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. `(pi)/(4) "radians"//sec`.
B. `(4)/(pi) "radians"//sec`.
C. `(0.025) "radians"//sec`.
D. `(pi)/(6) "radians"//sec`.

Answer

To solve the problem step by step, we will use the concepts of related rates in calculus.

### Step 1: Understand the scenario
We have a ladder of length \( L = 5 \) m leaning against a wall. The top of the ladder slides down the wall at a rate of \( \frac{dh}{dt} = -10 \) cm/sec (negative because it is sliding down). 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 \( x = 4 \) m away from the wall.

### Step 2: Convert units
First, convert the rate of descent of the ladder's top from cm/sec to m/sec:
\[ \frac{dh}{dt} = -10 \text{ cm/sec} = -0.1 \text{ m/sec} \]

### Step 3: Set up the relationship
Using the Pythagorean theorem, we have:
\[ x^2 + h^2 = L^2 \]
where \( x \) is the distance from the wall to the foot of the ladder, \( h \) is the height of the top of the ladder from the ground, and \( L = 5 \) m is the length of the ladder.

### Step 4: Differentiate with respect to time
Differentiating both sides with respect to time \( t \):
\[ \frac{d}{dt}(x^2) + \frac{d}{dt}(h^2) = \frac{d}{dt}(L^2) \]
Since \( L \) is constant, \( \frac{d}{dt}(L^2) = 0 \):
\[ 2x \frac{dx}{dt} + 2h \frac{dh}{dt} = 0 \]
Dividing through by 2:
\[ x \frac{dx}{dt} + h \frac{dh}{dt} = 0 \]

### Step 5: Find \( h \) when \( x = 4 \) m
Using the Pythagorean theorem:
\[ 4^2 + h^2 = 5^2 \]
\[ 16 + h^2 = 25 \implies h^2 = 9 \implies h = 3 \text{ m} \]

### Step 6: Substitute known values
Substituting \( x = 4 \) m, \( h = 3 \) m, and \( \frac{dh}{dt} = -0.1 \) m/sec into the differentiated equation:
\[ 4 \frac{dx}{dt} + 3(-0.1) = 0 \]
\[ 4 \frac{dx}{dt} - 0.3 = 0 \implies 4 \frac{dx}{dt} = 0.3 \implies \frac{dx}{dt} = \frac{0.3}{4} = 0.075 \text{ m/sec} \]

### Step 7: Relate angle \( \theta \) to the sides
Using the cosine function:
\[ \cos \theta = \frac{x}{L} = \frac{4}{5} \]
To find \( \frac{d\theta}{dt} \):
\[ \frac{d\theta}{dt} = -\frac{1}{\sqrt{1 - \cos^2 \theta}} \cdot \frac{dx}{dt} \cdot \frac{1}{L} \]
Calculating \( \sin \theta \):
\[ \sin^2 \theta = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \implies \sin \theta = \frac{3}{5} \]
Thus, \( \sqrt{1 - \cos^2 \theta} = \sin \theta = \frac{3}{5} \).

### Step 8: Substitute values into the angle rate formula
\[ \frac{d\theta}{dt} = -\frac{1}{\frac{3}{5}} \cdot 0.075 \cdot \frac{1}{5} \]
\[ = -\frac{5}{3} \cdot 0.075 \cdot \frac{1}{5} = -\frac{0.075}{3} = -0.025 \text{ radians/sec} \]

### Conclusion
The angle \( \theta \) is decreasing at the rate of \( 0.025 \) radians/sec.