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.

Question

Accepted Answer

>

> [!TLDR]
> Differentiate the curve implicitly to get $\dfrac{dy}{dx}$ in terms of $x$, set it equal to $8$ (since $\dfrac{dy}{dt}=8\dfrac{dx}{dt}$ means $\dfrac{dy}{dx}=8$), solve for $x$, then find the matching $y$ from the curve.
>
> Curve: $6y = x^3+2$
>
> **Step 1: Differentiate w.r.t. $x$.**
> $$6\frac{dy}{dx} = 3x^2 \implies \frac{dy}{dx} = \frac{x^2}{2}$$
>
> **Step 2: Translate the rate condition.** "$y$-coordinate changing $8$ times as fast as the $x$-coordinate" means $\dfrac{dy}{dt}=8\dfrac{dx}{dt}$. Dividing both sides by $\dfrac{dx}{dt}$:
> $$\frac{dy}{dx} = 8$$
>
> **Step 3: Solve for $x$.**
> $$\frac{x^2}{2} = 8 \implies x^2 = 16 \implies x = \pm4$$
> …

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 understanding — Related Rates

Related Rates

The idea: quantities that change together

The core mechanism: differentiate with respect to time

The standard procedure

Worked example

Q.A particle moves along the curve 6y=x3+26y=x^3+26y=x3+2. Find the points on the curve at which the yyy-coordinate is changing 8 times as fast as the xxx-coordinate.

🎓Unlock everything free for 14 days

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.

Unlock everything free for 14 days