## Calculus: Early Transcendentals 8th Edition

Given: $r(t)=(0,t^{2},4t)$ Here, $x=0,y=t^{2},z=4t$ As $z=4t$ gives $t=\frac{z}{4}$ and $y=t^{2}$ becomes $y=(\frac{z}{4})^{2}$ $y=\frac{z^{2}}{16}$ which represents a standard form of the equation of a parabola. Hence, the statement is true.