Question

At the point $(2,3)$ on the curve $y=x^3-2x+1,$ the gradient of the curve increases $k$ times as fast as its abscissa, Then the value of $k$ is

Answer 1

$y=x^3-2x+1$
Differentiating w.r.t $x$
$y_1=\frac{dy}{dx}=3x^2-2,(\text{gradient}=\frac{dy}{dx})$
Now, rate of change of gradient is,
$\Rightarrow \frac{d{y}_1}{dt}=6x\frac{dx}{dt}$
$(\text{given,}\frac{d}{dt}\left(\frac{dy}{dx}\right)=k\cdot \frac{dx}{dt})$
At $x=2$,$k=6\times 2=12$
Thus $k=12$