A ladder, metres long standing on a floor leans against a vertical wall . If the top of the ladder slides downwards at the rate of 10 cm/sec , then the rate at which the angel between the floor and the ladder is decreasing when the lower end of the ladder is 2 metres away from the wall is

Options

A. `(1)/(10)` radian/sec
B. `(1)/(20)` radina/sec
C. 5 radian/sec
D. 10 radian/ sec

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a ladder of length \( L = 5 \) meters leaning against a wall. The top of the ladder is sliding down at a rate of \( \frac{dy}{dt} = -10 \) cm/sec (negative because \( y \) is decreasing). We need to find the rate at which the angle \( \theta \) between the ladder and the floor is decreasing when the bottom of the ladder is \( x = 2 \) meters away from the wall.

### Step 2: Set Up the Relationship
We can use the right triangle formed by the ladder, the wall, and the floor. The lengths of the sides are:
- \( AB = x \) (horizontal distance from the wall)
- \( BC = y \) (vertical distance from the floor to the top of the ladder)
- \( AC = L = 5 \) meters (length of the ladder)

Using the Pythagorean theorem:
\[ x^2 + y^2 = L^2 \]

### Step 3: Convert Units
Convert the length of the ladder to centimeters:
\[ L = 5 \text{ m} = 500 \text{ cm} \]
Also, convert \( x \) to centimeters:
\[ x = 2 \text{ m} = 200 \text{ cm} \]

### Step 4: Find \( y \)
Using the Pythagorean theorem:
\[ 200^2 + y^2 = 500^2 \]
\[ 40000 + y^2 = 250000 \]
\[ y^2 = 250000 - 40000 = 210000 \]
\[ y = \sqrt{210000} \approx 458.26 \text{ cm} \]

### Step 5: Relate \( \theta \) to \( x \) and \( y \)
Using trigonometric relationships:
\[ \sin \theta = \frac{y}{L} \quad \text{and} \quad \cos \theta = \frac{x}{L} \]

### Step 6: Differentiate with Respect to Time
Using implicit differentiation:
\[ \frac{d}{dt}(\sin \theta) = \cos \theta \frac{d\theta}{dt} \]
\[ \frac{d}{dt}(\cos \theta) = -\sin \theta \frac{d\theta}{dt} \]

### Step 7: Find \( \frac{d\theta}{dt} \)
From the relationship:
\[ \frac{dy}{dt} = \frac{d}{dt}(L \sin \theta) \Rightarrow \frac{dy}{dt} = L \cos \theta \frac{d\theta}{dt} \]
Substituting known values:
\[ -10 = 500 \cos \theta \frac{d\theta}{dt} \]
Now we need to find \( \cos \theta \) when \( x = 200 \) cm and \( y \approx 458.26 \) cm:
\[ \cos \theta = \frac{x}{L} = \frac{200}{500} = 0.4 \]
Substituting \( \cos \theta \):
\[ -10 = 500 \cdot 0.4 \cdot \frac{d\theta}{dt} \]
\[ -10 = 200 \frac{d\theta}{dt} \]
\[ \frac{d\theta}{dt} = -\frac{10}{200} = -\frac{1}{20} \text{ rad/sec} \]

### Final Answer
The rate at which the angle \( \theta \) is decreasing is:
\[ \frac{d\theta}{dt} = \frac{1}{20} \text{ rad/sec} \]