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-coordiante at a certain time is 3 times the rate of change of its x-coordinate is/are

(1, 13) and (-1, 11)

The correct answer is Option (3) → (1, 13) and (-1, 11)

Given the curve: $y = x^3 + 12$

Let the car move along the curve, so both $x$ and $y$ are functions of time $t$.

Differentiate both sides with respect to $t$ using the chain rule:

$\frac{dy}{dt} = \frac{d}{dt}(x^3 + 12) = 3x^2 \cdot \frac{dx}{dt}$

Given: $\frac{dy}{dt} = 3 \cdot \frac{dx}{dt}$

Substitute into the derivative:

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

Cancel $\frac{dx}{dt}$ (assuming it's non-zero):

$3x^2 = 3 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$

Now, find the corresponding $y$ values:

If $x = 1$, then $y = 1^3 + 12 = 13$

If $x = -1$, then $y = (-1)^3 + 12 = -1 + 12 = 11$

Therefore, the required points are $(1, 13)$ and $(-1, 11)$