Find the curve of intersection for the following surfaces: $z= x + \frac{y}{2} + \frac{1}{2}$ and $z^2= -x^2 + y$
I keep trying and trying to set them equal to each other and just end up with a mess. I wonder if it's possible to set them up with sin and cos, but I honestly have no idea. This is the first step to a problem I have, and then I have to calculate the length, which I can do, but I just can't find the parametric equations.
Best Answer
I am following your own intuition on how to solve. First, we have that
$$z^2=-x^2+y \qquad \qquad z=x+\frac{1}{2}y+\frac{1}{2}$$
Note that both equations have $x,y,$ and $z$, but only $y$ has the same power. So it makes sense then to solve both of them for $y$.
$$y=x^2 +z^2 \qquad \qquad y=2z-2x-1$$
Thus
$$x^2+z^2=2z-2x-1 \quad \Rightarrow \quad (x+1)^2+(z-1)^2=1$$
This is probably where you got lost if I imagine. You need to complete squares to try and get the formula in terms of a conic perhaps. Note that the equation remaining is that of a circle centered at $(-1,1)$ with radius $1$. The fact that it is a circle means we can parametrize using trig functions. So let $x+1=\cos{t}$ and $z-1=\sin{t}$, so $x=\cos{t}-1$ and $z=\sin{t}+1$. If the radius was larger, say $r$, we'd need to change to $r\cos{t}$ and $r\sin{t}$.
Plugging them back into either equation gives us $y$ in terms of $t$. Therefore your three parametric equations are
$$x=\cos{t}-1, \qquad y=2\sin{t}-2\cos{t}-1, \qquad z=\sin{t}+1$$