A particle moves along the curve `6y = x^3 + 2`. Find the points on the curve at which y-co-ordinate is changing 8 times as fast as the x-co-ordinate.

Answer

To solve the problem, we need to find the points on the curve \(6y = x^3 + 2\) where the y-coordinate is changing 8 times as fast as the x-coordinate.

### Step-by-Step Solution:

1. **Understand the Relationship**:
We know that if the y-coordinate is changing 8 times as fast as the x-coordinate, we can express this relationship as:
\[ \frac{dy}{dt} = 8 \frac{dx}{dt} \]

2. **Differentiate the Curve Equation**:
The given equation of the curve is:
\[ 6y = x^3 + 2 \]
We will differentiate both sides with respect to \(t\):
\[ \frac{d}{dt}(6y) = \frac{d}{dt}(x^3 + 2) \]
This gives us:
\[ 6 \frac{dy}{dt} = 3x^2 \frac{dx}{dt} \]

3. **Substitute the Relationship**:
Now, substitute \(\frac{dy}{dt} = 8 \frac{dx}{dt}\) into the differentiated equation:
\[ 6(8 \frac{dx}{dt}) = 3x^2 \frac{dx}{dt} \]
Simplifying this, we get:
\[ 48 \frac{dx}{dt} = 3x^2 \frac{dx}{dt} \]

4. **Canceling \(\frac{dx}{dt}\)**:
Assuming \(\frac{dx}{dt} \neq 0\), we can divide both sides by \(\frac{dx}{dt}\):
\[ 48 = 3x^2 \]

5. **Solving for \(x\)**:
Now, we can solve for \(x\):
\[ x^2 = \frac{48}{3} = 16 \]
Taking the square root, we find:
\[ x = 4 \quad \text{or} \quad x = -4 \]

6. **Finding Corresponding \(y\) Values**:
Now we will substitute these \(x\) values back into the original curve equation to find the corresponding \(y\) values.

- For \(x = 4\):
\[ 6y = 4^3 + 2 = 64 + 2 = 66 \implies y = \frac{66}{6} = 11 \]

- For \(x = -4\):
\[ 6y = (-4)^3 + 2 = -64 + 2 = -62 \implies y = \frac{-62}{6} = -\frac{31}{3} \]

7. **Final Points**:
The points on the curve where the y-coordinate is changing 8 times as fast as the x-coordinate are:
\[ (4, 11) \quad \text{and} \quad (-4, -\frac{31}{3}) \]

### Summary of the Solution:
The points on the curve where the y-coordinate is changing 8 times as fast as the x-coordinate are \((4, 11)\) and \((-4, -\frac{31}{3})\).