anonymous
  • anonymous
Find the parametric equations for the normal line to the surface z=e^(2x^2+8y^2) at the point (0,0,1)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
normal line to the surface, is that anyting like a normal vector? perpendicular to the surface?
anonymous
  • anonymous
I think its the unit normal vector
anonymous
  • 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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anonymous
  • 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
  • amistre64
...do we find the normal plane by partial derivatives?
anonymous
  • anonymous
Yes.
amistre64
  • amistre64
i was reading up on some of the vector stuff last night as my eyes were glossing over :)
amistre64
  • amistre64
I would soo love to be able to answer this question as easily and stupididly as I do the algebra stuff lol
anonymous
  • anonymous
if i remember correctly, you don't need to actually find the plane, just the gradient
anonymous
  • anonymous
I think its the unit normal vector
anonymous
  • 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
  • anonymous
how would I find the equation of the tangent plane at that point?
amistre64
  • amistre64
you a good wizard or a bad wizard? I can never get inline equations....
anonymous
  • anonymous
it's all points perpendicular to that vector
anonymous
  • anonymous
dz/dx(x-x0) + dz/dy(y-y0) + -1(z-z0) = 0, where (x0,y0,z0) is your initial point
anonymous
  • anonymous
ok, so the unit tangent vector is the derivative over the magnitude of the derivative?
anonymous
  • 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
  • anonymous
so the normal vector is just the gradient of the tangent vector?
anonymous
  • anonymous
No. The normal is the gradient of the surface.
anonymous
  • anonymous
ok, so when i find the gradient where do i plug the coordinates of the point given into the parametric equation?
anonymous
  • anonymous
so using the normal vector, how do i convert it to the parametric equation? sorry i just have trouble grasping this stuff
anonymous
  • anonymous
use slope intercept form, y = mx + b, or x = mt + b
anonymous
  • anonymous
where m is the x coordinate of the gradient at that point, and b is the x coordinate at that oint
anonymous
  • anonymous
and do the same with y and b
anonymous
  • 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
  • anonymous
correct
anonymous
  • anonymous
that will give you the slope of the x coordinate for the parametric equation
anonymous
  • anonymous
that will give me x0 or will it give me a?
anonymous
  • anonymous
what is a?
anonymous
  • anonymous
x=x0 +at
anonymous
  • anonymous
the parametric equations will take the form of x = (dz/dx@x0)*t + x0
anonymous
  • anonymous
yes, that will give you a
anonymous
  • anonymous
ok thanks, and x0 is just equal to 0 correct?
anonymous
  • anonymous
since the point is (0,0,1)
anonymous
  • anonymous
correct
anonymous
  • anonymous
thank you very much
anonymous
  • anonymous
no problem
anonymous
  • anonymous
ok, so when i find the gradient where do i plug the coordinates of the point given into the parametric equation?
anonymous
  • anonymous
x = mt + b, where m is the gradient of x @ the point, and b is the x coordinate
anonymous
  • anonymous
so be would be 0, for (0,0,1)

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