A particle is in motion along a curve 12y=`x^(3)`. The rate of change of its ordinate exceeds that of abscissa in

Options

A. `-2 lt x lt 2`
B. ` x= pm 2`
C. `x lt -2`
D. `x gt2`

Answer

To solve the problem step by step, we will analyze the motion of the particle along the curve given by the equation \(12y = x^3\). We need to determine the conditions under which the rate of change of the ordinate (y-coordinate) exceeds that of the abscissa (x-coordinate).

### Step 1: Understand the relationship between ordinate and abscissa
The ordinate refers to the y-coordinate, and the abscissa refers to the x-coordinate. The rates of change of these coordinates with respect to time \(t\) are given by:
- \(\frac{dy}{dt}\) for the ordinate
- \(\frac{dx}{dt}\) for the abscissa

### Step 2: Differentiate the curve equation with respect to time
Given the curve \(12y = x^3\), we differentiate both sides with respect to time \(t\):
\[ \frac{d}{dt}(12y) = \frac{d}{dt}(x^3) \]
Using the chain rule, we get:
\[ 12 \frac{dy}{dt} = 3x^2 \frac{dx}{dt} \]

### Step 3: Isolate \(\frac{dy}{dt}\)
We can rearrange the equation to express \(\frac{dy}{dt}\):
\[ \frac{dy}{dt} = \frac{3x^2}{12} \frac{dx}{dt} = \frac{x^2}{4} \frac{dx}{dt} \]

### Step 4: Set up the inequality
We need to find when the rate of change of the ordinate exceeds that of the abscissa:
\[ \frac{dy}{dt} > \frac{dx}{dt} \]
Substituting our expression for \(\frac{dy}{dt}\):
\[ \frac{x^2}{4} \frac{dx}{dt} > \frac{dx}{dt} \]

### Step 5: Simplify the inequality
Assuming \(\frac{dx}{dt} \neq 0\) (since we are considering motion), we can divide both sides by \(\frac{dx}{dt}\):
\[ \frac{x^2}{4} > 1 \]

### Step 6: Solve the inequality
Multiplying both sides by 4 gives:
\[ x^2 > 4 \]
This can be factored as:
\[ (x - 2)(x + 2) > 0 \]

### Step 7: Analyze the inequality
To solve the inequality \( (x - 2)(x + 2) > 0 \), we find the critical points where the expression equals zero:
- \(x = -2\)
- \(x = 2\)

Now we test the intervals:
1. \(x < -2\): Both factors are negative, so the product is positive.
2. \(-2 < x < 2\): One factor is negative and the other is positive, so the product is negative.
3. \(x > 2\): Both factors are positive, so the product is positive.

Thus, the solution to the inequality is:
\[ x < -2 \quad \text{or} \quad x > 2 \]

### Final Answer
The rate of change of the ordinate exceeds that of the abscissa when:
\[ x < -2 \quad \text{or} \quad x > 2 \]