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Wednesday, 1 April 2015

Directional derivatives

The partial derivatives f_x(x_0,y_0) and f_y(x_0,y_0) are the rate of changes of z=f(x,y) at (x_0,y_0) in the positive x- and y- directions. Rate of change in other directions are given by directional derivatives. In fact, the partial derivatives of f with respect to x and y are just special cases of the directional derivative.

To find the directional derivatives of z=f(x,y) at (x_0,y_0) in the direction of the unit vector \vec{u}=\langle u_1, u_2 \rangle in the xy-plane, we introduce an s-axis with its origin at (x_0,y_0), with its positive direction in the direction of \vec{u}. Then the point at s on the s-axis has xy-coordinates x=x_0+su_1, y=y_0+su_2, and the value of z = f(x, y) at the point s on the s-axis is F(s)=f(x_0+su_1,y_0+su_2). We call z=F(s) the cross section through (x_0,y_0) of z=f(x,y) in the direction of \vec{u}.



If the origin is not (0,0), we introduce a second vertical z-axis (in blue) with its origin at the point (x_0, y_0,0) (the origin on the s-axis). Then the graph of z=F(s) is the intersection of the surface z=f(x,y) with the sz-plane. The directional derivative of z=f(x,y) is the slope of the tangent line to this curve in the positive s-direction at s = 0, which is at the point (x_0, y_0, f(x_0, y_0)).



The directional derivative [denoted D_{\vec{u}}f(x_0,y_0)] of z=f(x,y) at (x_0,y_0) in the direction of the unit vector \vec{u}=\langle u_1,u_2 \rangle is the derivative of the cross section function (1) at s=0:
D_{\vec{u}}f(x_0,y_0)=\frac{d}{ds}f(x_0+su_1,y_0+su_2)|_{s=0}

For any unit vector u=\langle u_1,u_2 \rangle, the (directional) derivative of z=f(x, y) at (x_0, y_0) in the direction of \vec{u} is D_{\vec{u}}f(x_0,y_0)=f_x(x_0,y_0)u_1+f_y(x_0,y_0)u_2

Proof:
F'(s)\\=\frac{d}{ds}[f(x_0+su_1,y_0+su_2)]\\=f_x(x_0+su_1,y_0+su_2)\frac{d}{ds}(x_0+su_1)+f_y(x_0+su_1,y_0+su_2)\frac{d}{ds}(y_0+su_2)\\=f_x(x_0+su_1,y_0+su_2)u_1+f_y(x_0+su_1,y_0+su_2)u_2\\=f_x(x_0,y_0)u_1+f_y(x_0,y_0)u_2 \:\:\text{[put s=0]}

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