Find the point on the curve

Solution:

Given:

The curve is $y=x^{3}$

$y=x^{3}$

Differentiating the above w.r.t $x$

$\Rightarrow \frac{d y}{d x}=3 x^{2}-1$

$\Rightarrow \frac{d y}{d x}=3 x^{2} \ldots(1)$

Also given the The Slope of the tangent is equal to the $x$ - coordinate,

$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x} \ldots(2)$

From $(1) \&(2)$, we get,

i.e, $3 x^{2}=x$

$\Rightarrow x(3 x-1)=0$

$\Rightarrow x=0$ or $x=\frac{1}{3}$

Substituting $x=0$ or $x=\frac{1}{3}$ this in $y=x^{3}$, we get,

when $x=0$

$\Rightarrow y=0^{3}$

$\Rightarrow y=0$

when $x=\frac{1}{3}$

$\Rightarrow y=\left(\frac{1}{3}\right)^{3}$

$\Rightarrow y=\frac{1}{27}$

Thus, the required point is $(0,0) \&\left(\frac{1}{3}, \frac{1}{27}\right)$