Let a function $y = y(x)$ be defined parametrically by $x = 2t - |t|,y = t^{2} + t|t|$. Then, $y^{'}(x),x 0$

Question

Let a function $y = y(x)$ be defined parametrically by $x = 2t - |t|,y = t^{2} + t|t|$. Then, $y^{'}(x),x > 0$

Dr. Abubacker · Accepted Answer

(d)
$\because x = 2t - |t| = \left\{ \begin{matrix}
t, & t \geq 0 \
3t, & t < 0
\end{matrix} ight.\ ,$
$	herefore t = \left\{ \begin{matrix}
x, & x \geq 0 \
x/3, & x < 0
\end{matrix} ight.\$
$	herefore y = t^{2} + t|t| = \left\{ \begin{matrix}
2t^{2}, & x \geq 0 \
0, & x < 0
\end{matrix} ight.\$
$= \left\{ \begin{matrix}
2x^{2}, & x \geq 0 \
0, & x < 0
\end{matrix} ight.\$
Hence, $y^{'}(x) = \left\{ \begin{matrix}
4x,\  & x > 0 \
0, & x < 0
\end{matrix} ight.\$
$	herefore$ We can’t find $\frac{dx}{dt}$ as the derivative does not exist at $t = 0$

Q. Let a function $y = y(x)$ be defined parametrically by $x = 2t - |t|,y = t^{2} + t|t|$. Then, $y^{'}(x),x > 0$

Explanation

Reviewed by Dr. Abubacker

Want more practice?