A satellite travels in a circular orbit of radius R. If its x- coordinate decreases at the rate of 2 units/s at the point (a, b ) how fast is the y-coordinate changing?

Answer

To solve the problem step by step, we will use the relationship between the coordinates of the satellite in a circular orbit and the rates of change of those coordinates.

### Step-by-Step Solution:

1. **Understand the Circular Motion**:
The satellite is moving in a circular orbit of radius \( R \). The equation of the circle can be expressed as:
\[ x^2 + y^2 = R^2 \]

2. **Differentiate the Equation**:
We need to differentiate the equation with respect to time \( t \). Using implicit differentiation, we get:
\[ \frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(R^2) \]
Since \( R \) is a constant, the derivative of \( R^2 \) is 0. Thus, we have:
\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]

3. **Simplify the Equation**:
We can simplify the equation by dividing everything by 2:
\[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \]

4. **Express \( \frac{dy}{dt} \)**:
Rearranging the equation gives us:
\[ y \frac{dy}{dt} = -x \frac{dx}{dt} \]
Therefore,
\[ \frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt} \]

5. **Substitute Known Values**:
We know that \( \frac{dx}{dt} = -2 \) units/s (the x-coordinate is decreasing). At the point \( (a, b) \), we substitute \( x = a \) and \( y = b \):
\[ \frac{dy}{dt} = -\frac{a}{b} (-2) = \frac{2a}{b} \]

6. **Final Result**:
Thus, the rate at which the y-coordinate is changing at the point \( (a, b) \) is:
\[ \frac{dy}{dt} = \frac{2a}{b} \text{ units/s} \]