Mathematics - Differentiation Question with Solution | TestHub

Given curve is $x=3t^2+1$


Second curve is $y=t^3-1$

$\therefore \frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=3t^2\times \frac{1}{6t}=\frac{t}{2}$
But from (i) when $x=1$
we have $1=3t^2+1\Rightarrow 3t^2=0\Rightarrow t=0$
$\therefore$ When $\mathrm{x}=1$ then $\mathrm{t}=0\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=0$
Hence, slope of the tangent to the curve $=0$