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Question about Gradient, Directional Calculus and Normal line

 
 
zollen
 
Reply Sun 22 Jan, 2017 01:49 pm
This question is about vector calculus, gradient, directional derivative and normal line.

If the gradient is the direction of the steepest ascent:

>> gradient(x, y) = [ derivative_f_x(x, y), derivative_f_y(x, y) ]

Then it really confuse me as when calculating the normal line perpendicular to the tangent plane, the formula would be:

>> normal line = (derivative_f_x(x, y), derivative_f_y(x, y), z),

But both derivative_f_x(x,y) & derivative_f_y(x,y) are gradient (the slope of the tangent plane). I don't think the steepest ascent/descent is the slope of the normal line perpendicular to the tangent plane!

For example
Find a vector function for the line normal to x^2 + 2y^2 + 4z^2 = 26 at (2, -3, -1).
Answer: (2 + 4t, -3 -12t, -1 - 8t).

Anyone care to give it a shot and show me the step??


Any information would be much appreciated.

Thanks.
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