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

Answer: (C)

$(11, -4), \left( -\frac{31}{3}, -4 \right)$

Answer 2

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:

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