Geometry of Partial Derivatives

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

Start with the surface $$z \ = \ f(x,\, y) \ = \ 3x^2 -y^2 -x^3 +2,$$ 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 understand what this means, fix a value of $y$, say $y=-1$. Equivalently, slice the surface $z=f(x,y)$ with the plane $y=-1$, as in the figure. The cubic curve of intersection is shown in orange and can be given the parametrization $${\bf r}(x)\, = \, \bigl\langle\, x,\, -1,\, f(x,\,-1)\,\bigl\rangle\,=\, \bigl\langle\, x,\, -1,\, 3x^2 -x^3 +1\,\bigl\rangle\,.$$ A vector tangent to this orange curve is $${\bf r}'(x)\ = \ \bigl\langle\, 1,\, 0,\, 6x -3x^2\,\bigl\rangle \ = \ \bigl\langle\, 1,\, 0,\, f_x\,\bigl\rangle\,,$$ Thus $f_x$ gives the slope of the surface $z = f(x,\,y)$ as we move in the $x$-direction.

Likewise, if we fix $x= 1$, we see that $f_y$ gives the slope of the other orange curve (obtained by intersecting our surface with the plane $x=1$) as we move 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 a Tangent Vector in the $x$-direction, while $$\frac{\partial f}{\partial y}\Bigl|_{(a,\,b)}\, \qquad \hbox{and} \qquad \Bigl\langle \, 0,\, 1,\, \frac{\partial f}{\partial y}\Bigl|_{a,\,b}\,\Bigl \rangle$$ give the slope and a tangent vector at $P$ in the $y$-direction.

Example 1: 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: Starting at $Q$, the surface slopes up for fixed $x$ as $y$ increases, so $f_y\bigl|_{Q} > 0$, while moving slightly in the $x$ direction takes us neither uphill nor downhill, so so $f_x\bigl|_{Q}= 0$. Likewise, $$f_x\bigl|_{R} \,=\,0,\quad f_y\bigl|_{R} \,>\, 0\,, \qquad f_x\bigl|_{P} \,<\, 0,\, \quad f_y\bigl|_{P} \,=\, 0 \,.$$ We leave the signs of $f_x \bigl|_{S}$ and $f_y\bigl|_S$ to you.

All the same ideas carry over in exactly the same way to functions $f(x,\,y,\,z)$ 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 $f(x,\,y)$ 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 $f(x,\,y)$ and relative size can be read off from the contour map of $f$.

Example 2: To the right is the contour map of the earlier function $$f(x,\, y) \ = \ 3x^2 -y^2 -x^3 +2\,,$$ 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 $R$, the colors are lighter (higher) to the right and darker (lower) to the left, so $f_x \bigl |_R > 0$. Likewise, the colors are lighter below $R$ than above $R$, so $f_y \bigl |_R<0$. At $P$ and at $S$, the ground is flat: $S$ is at a local maximum, and $P$ is at a saddle. In both cases, $f_x=f_y=0$. We leave the remaining points to you.

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