Question:
A particle moves along the curve $y=x^{3}$. Find the points on the curve at which the $y$-coordinate changes three times more rapidly than the $x$-coordinate.
Solution:
According to the question,
$\frac{d y}{d t}=3 \frac{d x}{d t}$
Now,
$y=x^{3}$
$\Rightarrow \frac{d y}{d t}=3 x^{2} \frac{d x}{d t}$
$\Rightarrow 3 \frac{d x}{d t}=3 x^{2} \frac{d x}{d t}$
$\Rightarrow x^{2}=1$
$\Rightarrow x=\pm 1$
Substituting $x=\pm 1$ in $y=x^{3}$, we get
$y=\pm 1$
So the points are $(1,1)$ and $(-1,-1)$.
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