Question

The slope of the tangent to the curve x = 3t² + 1, y = t³ − 1 at x = 1 is ________________.

Answer 1

The given curve is x = 3t² + 1, y = t³ − 1.

When x = 1, we have

3t² + 1 = 1

⇒ 3t² = 0

⇒ t = 0

Now,

x = 3t² + 1

Differentiating both sides with respect to t, we get

$\frac{dx}{dt}=6t$ .....(1)

y = t³ − 1

Differentiating both sides with respect to t, we get

$\frac{dy}{dt}=3t^2$ .....(2)

$\therefore \frac{dy}{dx}$$=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}$ $=\frac{3t^2}{6t}$ $=\frac{t}{2}$ [Using (1) and (2)]

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

So, the slope of the tangent to the given curve at t = 0 (or x = 1) is 0.

Thus, the slope of the tangent to the curve x = 3t² + 1, y = t³ − 1 at x = 1 is 0.


The slope of the tangent to the curve x = 3t² + 1, y = t³ − 1 at x = 1 is ___0___.