Q.A particle moves along the curve $6y = x^3 + 2$. Find the points on the curve at which the $y$-coordinate is changing $8$ times as fast as the $x$-coordinate.

Concept: Related Rates — we relate $\frac{dy}{dt}$ and $\frac{dx}{dt}$ using the derivative of the given relation.

Step 1: Differentiate $6y = x^3 + 2$ with respect to $t$:

Step 2: We are told $\frac{dy}{dt} = 8 \frac{dx}{dt}$. Substitute:

Step 3: Cancel $\frac{dx}{dt}$ (non-zero at the points of interest) and solve:

$48 = 3x^2 \quad\Rightarrow\quad x^2 = 16 \quad\Rightarrow\quad x = \pm 4$ …

This is a classic related rates problem. The key idea: when two quantities are linked by an equation, their rates of change are also linked. Here, $x$ and $y$ move together along the curve $6y = x^3 + 2$, and we are told that at some instant, the $y$-coordinate is changing 8 times faster than the $x$-coordinate. That means $\frac{dy}{dt} = 8 \frac{dx}{dt}$.

We don't know the time $t$ explicitly — we don't need to. We just differentiate the curve equation with respect to $t$, substitute the rate relationship, and solve for the coordinates.

1. Differentiate the curve equation with respect to time $t$.

   The curve is:

Differentiate both sides with respect to $t$ (remember $x$ and $y$ are functions of $t$):

This is the core related-rates equation linking $\frac{dy}{dt}$ and $\frac{dx}{dt}$.

2. Apply the given condition.

   We are told: the $y$-coordinate changes 8 times as fast as the $x$-coordinate. That means:

Substitute this into the differentiated equation:

So:

3. Solve for $x$.

   If $\frac{dx}{dt} = 0$, then the $x$-coordinate isn't changing at all — but then $\frac{dy}{dt} = 0$ as well, which would mean the rate condition $8 \times 0 = 0$ holds trivially. However, the problem asks for points where the $y$-coordinate is changing 8 times as fast as the $x$-coordinate, implying both rates are non-zero. So we assume $\frac{dx}{dt} \neq 0$ and divide both sides by it:

$x = \pm 4$ …

Method: Related Rates on a Curve — Finding WHERE a Given Rate Ratio Holds

This method applies when a particle moves along a curve $y=f(x)$ and you are told the ratio between $\frac{dy}{dt}$ and $\frac{dx}{dt}$ at some unspecified instant, and asked to find the point(s) on the curve where that ratio occurs.

Steps

Step 1: Differentiate the curve's equation with respect to time $t$

Treat both $x$ and $y$ as functions of $t$ and differentiate implicitly. Every term picks up its own rate via the chain rule — e.g. differentiating a term in $x^n$ gives $n x^{n-1}\frac{dx}{dt}$.

Step 2: Translate the given rate relationship into an equation

A phrase like "$y$-coordinate changing $k$ times as fast as the $x$-coordinate" means

Step 3: Substitute this relationship into the differentiated curve equation

This leaves an equation with a common factor of $\frac{dx}{dt}$ on both sides.

Step 4: Cancel $\frac{dx}{dt}$ and solve the resulting purely algebraic equation for $x$ …

Common Mistakes

Mistake 1: Keeping only the positive root and discarding $x=-4$

Why it's wrong: $x^2=16$ has TWO real solutions, $x=4$ and $x=-4$, and the curve $6y=x^3+2$ is defined for negative $x$ too — dropping the negative root misses a genuine point on the curve that satisfies the given rate condition. Correct approach: always list both roots of a squared equation like $x^2=16$, and check each one against the original curve equation.

Mistake 2: Reporting only the $x$-coordinates as the final answer

Why it's wrong: the question asks for "points on the curve", which means ordered pairs $(x,y)$ — stopping after finding $x=\pm4$ leaves the answer incomplete. Correct approach: substitute each $x$-value back into $6y=x^3+2$ to compute the matching $y$, and state the full coordinate pairs.

Mistake 3: Cancelling $\frac{dx}{dt}$ without justification …