac{\sqrt{3}}{2} \]
We can differentiate \(x\) with respect to time:
\[ \frac{dx}{dt} = \frac{\sqrt{3}}{2} \cdot \frac{dL}{dt} \]

### Step 5: Use the Pythagorean Theorem
Using the Pythagorean theorem:
\[ L^2 = x^2 + y^2 \]
Differentiating this with respect to time gives:
\[ 2L \frac{dL}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \]
Since \(\frac{dL}{dt} = 0\):
\[ 0 = 2x \frac{dx}{dt} + 2y \cdot 6 \]
\[ 0 = x \frac{dx}{dt} + 6y \]
Thus,
\[ x \frac{dx}{dt} = -6y \]

### Step 6: Substitute Known Values
We know \(y = \frac{L}{2}\) and \(x = L \cdot \frac{\sqrt{3}}{2}\). Substitute these into the equation:
\[ L \cdot \frac{\sqrt{3}}{2} \cdot \frac{dx}{dt} = -6 \cdot \frac{L}{2} \]
Cancelling \(L\) (assuming \(L \neq 0\)):
\[ \frac{\sqrt{3}}{2} \cdot \frac{dx}{dt} = -3 \]
Thus,
\[ \frac{dx}{dt} = -\frac{3 \cdot 2}{\sqrt{3}} = -2\sqrt{3} \]

### Step 7: Find the Rate at Which the Man is Approaching the Wall
The negative sign indicates that the distance \(x\) is decreasing, which means the man is approaching the wall. The rate at which the man is approaching the wall is:
\[ \frac{dx}{dt} = -2\sqrt{3} \approx -3.464 \text{ feet/sec} \]

### Final Answer
The rate at which the man is approaching the wall is approximately \(3.464\) feet/sec.

---