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

Answer

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, we can follow these steps:

### Step 1: Understand the relationship between dy and dx
Given that the y-coordinate is changing 8 times as fast as the x-coordinate, we can express this relationship as:
\[ \frac{dy}{dx} = 8 \]

### Step 2: Differentiate the curve equation
We start with the curve equation:
\[ 6y = x^3 + 2 \]
To find \(\frac{dy}{dx}\), we differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(6y) = \frac{d}{dx}(x^3 + 2) \]
This gives us:
\[ 6 \frac{dy}{dx} = 3x^2 \]

### Step 3: Solve for dy/dx
Now, we can isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{3x^2}{6} = \frac{x^2}{2} \]

### Step 4: Set dy/dx equal to 8
We know from Step 1 that \(\frac{dy}{dx} = 8\). Therefore, we set the equation from Step 3 equal to 8:
\[ \frac{x^2}{2} = 8 \]

### Step 5: Solve for x
To solve for \(x\), we multiply both sides by 2:
\[ x^2 = 16 \]
Taking the square root of both sides gives us:
\[ x = 4 \quad \text{or} \quad x = -4 \]

### Step 6: Find corresponding y-values
Now we will substitute \(x = 4\) and \(x = -4\) back into the original curve equation to find the corresponding \(y\)-coordinates.

1. For \(x = 4\):
\[ 6y = 4^3 + 2 = 64 + 2 = 66 \]
\[ y = \frac{66}{6} = 11 \]
So, one point is \((4, 11)\).

2. For \(x = -4\):
\[ 6y = (-4)^3 + 2 = -64 + 2 = -62 \]
\[ y = \frac{-62}{6} = -\frac{31}{3} \]
So, the other point is \((-4, -\frac{31}{3})\).

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

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