Question

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

• A. \( (11, 4), \left( -\frac{31}{3}, 4 \right) \)
• B. \( (-11, 4), \left( \frac{31}{3}, -4 \right) \)
• C. \( (11, -4), \left( -\frac{31}{3}, -4 \right) \)
• D. \( (11, 4), \left( -\frac{31}{3}, -4 \right) \)

Answer 1

The Correct Option is C

The curve is given as:

\[ 6x = y^3 + 2. \]

Differentiating both sides with respect to \(t\), we get:

\[ 6\frac{dx}{dt} = 3y^2\frac{dy}{dt}. \]

Rewriting the relation:

\[ \frac{dx}{dt} = \frac{y^2}{2}\frac{dy}{dt}. \]

We are given that the \(x\)-coordinate changes 8 times as fast as the \(y\)-coordinate, i.e., \(\frac{dx}{dt} = 8\frac{dy}{dt}\).
Substituting this condition:

\[ 8\frac{dy}{dt} = \frac{y^2}{2}\frac{dy}{dt}. \]

Cancel \(\frac{dy}{dt}\) (since \(\frac{dy}{dt} \neq 0\)):

\[ 8 = \frac{y^2}{2}. \]

Solve for \(y\):

\[ y^2 = 16 \implies y = \pm 4. \]

Substitute \(y = 4\) and \(y = -4\) back into the curve equation \(6x = y^3 + 2\) to find \(x\):

• For \(y = 4\):
• For \(y = -4\):

Thus, the points are:

\[ (11, 4) \quad \text{and} \quad \left(-\frac{31}{3}, -4\right). \]