On the curve `x^3=12 y ,` find the interval of values of `x` for which the abscissa changes at a faster rate than the ordinate?

Options

A. `(-2, 2)`
B. `(-oo, -2) uu (2,oo)`
C. `[-2, 2]`
D. none of these

Answer

To solve the problem of finding the interval of values of \( x \) for which the abscissa changes at a faster rate than the ordinate on the curve \( x^3 = 12y \), we will follow these steps:

### Step 1: Differentiate the equation with respect to \( t \)

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

### Step 2: Express \( \frac{dy}{dt} \) in terms of \( \frac{dx}{dt} \)

From the differentiated equation, we can express \( \frac{dy}{dt} \):
\[ \frac{dy}{dt} = \frac{3x^2}{12} \frac{dx}{dt} = \frac{x^2}{4} \frac{dx}{dt} \]

### Step 3: Set up the inequality for rates of change

We need to find when the rate of change of the abscissa ( \( \frac{dx}{dt} \) ) is greater than the rate of change of the ordinate ( \( \frac{dy}{dt} \) ):
\[ \frac{dx}{dt} > \frac{dy}{dt} \]
Substituting for \( \frac{dy}{dt} \):
\[ \frac{dx}{dt} > \frac{x^2}{4} \frac{dx}{dt} \]

### Step 4: Rearrange the inequality

Assuming \( \frac{dx}{dt} \neq 0 \), we can divide both sides by \( \frac{dx}{dt} \):
\[ 1 > \frac{x^2}{4} \]
This can be rearranged to:
\[ 4 > x^2 \]
or
\[ x^2 < 4 \]

### Step 5: Solve the inequality

Taking square roots gives:
\[ -2 < x < 2 \]

### Step 6: Write the final answer

Thus, the interval of values of \( x \) for which the abscissa changes at a faster rate than the ordinate is:
\[ x \in (-2, 2) \]