next up previous contents
Next: Functions of 3 variables. Up: Directional derivative and Gradient Previous: Directional derivative and Gradient

Functions of 2 variables.

Definition 6.9   Let f be a function of two real variables, defined on a neighborhhod of the point P0 (x0,y0). Suppopse that f has partial derivatives of first order at P0. The vector $\nabla_{P_0} f = \frac {\partial f}{\partial x}\Bigg\vert _{P_0} \overrightarrow{i} +\frac {\partial f}{\partial y} \Bigg\vert _{P_0} \overrightarrow{j} $ is called the gradient of f at P0.

Example 6.10   Let $f(x,y)=x \cos y$ and $P_0(1, \pi)$. Then $f_x=\cos y$ and $f_y= -x \sin y$. The gradient of f at P0 is the vector $\nabla_{P_0}f= - \overrightarrow{i} $.

Definition 6.11   We use the settings of Def  6.9; let $\overrightarrow{u} $ be a unit vector. The directional derivative of f at P0 in the direction of $\overrightarrow{u} $ is the number:

\begin{displaymath}(D_{\overrightarrow{u} }f)(P_0)=\nabla_{P_0} f \cdot \overrightarrow{u}\end{displaymath}

Example 6.12   Let f(x,y)=4- x2-y2, P0(1,1). The graph of f is displayed on Figure  4(a).

We have: fx=-2x and fy=-2y. At the point P0, the gradient of f is $\nabla_{P_0}f=2\overrightarrow{i} +2 \overrightarrow{j} $.

This result is not surprising as the graph of f is a paraboloid of revolution, i.e. the behaviour of f in any direction perpendicular to the z-axis is the same.
   
Figure 4: Paraboloids.
\begin{figure}
\mbox{\subfigure[A paraboloid of revolution]{\epsfig{file=z_-x2-y...
... elliptic paraboloid]{\epsfig{file=z_-x2-2y2-4.eps,height=5cm} } }\end{figure}

Example 6.13   Let f(x,y)=4- x2-2y2, P0(1,1). The graph of f is displayed on Figure  4(a).

We have: fx=-2x and fy=-4y. At the point P0, the gradient of f is $\nabla_{P_0}f=-2\overrightarrow{i} +2 \overrightarrow{j} $.

Compare this result to the previous example.

Proposition 6.14   Let f, P0, $\nabla_{P_0}f$ and $D_{\overrightarrow{u} }(P_0)$ be as above (see Def  6.11). We denote by $\theta$ the angle of the vectors $\nabla_{P_0}f$ and $\overrightarrow{u} $. Then:
1.
The directional derivative has its greatest value when $\cos \theta = 1$, i.e. when the direction of $\overrightarrow{u} $ and the direction of the gradient $\nabla_{P_0}f$are identical. This means that f increases most fastly at a point P0 in the direction of the gradient.
2.
The directional derivative has its least value when $\cos \theta = -1$, i.e. when the direction of $\overrightarrow{u} $ and the direction of the gradient $\nabla_{P_0}f$are opposite. This means that f decreases most fastly at a point P0 in the direction opposite to the gradient's direction.
3.
If $\overrightarrow{u} $ is orthogonal to the gradient $\nabla_{P_0}f$, then the directional derivative is equal to 0.

Example 6.15   Let f(x,y)=x2-y2. Then: fx=2x and fy=-2y. At the point P0(1,0), we have: $\nabla_{P_0}f = 2 \overrightarrow{i} $. Thus at P0, the greatest value of the directional derivative is afforded in the direction of $\overrightarrow{i} $, i.e. in the positive direction of the x-axis (it points to the right in Figure  5).
  
Figure 5: Two special directions.
\begin{figure}
\mbox{\epsfig{file=z-x2-y2_2.eps,height=4cm} }
\end{figure}

Proposition 6.16   At every point P0 in the domain of f, the vector $\nabla_{P_0}f$ is normal to the level curve through P0.

Example 6.17   Let f(x,y)=x2+y2 and P0(1,2). Then $\nabla f=2x \overrightarrow{i} +
2y \overrightarrow{j} $ and $\nabla_{P_0} f=2 \overrightarrow{i} +
4 \overrightarrow{j} $. This vector is normal to the level curve $\mathcal{C}_{P_0}$through P0, whose equation is x2+y2=5. The line Through P0 and normal to $\nabla_{P_0}$ is the tangent to the circle $\mathcal{C}_{P_0}$ at P0; its equation is 2(x-1)+4(y-2)=0, i.e. x+2y-3=0.
  
Figure 6: A tangent to a level curve.
\begin{figure}
\mbox{\epsfig{file=TangentLevelCurve.eps,height=6cm} }
\end{figure}


next up previous contents
Next: Functions of 3 variables. Up: Directional derivative and Gradient Previous: Directional derivative and Gradient
Noah Dana-Picard
2001-05-30
le: