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!