A particle is moving along the parabola `y^(2) = 4(x+2)`. As it passes through the point (7,6) its y-coordinate is increasing at the rate of 3 units per second. The rate at which x-coordinate change at this instant is (in units/sec)

Options

A. 4
B. 6
C. 8
D. 9

Answer

To solve the problem, we need to find the rate at which the x-coordinate of the particle is changing (dx/dt) as it moves along the parabola defined by the equation \( y^2 = 4(x + 2) \).

### Step-by-step solution:

1. **Differentiate the equation of the parabola with respect to time (t)**:
Given the equation:
\[ y^2 = 4(x + 2) \]
We differentiate both sides with respect to \( t \):
\[ \frac{d}{dt}(y^2) = \frac{d}{dt}(4(x + 2)) \]
Using the chain rule, we have:
\[ 2y \frac{dy}{dt} = 4 \frac{dx}{dt} \]

2. **Rearranging the equation**:
From the differentiated equation, we can express \( \frac{dx}{dt} \):
\[ 2y \frac{dy}{dt} = 4 \frac{dx}{dt} \]
Dividing both sides by 4 gives:
\[ \frac{dx}{dt} = \frac{2y}{4} \frac{dy}{dt} = \frac{y}{2} \frac{dy}{dt} \]

3. **Substituting the known values**:
We know that at the point (7, 6):
- \( y = 6 \)
- \( \frac{dy}{dt} = 3 \) units/second (given)

Substituting these values into the equation:
\[ \frac{dx}{dt} = \frac{6}{2} \cdot 3 \]
Simplifying this gives:
\[ \frac{dx}{dt} = 3 \cdot 3 = 9 \text{ units/second} \]

4. **Conclusion**:
Thus, the rate at which the x-coordinate changes at the instant the particle passes through the point (7, 6) is:
\[ \frac{dx}{dt} = 9 \text{ units/second} \]

### Final Answer:
The rate at which the x-coordinate changes is \( 9 \) units/second.

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