http://sites.science.oregonstate.edu/BridgeBook/ 2020-01-26T15:17:28-08:00 http://sites.science.oregonstate.edu/BridgeBook/ http://sites.science.oregonstate.edu/BridgeBook/lib/images/favicon.ico text/html 2015-08-28T14:17:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/2ndderivtest?rev=1440796620 The argument presented here requires some familiarity with the properties of eigenvalues and eigenvectors of symmetric matrices. Curvature Recall that the curvature of a parametric curve $\rr(t)$ is given by \begin{equation} \kappa = \frac{|\vv\times\aa|}{|\vv|^3} \end{equation} This expression is independent of the parameter used, so choose an arclength parameterization, in which case $\vv$ is just the unit tangent vector $\TT$ and $|\vv|=1$, so that \begin{equation} \kappa = |\TT\times\aa| \… text/html 2015-08-22T15:11:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/acknowledge?rev=1440281460 This book grew out of class notes for several different courses at Oregon State University (OSU) in both mathematics and physics. In mathematics, these courses included the two second-year calculus courses, covering Multivariable Calculus and Vector Calculus, respectively, which form part of the Bridging the Vector Calculus Gap project at OSU, begun in 2000 with funding from the National Science Foundation (NSF). In physics, these courses included the two third-year ``paradigms'' courses entit… text/html 2015-08-28T13:52:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/arclength?rev=1440795120 Consider the infinitesimal (3-d) version of the Pythagorean Theorem \begin{equation} ds^2 = dx^2 + dy^2 + dz^2 \end{equation} which implies that \begin{equation} \left({ds\over du}\right)^2 = \left({dx\over du}\right)^2 + \left({dy\over du}\right)^2 + \left({dz\over du}\right)^2 \end{equation} (just ``divide'' by $du^2$...) Thus, we see that the speed $v$ satisfies \begin{equation} v = {ds\over du} \end{equation} Since speed equals distance divided by time, the arc length $s$ between the poin… text/html 2015-08-27T22:25:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/basisvectors?rev=1440739500 Vectors are often expressed in terms of their components in rectangular coordinates. One common convention is to write these components as an ordered triple, namely $$\vv=(v_x,v_y,v_z)$$ or as a list of components, namely $$\vv=\langle v_x,v_y,v_z \rangle$$ Another popular convention is to call the basis vectors in the $x$, $y$, and $z$ directions $\{\ii,\jj,\kk\}$, respectively, leading to $$\vv=v_x\ii +v_y\jj+ v_z\kk$$ This convention is a historical hangover from attempts to describe electro… text/html 2015-08-22T08:47:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/basisvectorshint?rev=1440258420 A basis vector of the form $\widehat{coordinate}$ is the unit vector that points in the direction in which $coordinate$ is changing. For example, $\xhat$ is the unit vector that points in the direction that $x$ is changing. Figure 1: The definition of cylindrical and spherical coordinates, showing the associated basis vectors. text/html 2015-08-27T15:16:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/chain?rev=1440713760 One advantage of working with differentials is that the chain rule becomes automatic. For example, if you know the temperature $T$ of a metal girder as a function of position $x$, and you know your position as a function of time $t$, then you can of course obtain temperature as a function of time by substitution. The resulting expression could be differentiated to determine how quickly the temperature at your location is changing as you move along the girder. But you could also use the chain … text/html 2015-08-27T15:29:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/chainapp?rev=1440714540 When differentiating functions of several variables, it is essential to keep track of which variables are being held fixed. As a simple example, suppose \begin{equation} f = 2x+3y \end{equation} for which it seems clear that \begin{equation} \Partial{f}{x} = 2 \end{equation} But suppose we know that \begin{equation} y=x+z \end{equation} so that \begin{equation} f = 2x+3(x+z) = 5x+3z \end{equation} from which it seems equally clear that \begin{equation} \Partial{f}{x} = 5 \end{equation} In such … text/html 2016-04-21T10:45:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/chainddiag?rev=1461260700 Figure 1a:A tree diagram when $x$, $y$ are functions of $u$, $v$. Figure 1b:A tree diagram when $u$, $v$ are functions of $x$, $y$. Suppose $f=f(x,y)$. Then of course \begin{equation} df = \left(\Partial{f}{x}\right)_y dx + \left(\Partial{f}{y}\right)_x dy \end{equation} where the