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=
$f(r,\theta)=<2r\cos(\theta),-4r\sin(\theta),r>$

taking some derivatives we get

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

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

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

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

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

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

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

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

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

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

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)

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

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

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

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

taking some derivatives we get

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

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

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

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

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

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

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

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

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

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

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)

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

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

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

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

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

And then just used the fact that

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