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anonymous
 5 years ago
Find the parametric equations for the normal line to the surface z=e^(2x^2+8y^2) at the point (0,0,1)?
anonymous
 5 years ago
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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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0normal line to the surface, is that anyting like a normal vector? perpendicular to the surface?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think its the unit normal vector

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if it's parametric equations, then it's probably the line, not the vector, i.e., the line inthe direction of the vector

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0...do we find the normal plane by partial derivatives?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i was reading up on some of the vector stuff last night as my eyes were glossing over :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I would soo love to be able to answer this question as easily and stupididly as I do the algebra stuff lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if i remember correctly, you don't need to actually find the plane, just the gradient

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think its the unit normal vector

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That'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
 5 years ago
Best ResponseYou've already chosen the best response.0how would I find the equation of the tangent plane at that point?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you a good wizard or a bad wizard? I can never get inline equations....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it's all points perpendicular to that vector

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dz/dx(xx0) + dz/dy(yy0) + 1(zz0) = 0, where (x0,y0,z0) is your initial point

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, so the unit tangent vector is the derivative over the magnitude of the derivative?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0so the normal vector is just the gradient of the tangent vector?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No. The normal is the gradient of the surface.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, so when i find the gradient where do i plug the coordinates of the point given into the parametric equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so using the normal vector, how do i convert it to the parametric equation? sorry i just have trouble grasping this stuff

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0use slope intercept form, y = mx + b, or x = mt + b

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where m is the x coordinate of the gradient at that point, and b is the x coordinate at that oint

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and do the same with y and b

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0that will give you the slope of the x coordinate for the parametric equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that will give me x0 or will it give me a?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the parametric equations will take the form of x = (dz/dx@x0)*t + x0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, that will give you a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok thanks, and x0 is just equal to 0 correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since the point is (0,0,1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, so when i find the gradient where do i plug the coordinates of the point given into the parametric equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x = mt + b, where m is the gradient of x @ the point, and b is the x coordinate

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so be would be 0, for (0,0,1)
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