- #1
zollen
- 3
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Hi All,
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.
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.