Day 10 in Mat329

• Reminder: stow those rectangles....
• You have a homework due today (and Friday).
• Your quizzes are returned:
  • Most of you (but not all of you) know the "most important definition in calculus". Memorize it!
  • Clairaut's theorem has some requirements, which some of you neglected to mention (or mangled a bit):

    Suppose $f$ is defined on a disk $D$ that contains the point $(a,b)$. If the functions $\frac{\partial^2 f}{\partial x \partial y}$ and $\frac{\partial^2 f}{\partial y \partial x}$ are both continuous on $D$, then \[ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} \]

Just like derivatives are closely linked to tangent lines, which kiss a curve (osculate); about the slopes of those linear approximations (tangent lines) to a univariate function; so are partial derivatives are about slopes of univariate functions obtained by slicing multivariate functions with planes. If the function is differentiable at a point, then there is a tangent plane, which kisses the surface.

• An example from our author's TEC animations.

• Here's a Mathematica file for similar problems which I stole from someone at the Air Force Academy....

• The tangent plane, provided it exists, is given by three points on the curve. You can find three points using the point at $(a,b)$, and points obtained by moving along the cross-sections in the $x$ and $y$ directions.

  The slopes in those directions are given by $f_x(a,b)$ and $f_y(a,b)$.

  If you take a one-unit step in either of those directions, the $z$-value would rise by either slope: \[ T(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b) \] If you take partial derivatives of $T$ at $(a,b)$, you will see that the partials (and the function value) agree with those of $f$ there.

• #46, p. 944
  • Use the limit definition to show existence
  • Use two different approaches to show failure of continuity (limit doesn't exist)
  • We'll return to finish this up in a moment, requiring #45 (see the definition below -- "To prevent problems....") and some more machinary from below.

• #45: differentiability implies continuity; hence, if the function of #46 is not continuous at (0,0), then it is not differentiable at (0,0).
• Finally we show that the partials are not continuous at (0,0) (fail to have limits). If not, the function of #46 would be differentiable, and the linearization would be a tangent plane for the function. As it is, this function has no tangent plane at the origin.

To prevent problems, we introduce the following definition:

If $z=f(x,y)$, then $f$ is differentiable at $(a,b)$ if the increment $\Delta z$, which is \[ \Delta z = f(a+\Delta x,b+\Delta y) - f(a,b) \] be expressed in the form \[ \Delta z = f_x(a,b)(x-a) + f_y(a,b)(y-a) + \epsilon_1 \Delta x + \epsilon_2 \Delta y \] where $\epsilon_1$ and $\epsilon_2 \to 0$ as $(\Delta x, \Delta y) \to (0,0)$.

We can generally avoid using this definition, however, because of Theorem 8, p. 942:

Existence and continuity of partials $f_x$ and $f_y$ are sufficient to imply differentiability.

For a function $z=f(x,y)$, we define the differentials $dx$ and $dy$ to be independent variables; that is, they can be given any values. Then the differential $dz$, the total differential, is defined by \[ dz=f_x(x,y)dx + f_y(x,y)dy \] Differentials give us approximate changes to a function in the neighborhood of a point, generally when $dx$ and $dy$ are small, obtained by following along the tangent plane.

$\Delta{z}$ gives the actual change, so \[ \Delta{z}\approx dz=f_x(x,y)dx + f_y(x,y)dy \] The picture at the end of the section summarizes what we're up to nicely.

• Why don't you carry out any analysis with a partner or two, and then we can check results with
• Some Mathematica code