Question about gradient, tangent plane and normal line.

[zollen]

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! Any information would be much appreciated.

Thanks.

 

[HallsofIvy]

Hi All,

This question is about vector calculus, gradient, directional derivative and normal line.

If the gradient is the direction of the steepest ascent:

No, the gradient is a vector that points in the direction of 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),

No, this is a vector not a line!

But both derivative_f_x(x,y) & derivative_f_y(x,y) are gradient (the slope of the tangent plane).

No, they are each part of the gradient vector.

I don't think the steepest ascent/descent is the slope of

the normal line perpendicular to the tangent plane! Any information would be much appreciated.

Thanks.

You are confusing three or more dimension with two dimensions. The slope of a line in two dimensions, y= mx+ b, is the number "m". The corresponding
equation for a line in three dimensions would be z= m<x, y>+ b where m is the gradient vector and m<x, y> is the dot product of two vectors. If you have a function given as z= f(x,y,z) then the corresponding equation for the plane is \(\displaystyle z= \left(\frac{\partial f}{\partial x}\right)(x- a)+ \left(\frac{\partial f}{\partial y}\right)(y- b)+ f(a, b)\). The part of that that involves x and y is \(\displaystyle \left(\frac{\partial f}{\partial x}\right)(x- a)+ \left(\frac{\partial f}{\partial y}\right)(y- b)\) which can be written as \(\displaystyle \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\cdot\left(x- a, y- b\right)\) so that the gradient, \(\displaystyle \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\) is perpendicular to the vector \(\displaystyle \left(x, y\right)\) so perpendicular to the tangent plane.