three-dimensional parametric equations

[mathstresser]

Show that the curve with parametric equations x= sin t, y = cos t, z= (sin)^2 t is the curve of intersection of the surfaces z= x^2 and x^2 + y^2 =1. Use this fact to help sketch the curve.

I think that z=x^2 is a a paraboloid that goes up the z-axis, and along the x-axis.
I think that x^2 +y^2=1 is a continuous cylinder that has a radius of 1 and runs along the z-axis.

But, how do I show that they intersect?

 

[soroban]

Helo, mathstresser!

Show that the curve with parametric equations: \(\displaystyle \left\{\begin{array}{ccc}x\,=\,\sin t\\ y\,=\,\cos t \\ z\,=\,\sin^2t\end{array}\)

is the curve of intersection of the surfaces: $\displaystyle z\:=\:x^2$ and $\displaystyle x^2\,+\,y^2\:=\:1$
Use this fact to help sketch the curve.

I think that $\displaystyle z\:=\:x^2$ is a a paraboloid that goes up the z-axis,
and along the x-axis. . . . not quite

There is a parabola in the $\displaystyle xz$-plane ("left wall"), vertex at (0,0,0),
$\displaystyle \;\;$ opening in the $\displaystyle z$-direction ("upward")
$\displaystyle \;\;$ extending continously in the $\displaystyle y$-direction ("right and left").

There's a parabolic "rain gutter" running right and left.

I think that $\displaystyle x^2\,+\,y^2\:=\:1$ is a continuous cylinder
that has a radius of 1 and runs along the z-axis. . . . Right!

There is a vertical cylinder (one-unit radius) standing straight up.

But, how do I show that they intersect?

Solve the system: \(\displaystyle \begin{array}{cc}[1]\\[2]\end{array}\; \begin{array}{cc}z\:=\:x^2 \\ x^2\,+\,y^2\:=\:1\end{array}\)

Substitute [1] into [2]: $\displaystyle \,z \,+\,y^2\:=\:1\;\;\Rightarrow\;\;z\:=\:1\,-\,y^2$

This is a parabola in the $\displaystyle yz$-plane ("back wall"), vertex at (0,0,1), opening "down",
$\displaystyle \;\;$and extending in the $\displaystyle x$-direction ("in and out of the screen")

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We can show that those parametric equations satisfy the two surfaces.
$\displaystyle \;\;$Just plug them in . . .

We have: \(\displaystyle \:\left\{\begin{array}{cc}(1) \\ (2)\\(3)\end{array}\;\begin{array}{ccc}x\,=\,\sin t \\ y\,=\,\cos t \\ z\,=\,\sin^2t\end{array}\)

and: \(\displaystyle \:\left\{\begin{array}{cc}(a)\\(b)\end{array}\;\begin{array}z\:=\:x^2\\ x^2\,+\,y^2\:=\:1\end{array}\)


Substitute (2) and (3) into $\displaystyle (a):\;\;z\;=\;x^2$
. . . . . . . . . . . . . . . . . . . . . . . $\displaystyle \downarrow\;\;\;\;\downarrow$
. . . . . . . . . . . . . . . . . . . . . \(\displaystyle \sin^2t \:=\\sin t)^2\)


Substitute (1) and (2) into $\displaystyle (b):\;\;x^2\,+\,y^2\;=\;1$
. . . . . . . . . . . . . . . . . . . . . . . $\displaystyle \downarrow\;\;\;\downarrow$
. . . . . . . . . . . . . . . . . . . . . $\displaystyle \sin^2t\,+\,\cos^2t\:=\:1$

 

[mathstresser]

Thanks!