I'm struggling with the following problem:
Given $\, z = \sqrt{x^2 + y^2}\,$ and $\, z = y+1\,$ find the vector function represented by the curve of intersection of the surfaces using the parametrization $\, x = t$.
I've done problems like this before but this one seems different. The $\,x\,$ is contained in a function with two other variables. My attempt was equating the two functions of $\,z$. Then I solved for $\,y\,$ which got me:
$\,y=\left(t^2-1\right)\big/\,2$.
Not sure how to approach this problem.
Best Answer
First, equate $\,z\,$ and square both sides of obtained expression: $$ z = \sqrt{x^2 + y^2} = y+1 \implies x^2 + y^2 = y^2 + 2y + 1 \implies \boxed{y = \dfrac{x^2 - 1}{2}} $$ Second, apply parametrization $\, x = t$: \begin{align} \begin{cases} y = \dfrac{x^2 - 1}{2}\\ z = y + 1 \\ z = \sqrt{x^2 + y^2} \end{cases} \stackrel{x\, =\, t}{\implies} \begin{cases} x = t \\ y = \dfrac{t^2 - 1}{2} \\ z = y + 1 \end{cases} \implies \begin{cases} x = t \\ y = \dfrac{t^2 - 1}{2} \\ z = \dfrac{t^2 +1}{2} \end{cases} \end{align}
Therefore your vector function will look like $$ F\left(t\right) = \begin{pmatrix} t \\ \dfrac{t^2 - 1}{2} \\ \dfrac{t^2 + 1}{2} \end{pmatrix} = \frac 1 2\; \begin{pmatrix} 2t \\ \ t^2 - 1 \\ t^2 + 1 \end{pmatrix} $$