A ladder `5m` long rests against a vertical wall. If its top slides down at the rate of `10 cm //sec`, then, when the foot of the ladder is `4 m` away from the wall, the angle between the floor and the ladder is decreasing at the rate of

Options

A. `1/10 rad//s`
B. `1/20 rad//s`
C. `20 rad//s`
D. `10 rad//s`

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the scenario
We have a ladder of length \( L = 5 \) meters resting against a vertical wall. The top of the ladder slides down the wall at a rate of \( \frac{dh}{dt} = -10 \) cm/s (which is equivalent to \( -0.1 \) m/s). We need to find the rate at which the angle \( \theta \) between the ladder and the floor is decreasing when the foot of the ladder is \( b = 4 \) m away from the wall.

### Step 2: Set up the relationship using Pythagorean theorem
Using the Pythagorean theorem:
\[ L^2 = h^2 + b^2 \]
where \( h \) is the height of the ladder on the wall and \( b \) is the distance from the wall to the foot of the ladder. Given \( L = 5 \) m and \( b = 4 \) m, we can find \( h \):
\[ 5^2 = h^2 + 4^2 \\ 25 = h^2 + 16 \\ h^2 = 25 - 16 \\ h^2 = 9 \\ h = 3 \text{ m} \]

### Step 3: Relate the angle \( \theta \) to the sides of the triangle
From the right triangle formed by the ladder, wall, and ground:
\[ \sin \theta = \frac{h}{L} \]
Differentiating both sides with respect to time \( t \):
\[ \cos \theta \frac{d\theta}{dt} = \frac{1}{L} \frac{dh}{dt} \]

### Step 4: Find \( \cos \theta \)
Using \( b \) to find \( \cos \theta \):
\[ \cos \theta = \frac{b}{L} = \frac{4}{5} \]

### Step 5: Substitute values into the differentiated equation
Substituting \( \cos \theta \) and \( \frac{dh}{dt} \):
\[ \frac{4}{5} \frac{d\theta}{dt} = \frac{1}{5} \left(-0.1\right) \]
This simplifies to:
\[ \frac{4}{5} \frac{d\theta}{dt} = -0.02 \]

### Step 6: Solve for \( \frac{d\theta}{dt} \)
Now, solve for \( \frac{d\theta}{dt} \):
\[ \frac{d\theta}{dt} = -0.02 \cdot \frac{5}{4} = -0.025 \text{ radians/second} \]

### Final Answer
The angle between the floor and the ladder is decreasing at the rate of \( -0.025 \) radians/second.

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