A particle moves along the curve `12y=x^(3)`. . Which coordinate changes at faster rate at `x=10`?

Options

A. x-coordinate
B. y-coordinate
C. Both x and y-coordinate
D. Data insufficient

Explanation

To solve the problem of determining which coordinate changes at a faster rate for a particle moving along the curve \(12y = x^3\) at \(x = 10\), we will follow these steps:

### Step 1: Differentiate the equation of the curve
Given the equation of the curve:
\[ 12y = x^3 \]
We can differentiate both sides with respect to time \(t\) to find the rates of change of \(y\) and \(x\).

### Step 2: Apply implicit differentiation
Differentiating both sides with respect to \(t\):
\[ \frac{d}{dt}(12y) = \frac{d}{dt}(x^3) \]
Using the chain rule, we get:
\[ 12 \frac{dy}{dt} = 3x^2 \frac{dx}{dt} \]

### Step 3: Solve for \(\frac{dy}{dt}\)
Rearranging the equation to solve for \(\frac{dy}{dt}\):
\[ \frac{dy}{dt} = \frac{3x^2}{12} \frac{dx}{dt} = \frac{x^2}{4} \frac{dx}{dt} \]

### Step 4: Evaluate at \(x = 10\)
Now, substituting \(x = 10\) into the equation:
\[ \frac{dy}{dt} = \frac{10^2}{4} \frac{dx}{dt} = \frac{100}{4} \frac{dx}{dt} = 25 \frac{dx}{dt} \]

### Step 5: Compare the rates of change
Now we have:
- \(\frac{dy}{dt} = 25 \frac{dx}{dt}\)
- \(\frac{dx}{dt}\) is the rate of change of \(x\)

This indicates that the rate of change of \(y\) is 25 times the rate of change of \(x\).

### Conclusion
Since \(\frac{dy}{dt} = 25 \frac{dx}{dt}\), we conclude that the \(y\)-coordinate changes at a faster rate than the \(x\)-coordinate when \(x = 10\).

### Final Answer
The \(y\)-coordinate changes at a faster rate at \(x = 10\).

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