subscripts keep track of which variables are being held constant when taking partial derivatives. If $x=x(u,v)$, $y=y(u,v)$, then \begin{equation} dx = \left(\Partial{x}{u}\right)_v du + \left(\Partial{x}{v}\right)… text/html 2015-08-27T15:21:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/chaindiag?rev=1440714060 Figure 1a:A tree diagram when $x$, $y$ are functions of $u$, $v$. Figure 1b:A tree diagram when $u$, $v$ are functions of $x$, $y$. Suppose $f=f(x,y)$. Then of course \begin{eqnarray*} df = \left(\Partial{f}{x}\right)_y dx + \left(\Partial{f}{y}\right)_x dy \end{eqnarray*} where the subscripts keep track of which variables are being held constant when taking partial derivatives. If $x=x(u,v)$, $y=y(u,v)$, then \begin{eqnarray*} dx = \left(\Partial{x}{u}\right)_v du + \left(\Partial{x}{v}\rig… text/html 2012-09-14T12:23:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/choosing?rev=1347650580 Coming soon. text/html 2015-08-21T15:33:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/circulation?rev=1440196380 An important special case of a line integral occurs when the curve $C$ is closed, that is, when it starts and ends at the same point. In this case, we write the integral in the form \begin{equation} W = \oint\limits_C \FF\cdot d\rr \end{equation} and refer to it as the circulation of $\FF$ around $C$. Unless otherwise specified, it is assumed that the curve is oriented in the counterclockwise direction. As with all vector line integrals, reversing the orientation results in an overall minus s… text/html 2010-06-20T10:38:38-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/conservative?rev=1277055518 The fundamental theorem implies that vector fields of the form $\FF=\grad{f}$ are special; the corresponding line integrals are always independent of path. One way to think of this is to imagine the level curves of $f$; the change in $f$ depends only on where you start and end, not on how you get there. These special vector fields have a name: A vector field $\FF$ is said to be conservative if there exists a potential function $f$ such that $\FF=\grad{f}$. text/html 2010-06-20T10:38:38-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/convergence?rev=1277055518 If a function has a power series expansion around some point $a$, then the circle of convergence extends to the nearest point at which the function is not analytic. (Briefly, a function which is not analytic is singular in some way. A function is certainly not analytic at any point at which its value becomes infinite or at a branch point of a root.) text/html 2015-08-22T13:12:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/coords?rev=1440274320 Choosing an appropriate coordinate system for a given problem is an important skill. The most frequently used coordinate system is rectangular coordinates, also known as Cartesian coordinates, after Ren\'e D\'escartes. One of the great advantages of rectangular coordinates is that they can be used in any number of dimensions. text/html 2015-08-29T14:28:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/coords2?rev=1440883680 ---------- The whole point of using curvilinear coordinates is that they are better adapted to the symmetries of the given problem. Ideally, this means that the entire problem should be done in curvilinear coordinates, without converting between coordinate systems, although this is not always possible. From this point of view, while it is certainly worth learning how to convert between, say, rectangular and polar coordinates, it is also worth learning how to avoid doing so as much as possible. text/html 2015-08-28T14:16:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/cov?rev=1440796560 \begin{itemize}\item RECALL: $dA=dx\,dy=r\,dr\,d\phi$ \end{itemize} Where did the factor of $r$ come from in the above expression? Consider a ``coordinate rectangle'' bounded by curves of the form $r=\hbox{constant}$, $\phi=\hbox{constant}$. If this ``rectangle'' is small enough, its sides have length $dr$ and $r\,d\phi$, so that its area is the product $(dr)(r\,d\phi)$. text/html 2015-08-27T22:37:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/cross2?rev=1440740220 The cross product is fundamentally a directed area. The magnitude of the cross product is defined to be the area of the parallelogram whose sides are the two vectors in the cross product. In the figure above, if the horizontal vector is $\vv$ and the upward-pointing vector is $\ww$, then the height of the parallelogram is $|\ww|\sin\theta$, so its area is \begin{equation} |\vv\times\ww| = |\vv||\ww|\sin\theta \label{crossmagnitude} \end{equation} which is therefore the