Question

Calculus 3: Finding he equation of a tangent plane to the given parametric surface at the specific point: I know how to find my partials, Just a little confused in finding the normal vector.

Answer 1

\[x=u+v~ , ~ y=3u^2~ , ~ z=u-v ~ ;~ (2,3,0)\]

Answer 2

\[\vec r_{u} = \left<1~,~6u~,~ 1\right>\]\[\vec r_v = \left<1~,~0~,~-1\right>\]

Answer 3

\[\vec n= \left<-6u~,~ -(-1-1)~,~ 0-6u\right>= \left<-6u~,~ 2 ~,~6u \right>\]

Answer 4

i think you'll like this http://tutorial.math.lamar.edu/Classes/CalcIII/GradientVectorTangentPlane.aspx

Answer 5

But now I am not sure how to correspond the normal to \(u\) and \(v\)

Answer 6

To find my normal...

Answer 7

OK I shall look.

Answer 8

and this http://tutorial.math.lamar.edu/Classes/CalcIII/TangentNormalVectors.aspx

Answer 9

I don't get why you need the normal and not the gradient. I guess

Answer 10

I've read over these before... but i'm not understanding the connection between the point \(P_0 = (2,3,0)\) and \(u=1~ , ~ v=1\) to end up with the normal vector to the surface, \(\vec n = < -6~,~2~,~-6>\)

Answer 11

Hmm... i'm not EXACTLY sure.. I was following this example in the book and working my homework problem off of that, I can upload that if you like?

Answer 12

its ok I found out why

Answer 13

http://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx

Answer 14

then here is your specififc section http://tutorial.math.lamar.edu/Classes/CalcIII/ParametricSurfaces.aspx

Answer 15

Thanks, it was this I was looking for: http://prntscr.com/5gkvmc

Answer 16

The point gives the x,y,z values, right?

Answer 17

Yes.

Answer 18

You use it to solve for \(u,v\) then.

Answer 19

\(z\) equation\[
0=u-v\implies u=v
\]

Answer 20

\[
2=u+v\implies 2=2u\implies u=1=v
\]

Answer 21

I wanted to try it..but ty :)

Answer 22

you can do it! just make up a new vector :)

Answer 23

like <4, -2, 0>

Answer 24

Now I can apply this to the rest of my problems. Glad I got that cleared up.

Answer 25

Use the \(y\) equation then.

Answer 26

or something

Answer 27

Ok, one sec.

Answer 28

\[3u^2 = 3 \implies u^2=1 \implies u = \pm 1 \] Hrmm... NOw to cmopare it.

Answer 29

compare*

Answer 30

\[u-v=0 \implies v=u\]\[v=u=\pm 1\]That cannot be right.

Answer 31

you have to look at the x then

Answer 32

the plus or minus will be decided there

Answer 33

\[u+v=2\]\[v=2-u\] \[u=1: v=2-1=1\]\[u=-1: v=2-(-1)=2+1=3\]

Answer 34

So the only choice would be \(u=1\)

Answer 35

Right?... @wio @FibonacciChick666

Answer 36

yes

Answer 37

yep

Answer 38

thank you very much.

Answer 39

You finished?

Answer 40

Yeah. I think so. \[u=v=1 \therefore \vec n = <-6~,~2~,~-6>\]

Answer 41

\[
\vec n \cdot (\langle x,y,z\rangle - P) =0
\]

Answer 42

Equation of plane: \(-6(x-2)+2(y-3)-6(z-0)=0\)

Answer 43

I was just confused in understanding how to obtain my normal vector when I had u's and v's in there.