Question

Find the parametric equations for the normal line to the surface z=e^(2x^2+8y^2) at the point (0,0,1)?

Answer 1

normal line to the surface, is that anyting like a normal vector? perpendicular to the surface?

Answer 2

I think its the unit normal vector

Answer 3

You first need to find the equation for the tangent plane at that point. Then you can use the equation of the plane to find the normal to the plane.

Answer 4

if it's parametric equations, then it's probably the line, not the vector, i.e., the line inthe direction of the vector

Answer 5

...do we find the normal plane by partial derivatives?

Answer 6

Yes.

Answer 7

i was reading up on some of the vector stuff last night as my eyes were glossing over :)

Answer 8

I would soo love to be able to answer this question as easily and stupididly as I do the algebra stuff lol

Answer 9

if i remember correctly, you don't need to actually find the plane, just the gradient

Answer 10

I think its the unit normal vector

Answer 11

That's true. \(<\frac{\delta z}{\delta x}|_{x=x_0,y=y_0}, \frac{\delta z}{\delta y}|_{x=x_0,y=y_0}, -1> \) is the normal vector.

Answer 12

how would I find the equation of the tangent plane at that point?

Answer 13

you a good wizard or a bad wizard? I can never get inline equations....

Answer 14

it's all points perpendicular to that vector

Answer 15

dz/dx(x-x0) + dz/dy(y-y0) + -1(z-z0) = 0, where (x0,y0,z0) is your initial point

Answer 16

ok, so the unit tangent vector is the derivative over the magnitude of the derivative?

Answer 17

The tangent vector is the gradient. The unit tangent is just the gradient divided by its magnitude. Just like all unit vectors are the vector divided by its magnitude.

Answer 18

so the normal vector is just the gradient of the tangent vector?

Answer 19

No. The normal is the gradient of the surface.

Answer 20

ok, so when i find the gradient where do i plug the coordinates of the point given into the parametric equation?

Answer 21

so using the normal vector, how do i convert it to the parametric equation? sorry i just have trouble grasping this stuff

Answer 22

use slope intercept form, y = mx + b, or x = mt + b

Answer 23

where m is the x coordinate of the gradient at that point, and b is the x coordinate at that oint

Answer 24

and do the same with y and b

Answer 25

so if the gradient of x is \[4xe ^{{2x^2}+{8y^2}}\] i just plug in the point (0,0,1) and solve it for x0?

Answer 26

correct

Answer 27

that will give you the slope of the x coordinate for the parametric equation

Answer 28

that will give me x0 or will it give me a?

Answer 29

what is a?

Answer 30

x=x0 +at

Answer 31

the parametric equations will take the form of x = (dz/dx@x0)*t + x0

Answer 32

yes, that will give you a

Answer 33

ok thanks, and x0 is just equal to 0 correct?

Answer 34

since the point is (0,0,1)

Answer 35

correct

Answer 36

thank you very much

Answer 37

no problem

Answer 38

ok, so when i find the gradient where do i plug the coordinates of the point given into the parametric equation?

Answer 39

x = mt + b, where m is the gradient of x @ the point, and b is the x coordinate

Answer 40

so be would be 0, for (0,0,1)