magnitude of the … text/html 2010-08-19T16:06:05-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/csint?rev=1282259165 text/html 2015-08-28T14:15:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/curl2?rev=1440796500 Figure 1: Computing the horizontal contribution to the circulation around a small rectangular loop. Consider a small rectangular loop in the $yz$-plane, with sides parallel to the coordinate axes, as shown in Figure 1. What is the circulation of $\AA$ around this loop? text/html 2015-08-28T14:14:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/curvature?rev=1440796440 The tangent vector $\TT$ has constant magnitude, and changes only in direction. Thus, its derivative measures how much the curves bends --- this is the curvature $\kappa$: \begin{equation} \kappa = \left| {d\TT\over ds} \right| \end{equation} If $\kappa\ne0$, then the curve bends in a particular direction, the principal unit normal vector $\NN$, given by \begin{equation} \NN = {1\over\kappa} {d\TT\over ds} \end{equation} text/html 2015-08-28T14:14:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/curves?rev=1440796440 ! Figure 1:The graph of a function of 1 variable. There are many ways to describe a curve. Consider the following descriptions: \begin{itemize}\item the unit circle; \item $x^2+y^2=1$; \item $y=\sqrt{1-x^2}$; \item $r=1$; \item $x=\cos\phi$, $y=\sin\phi$; \item $\rr(\phi)=\cos\phi\,\xhat+\sin\phi\,\yhat$; \end{itemize} all of which describe (pieces of) the same curve. Here are some more: \begin{itemize}\item The graph of $y=x^2$; \item The figure shown at the right. \end{itemize} Which rep… text/html 2015-08-29T17:59:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/curvilinear?rev=1440896340 Rectangular Coordinates The arbitrary infinitesimal displacement vector in Cartesian coordinates is: $$d\rr=dx\,\xhat + dy\,\yhat +dz\,\zhat$$ Given the cube shown below, find $d\rr$ on each of the three paths, leading from $a$ to $b$. Path 1: $d\rr=$ text/html 2015-08-22T09:16:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/curvint?rev=1440260160 Evaluating multiple integrals in other coordinate systems involves the same idea as in rectangular coordiantes: chop and add. The difference is in the way one chops. In cylindrical coordinates, one chops a region into small pieces that look like ``pineapple chunks'', with volume given by \begin{equation} dV = r\,dr\,d\phi\,dz \end{equation} This construction is illustrated in the first figure below (which can also be used for polar coordinates if one ignores the vertical path). text/html 2010-08-19T16:06:05-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/cylsph?rev=1282259165 text/html 2012-09-15T10:17:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/da?rev=1347729420 The simplest surfaces are those given by holding one of the coordinates constant. Thus, the $xy$-plane is given by $z=0$. Its (surface) area element is $dA=(dx)(dy)=(dr)(r\,d\phi)$, as can easily be seen by drawing the appropriate small rectangle. The surface of a cylinder is nearly as easy, as it is given by $r=a$ in cylindrical coordinates, and drawing a small ``rectangle'' yields for the surface element \begin{eqnarray} \hbox{cylinder:} \qquad && \dS = (a\,d\phi)(dz) = a\, d\phi \, dz \q… text/html 2015-08-18T21:49:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/dadvcurvi?rev=1439959740 Using the general formula for the vector surface element, \begin{equation} d\SS = d\rr_1 \times d\rr_2 \label{dr1dr2} \end{equation} find explicit formulas for the vector surface element in each of the following cases: \begin{itemize}\item a plane in both rectangular and polar coordinates; \item the three surfaces (top, bottom, and curved side) of a cylinder with finite length; \item the surface of a sphere. \end{itemize} text/html 2015-08-28T14:14:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/dadvcurvihint?rev=1440796440 In (ss)~{Activity: Surface Elements on Planes, Cylinders and Spheres}, you should have obtained the following common surface elements: \begin{eqnarray*} \hbox{(plane) $\quad z={\rm const}$}: \qquad && d\SS=dx\,dy \>\zhat \\ \hbox{(plane) $\quad z={\rm const}$}: \qquad && d\SS=r\,dr\,d\phi \>\zhat \\ \hbox{(cylinder---top) $\quad z={\rm const}$}: \qquad && d\SS=r\,dr\,d\phi \>\zhat \\ \hbox{(cylinder---bottom) $\quad z={\rm const}$}: \qquad && d\SS=-r\,dr\,d\phi \>\zhat \\ \hbox{(cylinder--… text/html 2015-08-21T17:51:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/dadvs?rev=1440204660 Using the expressions for the infinitesimal elements of length derived and discussed in (ss)~{Activity: Infinitesimal Distance in Cylindrical and Spherical