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

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

Mathematics
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normal line to the surface, is that anyting like a normal vector? perpendicular to the surface?
I think its the unit normal vector
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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if it's parametric equations, then it's probably the line, not the vector, i.e., the line inthe direction of the vector
...do we find the normal plane by partial derivatives?
Yes.
i was reading up on some of the vector stuff last night as my eyes were glossing over :)
I would soo love to be able to answer this question as easily and stupididly as I do the algebra stuff lol
if i remember correctly, you don't need to actually find the plane, just the gradient
I think its the unit normal vector
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.
how would I find the equation of the tangent plane at that point?
you a good wizard or a bad wizard? I can never get inline equations....
it's all points perpendicular to that vector
dz/dx(x-x0) + dz/dy(y-y0) + -1(z-z0) = 0, where (x0,y0,z0) is your initial point
ok, so the unit tangent vector is the derivative over the magnitude of the derivative?
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.
so the normal vector is just the gradient of the tangent vector?
No. The normal is the gradient of the surface.
ok, so when i find the gradient where do i plug the coordinates of the point given into the parametric equation?
so using the normal vector, how do i convert it to the parametric equation? sorry i just have trouble grasping this stuff
use slope intercept form, y = mx + b, or x = mt + b
where m is the x coordinate of the gradient at that point, and b is the x coordinate at that oint
and do the same with y and b
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?
correct
that will give you the slope of the x coordinate for the parametric equation
that will give me x0 or will it give me a?
what is a?
x=x0 +at
the parametric equations will take the form of x = (dz/dx@x0)*t + x0
yes, that will give you a
ok thanks, and x0 is just equal to 0 correct?
since the point is (0,0,1)
correct
thank you very much
no problem
ok, so when i find the gradient where do i plug the coordinates of the point given into the parametric equation?
x = mt + b, where m is the gradient of x @ the point, and b is the x coordinate
so be would be 0, for (0,0,1)

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