## Vector calculus summary

Advanced Calculus Demystified – David Bachman

Level curves of a function allow exploration by holding one variable constant (essentially contour plots of a surface).  A function $f$ has a limit at (a,b) only when it approaches the same value $L$ no matter how (x,y) approach (a,b).  A function is continuous at (a,b) if its limit at (a,b) equals the value of the function there.

$\lim_{(x,y)to (a,b)}f(x,y) = f(a,b)$

Functions don’t exist in their domain when 1) there is division by zero, 2) the square root of a negative, 3) the log of a non-positive, or 4) the tan of odd multiples of $pi/2$.

The multivariable calculus version of the chain rule can be given by

$\frac{d}{dt} f(\phi(t)) = \frac{ \partial f}{ \partial x} \frac{\partial x}{\partial t} + \frac{ \partial f}{ \partial y} \frac{\partial y}{\partial t}$

Cylindrical coordinates

$x = r \cos \theta$

$y = r \sin \theta$

$z = z$

Spherical coordinates

$x = \rho \cos \theta \sin \phi$

$y = \rho \sin \theta \sin \phi$

$z = \rho \cos \phi$

In parametrizations, always specify the ranges.

The vector dot product is given by $v \cdot w = |v| |w| \cos \theta$

Critical points in f are where $\nabla f = <0,0,0>$.  The second derivative is actually four derivatives, best organized in a matrix

$\left| \begin{array}{cc} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\ \frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial y^2} \end{array} \right|$

The determinant of that matrix at $(x_0, y_0)$  where $\nabla f(x_0, y_0) = <0,0>$ is $D$.

$D(x_0, y_0) > 0 \Rightarrow f(x,y) \textrm{has a local min or max at} (x_0,y_0)$

$D(x_0, y_0) < 0 \Rightarrow f(x,y) \textrm{has a saddle at} (x_0, y_0)$

$D(x_0, _0) = 0 \Rightarrow \textrm{no information at} f(x_0, y_0)$

if $\frac{\partial^2 f}{\partial x^2}(x_0,y_0) > 0$ then $f(x_0, y_0)$ is a local minimum.  If $<0$ then a local maximum.

It is customary to denote the points around the boundary of a domain $D$ as $\partial D$ (not to be confused with a partial derivative).

The method of Lagrange multipliers allows determination of extrema on $D \cup \partial D$.   Let $f$ be defined over $D$ and $\partial D$.  Find a new function G(x,y) for which $\partial D$ is a level curve.  $\nabla g$ is always perpendicular to g’s level curves.  $latex\nabla f(x_0,y_0)$ must be parallel to $\nabla g(x_0,y_0)$ and so one must be a scalar multiple of the other (by $\lambda$).  Those points that satisfy the relation $\nabla g = \lambda \nabla f$ must be included in the evaluation along with the normal critical points.

If $V$ and $W=$ then $D = \begin{array}{cc} a & b \\ c & d \end{array}$ is the area of a parallelogram.  The 3D equivalent is the volume of a parallelepiped.   The determinant of the cross product is the area of a parallelogram.

Line integral

$\int_C f(x,y) ds = \int_a^b f(\psi(t)) | \frac{\partial \psi}{\partial t} | dt$

Surface integral

$\int_S f(x,y,z) dS = \int\int_R f(\psi(u,v)) | \frac{\partial \psi}{\partial u} \times \frac{\partial \psi}{\partial v} du dv$

3D volume

$\int\int\int_V f(x,y,z) dx dy dz = \int \int \int_Q f(\psi(u,v,w)) | \frac{\partial \psi}{\partial u} \frac{\partial \psi}{\partial v} \frac{\partial \psi}{\partial w}| du dv dw$

Path independence

$\int_C \nabla f \cdot ds = f(\psi(b)) - f(\psi(a))$

Green’s Theorem

$\int \int_Q \frac{\partial g}{\partial x} -\frac{\partial f}{\partial y} dx dy = \int_{\partial Q} W \cdot ds$

Stoke’s Theorem

$\int_S (\nabla \times W) \cdot dS = \int_{\partial S} W \cdot ds$

Gauss’ Theorem

$\int \int \int_V \nabla \cdot W dx dy dz = \int_{\partial V} W \cdot dS$