For the curve $x = t^{2} + 1$ , $y = t^{3} - 3t$ , at which values of $t$ is the tangent horizontal?

t = 0

onlyCorrect answerYour answer

t = 1

onlyCorrect answerYour answer

t = \sqrt{3}

and

t = -\sqrt{3}

Correct answerYour answer

t = 1

and

t = -1

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

The tangent is horizontal where $\frac{dy}{dx} = 0$ : $\frac{dy}{dt} = 0$ while $\frac{dx}{dt} \ne 0$ .
$3t^{2} - 3 = 0$ gives $t = \pm 1$ , and there $\frac{dx}{dt} = 2t \ne 0$ .

Worth noting

At $t = 0$ it is $\frac{dx}{dt}$ that is 0: the tangent there is vertical, not horizontal. $t = \pm\sqrt{3}$ is where $y = 0$ .

Ten questions from Derivatives & Second-Order Derivatives, at your own pace.

For the curve x=t2+1x = t^{2} + 1x=t2+1, y=t3−3ty = t^{3} - 3ty=t3−3t, at which values of ttt is the tangent horizontal?