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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.

Toño Bobadilla
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    Have you tried with polar co-ordinates? – Saket Gurjar May 21 '20 at 20:00
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    One is a plane and the other a quadric, so their intersection is some sort of conic. Project it onto one of the coordinates planes by eliminating a variable ($y$ looks like an easy one to try first) and parameterize that conic. – amd May 21 '20 at 20:02
  • As the 3d-figure is infinite paraboloid, I'd change the variables to $y=h,,r^2=x^2+z^2,,x=r\cos\varphi,,z=r\sin\varphi$. btw, @amd 's approach works fine – Alexey Burdin May 21 '20 at 20:02

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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

$$x=\cos{t}-1, \qquad y=2\sin{t}-2\cos{t}-1, \qquad z=\sin{t}+1$$

Mando
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Not an answer....but just a visualisation for the intersection.

enter image description here

But I would start with

$$x=r\cos\theta , \ \ \ z=r\sin\theta \ \ \ y=r^2$$and put it in equation of plane.

(Just a thought though)

EDIT: Extending the thought:

$$r\sin\theta=r\cos\theta+\frac{r^2}{2}+\frac{1}{2}$$

$$2\sqrt2r\sin{\left( \theta-\frac{\pi}{4} \right)}=r^2+1$$

Use quadratic formula to get $r$ in terms of $\theta$.

Substitue $r(\theta)$ everywhere where $r$ occurs

Saket Gurjar
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  • What program do you use to generate your graphs? – Mando May 21 '20 at 22:23
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    @Mando I use Geogebra....its free on PC. – Saket Gurjar May 22 '20 at 16:35
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    It renders nicely. I usually use Mathematica, but my coding skills for visualizing the 3D models aren't great, and liked yours. – Mando May 22 '20 at 18:37
  • @Mando I see......I don't know about Mathematica a lot, but Geogebra doesn't involve any sorts of coding....its simply a graphing calculator....I don't know how to code....but I would love to learn....can you suggest anywhere I could start ? – Saket Gurjar May 22 '20 at 18:53
  • Its not really "coding" per se. Just learning a lot of the functions that go into mathematica and using other integrated software commands to change the look of the rendering. Honestly, I learned the basics in a Calc 3 class years ago, and the rest was me just looking up the other information on Wolfram's page. Wolfram's database of their mathematica functions is pretty large and there are great explanation on how to use it. – Mando May 22 '20 at 19:06