Chapter 11.3, Problem 2QQ

To determine

To determine: The correct answer for the values of t for which the curve given by the parametric equations $x=t^3-t^2-1$ and $y=t^4+2t^2-8t$ have a vertical tangent.

Answer to Problem 2QQ

The required correct option is (C).

Explanation of Solution

Given information:

The options:

(A) 0 only

(B) 1 only

(C) 0 and $\frac{2}{3}$ only

(D) 0, $\frac{2}{3}$ and 1

(E) No value

Calculation:

The given parametric equations are $x=t^3-t^2-1$ and $y=t^4+2t^2-8t$ .

Differentiate the expression $x=t^3-t^2-1$ with respect to t .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left(t^3-t^2-1\right)\\ =\frac{d}{dt}\left(t^3\right)-\frac{d}{dt}\left(t^2\right)-\frac{d}{dt}\left(1\right)\\ =3t^2-2t-0\\ =3t^2-2t\end{array}$

To find the values of t for which the curves have a parametric equations have a vertical tangent, use $\frac{dx}{dt}=0$ .

  $\begin{array}{c}\frac{dx}{dt}=0\\ 3t^2-2t=0\\ t\left(3t-2\right)=0\end{array}$

Now, solve the above factors for t .

  $t=0$

And, $\begin{array}{c}3t-2=0\\ \frac{\cancel{3}t}{\cancel{3}}=\frac{2}{3}\\ t=\frac{2}{3}\end{array}$

Thus, the values of t are 0 and $\frac{2}{3}$ .

Hence, the required correct option is (C).