I can find the coordinates of point B on the slope line in 2 dimensions. How do I find a similar point on the steepest tangent line in 3 dimensions?
Starting with a point A with coordinates $\left(x_0,f(x_0)\right)$ that lies on a function f I can find an other point B.
B lies on the tangent line at A. $$ \left(x_0+ h, f(x_0)+h\frac{df}{dx} \right)$$
What is the 3 dimensional analogue of this?
I suspect it is something similar to B3 below:
$$A_3\rightarrow\left(x_0,y_0,f(x_0,y_0)\right)$$ $$B_3\rightarrow\left(x_0+h,y_0+g,f(x_0,y_0)+h \frac{\partial f}{\partial x}+g\frac{\partial f}{\partial y}\right)$$
As an example I used the function $$f_3(x,y)=\sin(x)\cos(y)$$
The line segment between the two points should be perpendicular to the contours. But my incorrect B lands to the left of where it should be for A=$\left(0,0,f_3(0,0) \right)$
Gradient Ascent computes the correct direction of steepest ascent.
$$A\rightarrow B$$ $$\left(x_0,y_0,f\left(x_0,y_0\right)\right)\rightarrow \left(x_0+h\frac{\partial f}{\partial x},y_0+h\frac{\partial f}{\partial y},f\left(x_0+h\frac{\partial f}{\partial x},y_0+h\frac{\partial f}{\partial y}\right)\right)$$ $$X\rightarrow h\nabla X$$
Gradient Ascent doesn't give you another point on the same line.
Instead it gives you another point in the correct direction but below the line on the surface of $f$.
Below is a modified gradient ascent that gives you another point B on the steepest tangent line. Unfortunate it varies its position on the tangent line as a function of the slope.
$$\left(x_0,y_0,f\left(x_0,y_0\right)\right)\rightarrow \left(x_0+h\frac{\partial f}{\partial x},y_0+h\frac{\partial f}{\partial y},f\left(x_0\right)+h\frac{\partial f}{\partial x}\frac{\partial f}{\partial x}+h\frac{\partial f}{\partial y}\frac{\partial f}{\partial y}\right)$$