Investigate the interactive figure to the right. Use the Triangle Law to interpret ${\bf r}(t+h) - {\bf r}(t)$ as the green vector.   (i) How does this green vector correspond to the 'rise' in the scalar-valued case?   (ii) What properties of vectors are needed to identify the Newtonian Quotient with the brown vector? How does the brown vector correspond to slope?   (iii) The Newtonian Quotient approaches the tangent vector to the graph of ${\bf r}(t)$ at $P$ as $h \to 0$.

    To understand partial derivatives geometrically, we need to interpret the algebraic idea of fixing all but one variable in terms of the by now familiar idea of slicing a surface by a plane to produce a curve in space.

Start with $$z \ = \ f(x,\, y) \ = \ 3x^2 -y^2 -x^3 +2$$ whose graph is shown to the right. Then $$f_x \ = \ \frac{\partial f}{\partial x} \ = \ \frac{\partial } {\partial x}\, \bigl( 3x^2 -y^2 -x^3 +2\bigl) \ = \ 6x -3x^2 \,.$$ To exploit interactivity, fix $y = -1$ and use the 'Fix $y$'-slider to intersect the graph of $f$ by the plane $y = -1$. The cubic curve of intersection shown in orange is the graph of the vector function $${\bf r}(x)\, = \, \bigl\langle\, x,\, -1,\, f(x,\,-1)\,\bigl\rangle\,=\, \bigl\langle\, x,\, -1,\, 3x^2 -x^3 +1\,\bigl\rangle\,,$$ (use 'Curve 1'-slider). The tangent vector to this orange curve is $${\bf r}'(x)\ = \ \bigl\langle\, 1,\, 0,\, 6x -3x^2\,\bigl\rangle \ = \ \bigl\langle\, 1,\, 0,\, f_x\,\bigl\rangle\,,$$ (use 'Tangent Vector 1' button). Thus $f_x$ gives the slope of the graph of $z = f(x,\,y)$ in the $x$-direction.   If we fix $x= 1$, say, and use the other sliders and button, we see that $f_y$ gives the slope of the graph of $z = f(x,\,y)$ in the $y$-direction. Thus

At a point $P=(a,\,b,\, f(a,\,b))$ on the graph of $z = f(x,\, y)$,     the value $\displaystyle\frac{\partial f}{\partial x}\Bigl|_{(a,\,b)}$ is the Slope in the $x$-direction,     $\displaystyle \Bigl\langle \, 1,\, 0,\, \frac{\partial f}{\partial x}\Bigl|_{a,\,b}\,\Bigl \rangle$ is the Tangent Vector in the $x$-direction, while $$\frac{\partial f}{\partial y}\Bigl|_{(a,\,b)}\,, \qquad \qquad \Bigl\langle \, 0,\, 1,\, \frac{\partial f}{\partial y}\Bigl|_{a,\,b}\,\Bigl \rangle$$ give the respective slope and tangent vector at $P$ in the $y$-direction.

Example 4: determine whether $f_x ,\ f_y$ are positive, negative, or zero at the points $P,\ Q, \ R,$ and $S$ on the surface to the right. Solution: at $Q$, for instance, the surface slopes up for fixed $x$ as $y$ increases, so $f_y\bigl|_{Q} > 0$, while the surface remains at a constant level at $Q$ in the $x$ direction for fixed $y$, so $f_x\bigl|_{Q}= 0$. On the other hand, $$f_x\bigl|_{R} \,=\,0,\quad f_y\bigl|_{R} \,>\, 0\,, \qquad f_x\bigl|_{P} \,<\, 0,\, \quad f_y\bigl|_{P} \,=\, 0 \,.$$ But what happens at $S$?

    All the same ideas carry over in exactly the same way to functions $w = f(x,\,y,\,z)$ of three or more variables - just don't expect lots of pictures!! The partial derivative $f_z$, for instance, is simply the derivative of $f(x,\,y,\,z)$ with respect to $z$, keeping the variables $x$ AND $y$ fixed now.

    Information about the partial derivatives of a function $z = f(x,\,y)$ can be detected also from the contour map of $f$. Indeed, as one knows from using contour maps to learn whether a path on a mountain is going up or down, or how steep it is, so the sign of the partial derivatives of $z = f(x,\,y)$ and relative size can be read off from the contour map of $f$.

Example 5: To the right is the contour map of the earlier function $$z \ = \ f(x,\, y) \ = \ 3x^2 -y^2 -x^3 +2\,,$$ with 'higher ground' being shown in lighter colors and 'lower ground' in darker colors. Determine whether $f_x,\, f_y$ are positive, negative, or zero at $P,\, Q,\, R,\, S$, and $T$.     At $R$, for instance, are the contours increasing or decreasing as $y$ increases for fixed $x$? That will indicate the sign of $f_y$. But what happens at $P$ or at $S$. Don't forget that the graph of $f$ appears in the earlier interactive animation!