Coordinates} and (ss)~{Wrap-Up: Infinitesimal Distance in Cylindrical and Spherical Coordinates}, you should find explicit formulas for the surface or volume elements in each of the following cases: \begin{itemize}\item the scalar surface elements for the three surfaces (top, bottom, and curved side) of a cylinder with finite length; \item t… text/html 2015-08-27T09:44:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/dadvshint?rev=1440693840 Evaluating surface integrals in other coordinate systems involves the same idea as double integrals in rectangular coordiantes: chop and add. The difference is in the way one chops. In polar coordinates, one chops a region into pie shaped regions using radial lines and circles. These lines are orthogonal to each other, so that a small enough piece is nearly rectangular, which means that its area is just its length times its width. As discussed in (ss)~{Double Integrals in Polar Coordinates… text/html 2015-08-28T14:14:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/davec?rev=1440796440 Since surfaces are two-dimensional, chopping up a surface is usually done by drawing two families of curves on the surface. Then you can compute $d\rr$ on each family and take the cross product, to get the vector surface element in the form \begin{equation} d\SS = d\rr_1 \times d\rr_2 \label{Surface} \end{equation} In order to determine the area of the vector surface element, we need the magnitude of this expression, which is \begin{equation} \dS = |d\SS| = |d\rr_1\times d\rr_2| \label{Scala… text/html 2015-08-19T14:32:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/delta3d?rev=1440019920 The three-dimensional delta function must satisfy: \begin{equation} \int\limits_{\hbox{$\scriptstyle all~space$}} \delta^3(\rr-\rr_0)\,d\tau=1 \end{equation} where $\rr=x\,\xhat +y \,\yhat +z\,\zhat$ is the position vector and $\rr_0=x_0\,\xhat +y_0 \,\yhat +z_0\,\zhat$ is the position at which the ``peak'' of the delta function occurs. In rectangular coordinates, it is just the product of three one-dimensional delta functions: \begin{equation} \delta^3(\rr-\rr_0)=\delta(x-x_0)\,\delta(y-y_0)\,… text/html 2012-10-31T17:46:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/deltadensity?rev=1351730760 The total charge/mass in space should be the same whether we consider it to be distributed as a volume density or idealize it as a surface or line density. See (ss)~{Densities}. The Dirac delta function relates line and surface charge densities (which are really idealizations) to volume densities. For example, if the surface charge density on a rectangular surface is $\sigma(x,y)$, with dimensions $C/L^2$, then the total charge on the slab is obtained by chopping up the surface into infinites… text/html 2019-02-07T16:43:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/deltaexp?rev=1549586580 As discussed in (ss)~{Representations of the Dirac Delta Function}, the Dirac delta function can be written in the form \begin{equation} \delta(x) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{ikx}\, dk .\\ \end{equation} We outline here the derivation of this representation. text/html 2015-08-21T17:45:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/deltaintro?rev=1440204300 The Dirac delta function $\delta(x)$ is not really a ``function''. It is a mathematical entity called a distribution which is well defined only when it appears under an integral sign. It has the following defining properties: \begin{equation} \delta(x) = \cases{0, \qquad &if $x\not= 0$\cr \infty, \qquad &if $x=0$\cr} \end{equation} \begin{equation} \int_b^c \delta(x)\, dx = 1 \qquad\qquad b<0<c \end{equation} \begin{equation} x\,\delta(x) \equiv 0 \end{equation} text/html 2012-10-28T11:48:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/deltaproperties?rev=1351450080 There are many properties of the delta function which follow from the defining properties in (ss)~{The Dirac Delta Function}. Some of these are: \begin{eqnarray} \delta(x) &=& \delta(-x) \\ \frac{d}{dx}\,\delta(x) &=& -\frac{d}{dx}\,\delta(-x) \\ \int_b^c f(x)\, \delta'(x-a)\, dx &=& -f'(a) \\ \delta(ax) &=& {1\over \vert a \vert}\,\delta(x) \\ \delta\bigl(g(x)\bigr) &=& \sum_i {1 \over \vert g'(x_i) \vert} \,\delta(x-x_i) \\ \delta(x^2-a^2) &=& \vert 2a \vert^{-1} \left[ \delta(x-a) + \delt… text/html 2017-04-04T09:51:55-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/deltareps?rev=1491324715 Some other useful representations of the delta function are: \begin{eqnarray} \delta(x) &=& {1 \over 2\pi}\int_{-\infty}^{\infty} e^{ixt}\, dt\\ \noalign{\medskip} \delta(x) &=& \lim_{\epsilon\rightarrow 0}\, {1 \over 2\epsilon} \left[ \Theta(x+\epsilon) - \Theta(x-\epsilon)\right]\\ \noalign{\medskip} \delta(x) &=& \lim_{\epsilon\rightarrow 0}\, {1\over \sqrt{2\pi}\epsilon}\exp\left(-{x^2 \over 2\epsilon^2}\right)\\ \noalign{\medskip} \delta(x) &=& {1 \over \pi} \,\lim_{\epsilon\rightarro… text/html 2012-09-30T20:28:22-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/densities?rev=1349062102 If a charge is distributed evenly throughout a region of space, then we can calculate the volume charge density $\rho$ by dividing the total charge $q$ by the volume $V$, obtaining $\rho=q/V$, with dimensions $Q/L^3$. If the charge is distributed unevenly, then we need to find out how the charge density depends on position, either through careful measurement or through theoretical arguments. In this case, we often write $\rho=\rho(\rr)$ as a reminder that $\rho$ is not constant. We can imagine… text/html 2012-09-20T10:58:56-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/densitystep?rev=1348163936 Typically, charge and mass densities do not extend throughout all of space, rather they are limited to the inside of some physical object. In some instances, it is useful to have an algebraic way of describing such a charge density that shows explicity where it turns on and turns off. You should be able to use step functions to write the charge density inside an insulating spherical shell (with finite thickness). text/html 2015-08-29T14:04:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/diffreview?rev=1440882240 Theory Differentiation is about how small changes in one quantity influence other quantities. This ``ratio of small changes viewpoint is often helpful in setting up problems involving differentiation, and will be especially useful later for partial derivatives. text/html 2015-08-27T18:03:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/directionderiv?rev=1440723780 Differentials such as $df$ are rarely themselves the answer to any physical question. So what good is the Master Formula? The short answer is that you can use it to answer any question about how $f$ changes. Here are some examples. \begin{enumerate}\item Suppose you are an ant walking in a puddle on a flat table. The depth of the puddle is given by $h(x,y)$. You are given $x$ and $y$ as functions of time $t$. How fast is the depth of water through which you are walking changing per unit t… text/html 2015-08-22T08:38:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/divcoord?rev=1440257880 Figure 1: Computing the radial contribution of the flux through a small box in spherical coordinates. The divergence is defined in terms of flux per unit volume. In (ss)~{Divergence}, we used this geometric definition to derive an expression for $\grad\cdot\EE$ in rectangular coordinates, namely \begin{eqnarray*} \grad\cdot\EE = \frac{\rm flux}{\rm unit~volume} = \Partial{E_x}{x} + \Partial{E_y}{y} + \Partial{E_z}{z} \end{eqnarray*} text/html 2015-08-22T08:49:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/divergence?rev=1440258540 Figure 1: Computing the vertical contribution of the flux through a small rectangular box. Consider a small closed box, with sides parallel to the coordinate planes, as shown in Figure 1. What is the flux of $\EE$ out of the box? Consider first the vertical contribution, namely the flux up through the top plus the flux down through the bottom. These two sides each have area element $dA=dx\,dy$, but the outward normal vectors point in opposite directions, that is \begin{eqnarray*} \nn_{\hbo… text/html 2015-08-19T10:26:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/divgradcurl?rev=1440005160 Rectangular Coordinates \begin{eqnarray*} d\rr &=& dx\,\xhat + dy\,\yhat + dz\,\zhat \\ \FF &=& F_x\,\xhat + F_y\,\yhat + F_z\,\zhat \end{eqnarray*} \begin{eqnarray*} \grad f &=& \Partial{f}{x}\,\xhat + \Partial{f}{y}\,\yhat + \Partial{f}{z}\,\zhat \\ \grad\cdot\FF &=& \Partial{F_x}{x} + \Partial{F_y}{y} + \Partial{F_z}{z} \\ \grad\times\FF &=& \left(\Partial{F_z}{y}-\Partial{F_y}{z}\right)\xhat + \left(\Partial{F_x}{z}-\Partial{F_z}{x}\right)\yhat + \left(\Partial{F_y}{x}-\Partial{F_x}{y}\r… text/html 2015-08-22T08:38:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/divthm?rev=1440257880 The total flux of the electric field out through a small rectangular box is \begin{eqnarray*} {\rm flux} = \sum_{\rm box} \EE \cdot d\AA = \grad\cdot\EE \> dV \end{eqnarray*} Figure 1a:The geometry of the Divergence Theorem: Chopping a region into small boxes. text/html 2015-08-21T17:06:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/dot?rev=1440201960 Before you read this section, you should take a few minutes to jot down