A rod of length 13 metres has one end P on the X-axis and the other end Q on the y-axis. If P moves on the X-axis with a speed of 12 m/sec, then the speed of the other end Q when it is 12 m from the origin is

Options

A. 3 m/sec
B. 5 m/sec
C. `-5` m/sec
D. 4 m/sec

Answer

To solve the problem, we need to analyze the situation involving a rod of length 13 meters with one end on the x-axis and the other on the y-axis. Let's denote the coordinates of point P on the x-axis as (x, 0) and point Q on the y-axis as (0, y).

### Step-by-Step Solution:

1. **Understanding the Geometry**:
The length of the rod is constant and given as 13 meters. According to the Pythagorean theorem, we can express this relationship as:
\[ x^2 + y^2 = 13^2 \]
or
\[ x^2 + y^2 = 169 \]

2. **Differentiating with Respect to Time**:
To find the relationship between the rates of change of x and y, we differentiate both sides of the equation with respect to time t:
\[ \frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(169) \]
Using the chain rule, we get:
\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]
Dividing through by 2 simplifies this to:
\[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \]

3. **Substituting Known Values**:
We know that the speed of point P (which is moving along the x-axis) is given as:
\[ \frac{dx}{dt} = 12 \text{ m/s} \]
We need to find \(\frac{dy}{dt}\) when point Q is 12 meters from the origin, which means \(y = 12\).

4. **Finding the Value of x**:
We can substitute \(y = 12\) into the original equation to find \(x\):
\[ x^2 + 12^2 = 169 \]
This simplifies to:
\[ x^2 + 144 = 169 \]
\[ x^2 = 169 - 144 = 25 \]
Therefore, \(x = 5\).

5. **Substituting into the Rate Equation**:
Now we substitute \(x = 5\), \(y = 12\), and \(\frac{dx}{dt} = 12\) into the differentiated equation:
\[ 5(12) + 12 \frac{dy}{dt} = 0 \]
This simplifies to:
\[ 60 + 12 \frac{dy}{dt} = 0 \]
Rearranging gives:
\[ 12 \frac{dy}{dt} = -60 \]
Dividing both sides by 12 results in:
\[ \frac{dy}{dt} = -5 \text{ m/s} \]

### Final Answer:
The speed of the other end Q when it is 12 m from the origin is \(-5\) m/s.