Thank you for reporting, we will resolve it shortly
Q.
The slope of the tangent to the curve $x = 3t^2 + 1, y= t^3 -1$ at x = 1 is:
Application of Derivatives
Solution:
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