Day 9 in Mat329

• Reminder: stow those rectangles....
• You have a homework due next time.

I left you with a little exercise that will lead us into section 14.4. (Some mathematica, as well.)

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.