Answer
$r(t)=ti+(\dfrac{t^2-1}{2})j+(\dfrac{t^2+1}{2})k$
Work Step by Step
Given: $z=\sqrt {x^2+y^2}$ and $z=1+y$
Equate the two given equations.
$x^2+y^2=1+2y+y^2$
After solving, we get $y=\dfrac{x^2-1}{2}$
Plug $x=t$, into the above equation
This gives: $y=\dfrac{t^2-1}{2}$
The value of $z$ is: $z=\dfrac{t^2-1}{2}+1=\dfrac{t^2+1}{2}$
Thus, $r(t)=ti+(\dfrac{t^2-1}{2})j+(\dfrac{t^2+1}{2})k$
