A 10cm long rod AB moves with its ends on two mutually perpendicular straight lines OX and OY . If the end A be moving at the rate of `2cm//sec`, then when the distance of A from O is 8cm, the rate at which the end B is moving is

Options

A. `8/3`
B. `3/8`
C. `(-8)/(3)`
D. None of these

Answer

To solve the problem step-by-step, we can follow these steps:

### Step 1: Understand the Geometry
We have a rod AB of length 10 cm, with point A moving along the x-axis (OX) and point B moving along the y-axis (OY). The coordinates of point A can be represented as (x, 0) and the coordinates of point B as (0, y).

### Step 2: Set Up the Relationship
Using the Pythagorean theorem, we can establish the relationship between the coordinates of points A and B:
\[ AB^2 = OA^2 + OB^2 \]
Given that the length of the rod AB is 10 cm, we have:
\[ x^2 + y^2 = 10^2 = 100 \]

### Step 3: Differentiate with Respect to Time
To find the rates of change, we differentiate the equation with respect to time \(t\):
\[ \frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(100) \]
Using the chain rule, we get:
\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]
Dividing through by 2 gives:
\[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \]

### Step 4: Substitute Known Values
We know that point A is moving at a rate of \(\frac{dx}{dt} = 2 \, \text{cm/sec}\) and we need to find \(\frac{dy}{dt}\) when the distance of A from O is \(x = 8 \, \text{cm}\).

First, we need to find \(y\) when \(x = 8\):
\[ 8^2 + y^2 = 100 \implies 64 + y^2 = 100 \implies y^2 = 36 \implies y = 6 \, \text{cm} \]

### Step 5: Substitute Values into the Rate Equation
Now we substitute \(x = 8\), \(y = 6\), and \(\frac{dx}{dt} = 2\) into the differentiated equation:
\[ 8(2) + 6 \frac{dy}{dt} = 0 \]
This simplifies to:
\[ 16 + 6 \frac{dy}{dt} = 0 \]
Solving for \(\frac{dy}{dt}\):
\[ 6 \frac{dy}{dt} = -16 \implies \frac{dy}{dt} = -\frac{16}{6} = -\frac{8}{3} \, \text{cm/sec} \]

### Conclusion
Thus, the rate at which point B is moving is:
\[ \frac{dy}{dt} = -\frac{8}{3} \, \text{cm/sec} \]