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

- anonymous

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- amistre64

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

- anonymous

I think its the unit normal vector

- anonymous

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.

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## More answers

- anonymous

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

- amistre64

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

- anonymous

Yes.

- amistre64

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

- amistre64

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

- anonymous

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

- anonymous

I think its the unit normal vector

- anonymous

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.

- anonymous

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

- amistre64

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

- anonymous

it's all points perpendicular to that vector

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

No. The normal is the gradient of the surface.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

and do the same with y and b

- anonymous

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?

- anonymous

correct

- anonymous

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

- anonymous

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

- anonymous

what is a?

- anonymous

x=x0 +at

- anonymous

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

- anonymous

yes, that will give you a

- anonymous

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

- anonymous

since the point is (0,0,1)

- anonymous

correct

- anonymous

thank you very much

- anonymous

no problem

- anonymous

- anonymous

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

- anonymous

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

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