Calculus: Early Transcendentals 8th Edition

The first pair of equations only gives the right half of the parabola, that is, $y=x^{2}$. Contrarily, the second pair of equations traces out the whole parabola. For example: First case: when $x=t^{2}, y=t^{4}$ Then $(x,y)=(1,1)$ at $t=1$ and $(x,y)=(1,1)$ at $t=-1$ Second case: when $x=t^{3}, y=t^{6}$ Then $(x,y)=(1,1)$ at $t=1$ and $(x,y)=(-1,1)$ at $t=-1$ Thus, the second pair of equations traces out the whole parabola, that is, $y=x^{2}$. Hence, the given statement is false.