A car is moving along the curve $y = x^3 + 12$. The point(s) on the curve at which the rate of change of its y-coordinate at a certain time is 3 times the rate of change of its x-coordinate is/are

The car is moving along the curve $y = x^3 + 12$.

The rate of change of the y-coordinate is $\frac{dy}{dt}$ and the rate of change of the x-coordinate is $\frac{dx}{dt}$.

The condition states: $\frac{dy}{dt} = 3 \times \frac{dx}{dt}$

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Given: $y = x^3 + 12$

Using the chain rule:

$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$

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Differentiating $y = x^3 + 12$ with respect to $x$:

$\frac{dy}{dx} = 3x^2$

Therefore:

$\frac{dy}{dt} = 3x^2 \times \frac{dx}{dt}$

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Applying the given condition $\frac{dy}{dt} = 3 \times \frac{dx}{dt}$:

$3x^2 \times \frac{dx}{dt} = 3 \times \frac{dx}{dt}$

Since the car is moving, $\frac{dx}{dt} \neq 0$:

$3x^2 = 3$

$x^2 = 1$

$x = 1$ or $x = -1$

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Using $y = x^3 + 12$ to find the corresponding y-coordinates:

When $x = 1$:

$y = (1)^3 + 12$

$y = 1 + 12$

$y = 13$

Point: $(1, 13)$

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When $x = -1$:

$y = (-1)^3 + 12$

$y = -1 + 12$

$y = 11$

Point: $(-1, 11)$

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Therefore, the points on the curve are $(1, 13)$ and $(-1, 11)$.