Find an equation of the tangent plane to the parametric surface x=2rcos(theta) y=-4rsin(theta) z=r at the point (2sqrt(2),-4sqrt(2),2) when r=2, theta=pi/4
z=

2. Originally Posted by karnold9
Find an equation of the tangent plane to the parametric surface x=2rcos(theta) y=-4rsin(theta) z=r at the point (2sqrt(2),-4sqrt(2),2) when r=2, theta=pi/4
z=
$\displaystyle f(r,\theta)=<2r\cos(\theta),-4r\sin(\theta),r>$

taking some derivatives we get

$\displaystyle \frac{\partial f}{\partial r}=<2\cos(\theta),-4\sin(\theta),1>$

$\displaystyle \frac{\partial f}{\partial \theta}=<-2r\sin(\theta),-4r\cos(\theta),0>$

Evaluating them at the point $\displaystyle (2,\frac{\pi}{4})$

$\displaystyle f(2,\frac{\pi}{4})=<2(2)\cos(\pi/4),-4(2)\sin(\pi/4),2>= <2\sqrt{2},-4\sqrt{2},2>$

$\displaystyle \frac{\partial f}{\partial r}=<2\cos(\pi/4),-4\sin(\pi/4),1>=<\sqrt{2},-2\sqrt{2},1>$

$\displaystyle \frac{\partial f}{\partial \theta}=<-2(2)\sin(\pi/4),-4(2)\cos(\pi/4),0>=<-2\sqrt{2},-4\sqrt{2},0>$

Now if we take the cross product of the partials we will get a normal vector to the parameteric surfact at the point.

$\displaystyle \begin{vmatrix} \vec i && \vec j && \vec k \\ \sqrt{2} && -2\sqrt{2} && 1 \\ -2\sqrt{2} && -4\sqrt{2} && 0 \\ \end{vmatrix}= (0+4\sqrt{2})\vec i -(0+2\sqrt{2}) \vec j + (-8-8) \vec k$

So the normal vector is $\displaystyle \vec n = <4\sqrt{2},-2\sqrt{2},-16>$

Now let (x,y,z) be any point in the plane. Using the other point we calculated before $\displaystyle (2\sqrt{2},-4\sqrt{2},2)$ we can create a vector in the plane $\displaystyle \vec v_p=<x-2\sqrt{2},y+4\sqrt{2},z-2>$

Since the above vector lies in the plane and we have a normal vector to the plane the dot product of these two vectors must be zero.(Becuase they are perpendicular)

$\displaystyle \vec n \cdot \vec v_p=0$

$\displaystyle (4\sqrt{2})(x-2\sqrt{2})+(-2\sqrt{2})(y+4\sqrt{2})+(-16)(z-2)=0$
$\displaystyle 4\sqrt{2}x-16-2\sqrt{2}y-16-16z+32=0$ simplifying

$\displaystyle 2\sqrt{2}x-\sqrt{2}y-8z=0 \iff z=\frac{\sqrt{2}}{4}x-\frac{\sqrt{2}}{8}y$

3. Originally Posted by TheEmptySet
$\displaystyle f(r,\theta)=<2r\cos(\theta),-4r\sin(\theta),r>$

taking some derivatives we get

$\displaystyle \frac{\partial f}{\partial r}=<2\cos(\theta),-4\sin(\theta),1>$

$\displaystyle \frac{\partial f}{\partial \theta}=<-2r\sin(\theta),-4r\cos(\theta),0>$

Evaluating them at the point $\displaystyle (2,\frac{\pi}{4})$

$\displaystyle f(2,\frac{\pi}{4})=<2(2)\cos(\pi/4),-4(2)\sin(\pi/4),2>= <2\sqrt{2},-4\sqrt{2},2>$

$\displaystyle \frac{\partial f}{\partial r}=<2\cos(\pi/4),-4\sin(\pi/4),1>=<\sqrt{2},-2\sqrt{2},1>$

$\displaystyle \frac{\partial f}{\partial \theta}=<-2(2)\sin(\pi/4),-4(2)\cos(\pi/4),0>=<-2\sqrt{2},-4\sqrt{2},0>$

Now if we take the cross product of the partials we will get a normal vector to the parameteric surfact at the point.

$\displaystyle \begin{vmatrix} \vec i && \vec j && \vec k \\ \sqrt{2} && -2\sqrt{2} && 1 \\ -2\sqrt{2} && -4\sqrt{2} && 0 \\ \end{vmatrix}= (0+4\sqrt{2})\vec i -(0+2\sqrt{2}) \vec j + (-8-8) \vec k$

So the normal vector is $\displaystyle \vec n = <4\sqrt{2},-2\sqrt{2},-16>$

Now let (x,y,z) be any point in the plane. Using the other point we calculated before $\displaystyle (2\sqrt{2},-4\sqrt{2},2)$ we can create a vector in the plane $\displaystyle \vec v_p=<x-2\sqrt{2},y+4\sqrt{2},z-2>$

Since the above vector lies in the plane and we have a normal vector to the plane the dot product of these two vectors must be zero.(Becuase they are perpendicular)

$\displaystyle \vec n \cdot \vec v_p=0$

$\displaystyle (4\sqrt{2})(x-2\sqrt{2})+(-2\sqrt{2})(y+4\sqrt{2})+(-16)(z-2)=0$
$\displaystyle 4\sqrt{2}x-16-2\sqrt{2}y-16-16z+32=0$ simplifying

$\displaystyle 2\sqrt{2}x-\sqrt{2}y-8z=0 \iff z=\frac{\sqrt{2}}{4}x-\frac{\sqrt{2}}{8}y$
Wow empty set...that was a lot of work.

Couldn't ou have said that since $\displaystyle r>0$ that

$\displaystyle \sqrt{\frac{x^2}{4}+\frac{y^2}{16}}=z$

And then just used the fact that

$\displaystyle z_{\text{tangent plane}}=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)$??