everything you know about the dot product. Many students feel that if they know one representation for a concept, that is sufficient. As you will discover throughout this book, the professional scientist's ability to solve problems often comes from the ability to play off several different representations against each other. text/html 2015-08-26T17:43:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/double?rev=1440636180 The same principles apply when integrating in higher dimensions: chop and add. For example, to find the amount of chocolate on a rectangular wafer given its (surface) mass density $\sigma$, first chop the plate into small rectangular pieces, as indicated symbolically in Figure~1. How big are the pieces? Surely \begin{equation} dA = dx \,dy = dy\,dx \end{equation} text/html 2015-08-17T16:03:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/drcoordhint?rev=1439852580 As outlined in the worksheet, expressions can be derived for $d\rr$ in other coordinate systems. When completing the worksheet, make sure you think about what the correct lengths are in each coordinate direction. Angles do not measure lengths. Also make sure you know where the center of a circle of constant latitude is. Upon completing the worksheet, you should have obtained the expressions \begin{eqnarray} d\rr &=& dx\,\xhat + dy\,\yhat + dz\,\zhat \\ &=& dr\,\rhat + r\,d\phi\,\phat + … text/html 2015-08-19T14:52:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/drpath?rev=1440021120 Suppose you want to find the work done by the force $\FF=y^2\,\xhat+y\,\yhat$ when moving along a given curve $C$. Curves can be specified in several different ways; let us consider some examples, all of which refer to the same curve, starting at $(1,0)$ and ending at $(0,1)$. text/html 2015-08-22T12:47:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/drpolar?rev=1440272820 Figure 1a:The infinitesimal vector version of the Pythagorean Theorem in rectangular coordinates. It is important to realize that $d\rr$ and $ds$ are defined geometrically, not by the component expressions in equations (2) and (3) of (ss)~{The Vector Differential}. Because of this coordinate-independent nature of $d\rr$, it is possible and useful to study $d\rr$ in another coordinate system, such as polar coordinates ($r$,$\phi$) in the plane. It is then natural to use basis vectors $\{\rh… text/html 2015-08-22T12:44:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/drvec?rev=1440272640 Figure 7: The infinitesimal displacement vector $d\rr$ along a curve, shown in an ``infinite magnifying glass''. In this and subsequent figures, artistic license has been taken in the overall scale and the location of the origin in order to make a pedagogical point. text/html 2015-08-22T12:53:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/dscoord?rev=1440273180 Cylindrical Coordinates You will now derive expressions for infintesimal distances in coordinate directions in cylindrical coordinates. Geometrically determine the length $ds$ of each of the three paths shown below. Notice that, along any of these three paths, only one coordinate $r$, $\phi$, or $z$ is changing at a time (i.e.~along path 1, $dz\ne0$, but $d\phi=0$ and $dr=0$). text/html 2012-09-30T18:46:18-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/dscoordhint?rev=1349055978 An infinitesimal element of length in the $z$-direction is simply $dz$, and similarly an infinitesimal element of length in the $r$ direction is simply $dr$. But, an infinitesimal element of length in the $\phi$ direction in cylindrical coordinates is not just $d\phi$, since this would be an angle and does not even have the units of length. text/html 2010-06-20T10:38:38-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/exdelta1?rev=1277055518 Prove that the derivative of the step function is the delta function, i.e.~show that \begin{equation} \frac{d}{dx}\,\Theta(x-a) \end{equation} satisfies the property of the delta function in equation~(\ref{fdelta}) in Section~\ref{deltaintro}. We need to show that \begin{equation} \Int_{b}^{c} f(x)\,\delta(x-a) \,dx = f(a) \end{equation} for $b<a<c$, where \begin{equation} \delta(x-a)=\frac{d}{dx}\,\Theta(x-a) \end{equation} The main strategy is to use integration by parts, paying strict at… text/html 2012-10-31T17:47:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/exdelta2?rev=1351730820 Prove some or all of the properties of the Dirac delta function listed in (ss)~{Properties of the Dirac Delta Function}. text/html 2015-08-28T14:13:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/fields?rev=1440796380 It is time to distinguish between several different