Question

Find the point on the curve $y=x^2$, where rate of change of $x$-co ordinate is equal to the rate of change of $y$-co ordinate.

Answer 1

Let the point be $(x_1,y_1)$

$y_1={x_1}^2⟹\frac{d{y}_1}{dt}=2x_1\frac{d{x}_1}{dt}$

But given $\frac{d{y}_1}{dt}=\frac{d{x}_1}{dt}⟹x_1=\frac{1}{2}$

$y_1=(\frac{1}{2}{)}^2=\frac{1}{4}$

Point is $(\frac{1}{2},\frac{1}{4})$