Question:
A particle moves along the curve $y=(2 / 3) x^{3}+1$. Find the points on the curve at which the $y$-coordinate is changing twice as fast as the $x$-coordinate.
Solution:
Here,
$y=\frac{2}{3} x^{3}+1$
$\Rightarrow \frac{d y}{d t}=2 x^{2} \frac{d x}{d t}$
$\Rightarrow 2 \frac{d x}{d t}=2 x^{2} \frac{d x}{d t}$ $\left[\because \frac{d y}{d t}=2 \frac{d x}{d t}\right]$
$\Rightarrow x=\pm 1$
Substituting the value of $x=1$ and $x=-1$ in $y=\frac{2}{3} x^{3}+1$, we get
$\Rightarrow y=\frac{5}{3}$ and $y=\frac{1}{3}$
So, the points are $\left(1, \frac{5}{3}\right)$ and $\left(-1, \frac{1}{3}\right)$.