A particle moves along the curve

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 $\left(-1, \frac{1}{3}\right)$.

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