OpenStudy (anonymous):

find a vector function that parametrizes the curve C of intersection of the cylinder y^2+4^2=9 and the plane 6x-y+2z-3=0, oriented clockwise when viewed from the positive x-axis.

5 years ago
OpenStudy (blockcolder):

Is that supposed to be y^2+x^2=9 or y^2+z^2=9?

5 years ago
OpenStudy (anonymous):

y^2+4 * z^2=9

5 years ago
OpenStudy (blockcolder):

You can rewrite the elliptical cylinder as \[\large \frac{y^2}{3^2}+\frac{z^2}{(\frac{3}{2})^2}=1\] You can parametrize the elliptical cylinder by using \(y=3\cos\theta, z=\frac{3}{2}\sin\theta\) (You can check that this works by substituting these into the cylinder.) You then substitute these into the plane: \[\begin{align} 6x-y+2z-3=0&\Rightarrow6x-3\cos\theta+3\sin\theta-3=0\\ &\Rightarrow 6x=3\cos\theta-3\sin\theta+3\\ &\Rightarrow x=\frac{1}{2}(\cos\theta-\sin\theta+1) \end{align}\] So that you can parametrize the intersection as: \[\mathbf{r}(\theta)=\left<\frac{1}{2}(\cos\theta-\sin\theta+1),3\cos\theta, \frac{3}{2}\sin\theta \right>\]

5 years ago
OpenStudy (blockcolder):

Oh wait. If the direction should be clockwise, then z should be \(z=-\frac{3}{2}\sin\theta\), and the final parametrization: \[\mathbf{r}(\theta)=\left<\frac{1}{2}(\cos\theta+\sin\theta+1),3\cos\theta, -\frac{3}{2}\sin\theta \right>\]

5 years ago