Question

Let w=f(x,y) be a function of two independent variables x,y. Write sown a vector which is perpendicular to the tangent plane at the point (x0,y0,z0). What formula could I use here? Or how would I go about solving this, I know that the result would be that the vector which is normal to the surface and so the normal vector dotted with any vector in the plane is 0

Answer 1

looks like gradient problem.

Answer 2

let's try for a sphere ... problem will be greatly simplified.

Answer 3

ooh er sphere is an implicit isnt it?

Answer 4

how about a parabaloid:

z = x^2 + y^2

Answer 5

okay that should do fine ... wow ... it's called parabaloid!!

Answer 6

I have the equation for the tangent plane;
\[z=f(x _{0},y _{0}) +[df/dx (x _{0},y _{0})] (x-x_{0})+[df/dy (x _{0},y _{0})] (y-y_{0})\]
and I read somewhere that the normal vector to a tangent plane is the gradient of the tangent plane which would give;
\[([df/dx(x_{0},y_{0})],[df/dy(x_{0},y_{0})],-1)\]
but that doesn't make sense

Answer 7

i know right! i found out the other day and i cant stop saying it

Answer 8

parabaloid

Answer 9

V = 2x i + 2 y j +k

is this going to be normal to z = x^2 + y^2 or not ..??

Answer 10

I honestly don't know :(

Answer 11

anyone know matlab??

Answer 12

answer mark 3

let us have function f(x,y)

let partial derivatives be
\[f'_x(x) \text{ , } f'_y(y)\]

so we can find tangents parallel to x axis and y axis
eg.
x tangent vector, \[v_x\] will be in the y=0 plane and will have gradient of\[ f'_x(x_0)\]
so will be parrallel to
\[\left(\begin{matrix}1 \\ 0 \\f'_x(x_0)\end{matrix}\right)\]
likewise \[v_y = \left(\begin{matrix}0 \\ 1 \\f'_y(y_0)\end{matrix}\right)\]

so computing their cross product we get

\[\pm\left(\begin{matrix}f'_x(x_0) \\ f'_y(y_0)\\1\end{matrix}\right)\]
as a normal (plus or minus for the order you cross them)

Answer 13

i know of matlab, i dont have it. is it good?

Answer 14

I know matab

Answer 15

matlab*

Answer 16

know how to plot 3d vector lines at points??

Answer 17

for z = x^2 + y^2

we have \[2x_0i + 2y_0j - k\]

i think

Answer 18

as with all my answers on this these are just educated guesses at best

Answer 19

Thats what I had @eigenschmeigen and the way you've explained it makes sense, @experimentX I don't know how to do that, I've only done one course in it which was mostly project based

Answer 20

yeah ... @eigenschmeigen thats amaizing.

Answer 21

yeah, @eigenschmeigen has the right answer, and it's not too hard
as experiment said in the beginning it's a gradient problem

Answer 22

my vector should have been V = 2x i + 2 y j - k
I realize now ... lol

Answer 23

@TuringTest it was quite a worthy to prove that gradient will be perpendicular to surface.

Answer 24

just define\[f(x,y,z)=x^2+y^2-z\]\[\nabla f=<2x,2y,-1>\]and evaluate it at whatever point...
and yeah @experimentX that's a nice thing to be able to prove

Answer 25

whats the name of the delta symbol here? does it have a name?

Answer 26

"del" it is usually called

Answer 27

is that delta?

Answer 28

"del of f"

Answer 29

actually i was looking for it myself.

It's called Del operator and it's equitvalent to (id/dx+jd/dy+kd/dz)

Answer 30

oh cool

Answer 31

\[\nabla f\]\[\nabla\times f\]\[\nabla\cdot f\]gradient, curl, and divergence respectively

Answer 32

though in latex it's called a "nabla"
I don't know why...

Answer 33

wait its called gradient but it gives you the normal vector? weird

Answer 34

Yup ... there's an interesting theorem on this.

Answer 35

\[\nabla f\] is the fastest possible rate change direction of f
if f describes a surface, like in this problem, it should make sense that the fastest possible rate change of f is directly out of the shape, i.e. perpendicular to it
if we instead took\[\nabla z\](imagine it's the equation f a hill or something)
we would get a vector that points along the parabaloid in the "steepest" direction possible

Answer 36

ahh i see

Answer 37

GRAONFKAMKASMDKLASMl just found a mistype in my answer

Answer 38

I didn't o-0

Answer 39

the cross product, the k component should be negative..

Answer 40

its really really annoying me

Answer 41

z component

Answer 42

oh yeah, I see now
signs get lost easily in cross-products :P
it happens...

Answer 43

\[\left(\begin{matrix}f'_x(x_0) \\ f'_y(y_0)\\-1\end{matrix}\right)\]