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 $\vec{u}$.
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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 $(x_0, y_0, f(x_0, y_0))$.
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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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