Question

The derivative of the function represented parametrically as $x=2t-\left|t\right|,y=t^3+t^2\left|t\right|$ is

• A. $0$
• B. $1$
• C. $-1$
• D. does not exist

Answer 1

The correct option is A $0$

We have,

$x=2t-\left|t\right|,y=t^3+t^2\left|t\right|$

$x=3t,y=0$ when $t<0$

$x=t,y=2t^3$ when $t\ge 0$

Eliminating the parameter $t,$ we get

$y=\{\begin{array}{c}\begin{array}{cc}0,& x<0\end{array}\\ \begin{array}{cc}2x^3,& x\ge 0\end{array}\end{array}$

Differentiating w.r.t $x$, we get

$\frac{dy}{dx}=\{\begin{array}{c}\begin{array}{cc}0,& x<0\end{array}\\ \begin{array}{cc}6x^2,& x\ge 0\end{array}\end{array}$

Hence the function is differentaible everywhere and its derivative at $x=0(t=0)$ is $0$