Question

The slope of the tangent to the curve $x=3t^2+1,y=t^3-1$ at x = 1 is:

• A. $\frac{1}{2}$
• B. 0
• C. -2
• D. $\infty$

Answer 1

Answer: (B)

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