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)

$0$

Given curves are,

$x=3t^2+1$ ---- ( 1 )

$y=t^3-1$ ---- ( 2 )

Substituting $x=1$ in ( 1 ) we get,

$\Rightarrow$ $3t^2+1=1$

$\Rightarrow$ $3t^2=0$

$\Rightarrow$ $t=0$

Differentiate ( 1 ) w.r.t. $t,$ we get

$\Rightarrow$ $\frac{dx}{dt}=6t$

Differentiate ( 2 ) w.r.t. $t,$ we get

$\Rightarrow$ $\frac{dy}{dt}=3t^2$

$\Rightarrow$ $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3t^2}{6t}$

$\therefore$ $\frac{dy}{dx}=\frac{t}{2}$

Slope of the tangent $={\left(\frac{dy}{dx}\right)}_{t=0}=\frac{0}{2}=0$