vector-like objects. The arrow pointing from the origin to the point with (Cartesian) coordinates $(x,y,z)$ is $$\rr = x\,\xhat + y\,\yhat + z\,\zhat$$ This is a vector, which is said to have its tail at the origin, or to live at the origin. There is nothing special about the origin; vectors can live at any point. text/html 2015-08-28T14:13:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/flux?rev=1440796380 At any give point along a curve, there is a natural vector, namely the (unit) tangent vector~$\TT$. Therefore, it is natural to add up the tangential component of a given vector field along a curve. When the vector field represents force, this integral represents the work done by the force along the curve. But there is no natural tangential direction at a point on a surface, or rather there are too many of them. The natural vector at a point on a surface is the (unit) normal vector $\nn$, so… text/html 2015-08-28T14:13:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/fluxshort?rev=1440796380 Consider a problem typical of those in calculus textbooks, namely finding the flux of the vector field $\FF=z\,\zhat$ up through the part of the plane $x+y+z=1$ lying in the first octant. We begin with the infinitesimal vector displacement in rectangular coordinates in 3 dimensions, namely \begin{equation} d\rr = dx\,\xhat + dy\,\yhat + dz\,\zhat \end{equation} A natural choice of curves in this surface is given by setting $y$ or $x$ constant, so that $dy=0$ or $dx=0$: \begin{eqnarray} d\rr_1 &… text/html 2015-08-28T14:12:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/geocurl?rev=1440796320 Put a paddlewheel into a moving body of water. Depending on the details of the flow, the paddlewheel might spin. Keeping its center fixed, change the orientation of the paddlewheel. There will be a preferred orientation, in which the paddlewheel spins the fastest. text/html 2015-08-28T12:14:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/geograd?rev=1440789240 How do you compute a derivative of a quantity that depends on a single variable? By taking the ratio of small changes in the quantity to small changes in the variable. But what if the quantity depends on several variables, such as the temperature in the room? Use the same strategy --- but the result will depend on which direction you go. text/html 2015-08-28T12:10:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/gradient?rev=1440789000 As discussed in (ss)~{The Multivariable Differential}, the chain rule for a function of several variables, written in terms of differentials, takes the form: \begin{equation} df = \Partial{f}{x}\,dx + \Partial{f}{y}\,dy + \Partial{f}{z}\,dz \end{equation} Each term is a product of two factors, labeled by $x$, $y$, and $z$. This looks like a dot product. Separating out the pieces, we have \begin{equation} df = \left( \Partial{f}{x}\,\xhat + \Partial{f}{y}\,\yhat + \Partial{f}{z}\,\zhat \righ… text/html 2015-08-28T12:17:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/gradientcurvilinear?rev=1440789420 The master formula can be used to derive formulas for the gradient in other coordinate systems. We illustrate the method for polar coordinates. In polar coordinates, we have $$ df = \Partial{f}{r}\,dr + \Partial{f}{\phi}\,d\phi $$ and of course $$ d\rr = dr\,\rhat + r\,d\phi\,\phat $$ which is~(1) of (ss)~{Other Coordinate Systems}. Comparing these expressions with the Master Formula~(4) of (ss)~{Gradient}, we see immediately that we must have \begin{equation} \grad f = \Partial{f}{r}\,\rhat… text/html 2015-08-22T08:29:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/gradprop?rev=1440257340 What does the gradient mean geometrically? Along a particular path, $df$ tells us something about how $f$ is changing. But the Master Formula tells us that $df=\grad f\cdot d\rr$, which means that the dot product of $\grad f$ with a vector tells us something about how $f$ changes along that vector. So let $\Hat w$ be a unit vector, and consider \begin{equation} \grad{f} \cdot \Hat w = |\grad{f}| \> |\Hat w| \cos\theta = |\grad{f}| \cos\theta \end{equation} which is clearly maximized by $\… text/html 2015-08-28T14:12:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/graphs?rev=1440796320 We can apply these techniques to surfaces which are the graphs of functions. Suppose $z=f(x,y)$. We slice the surface using curves along which either $y$ is constant or $x$ is constant. Since \begin{equation} dz = df = \Partial{f}{x}\,dx + \Partial{f}{y}\,dy \end{equation} we obtain \begin{eqnarray} d\rr_1 &=& dx\,\xhat + dz\,\zhat = \left(\xhat + \Partial{f}{x}\,\zhat\right) dx \\ d\rr_2 &=& dy\,\yhat + dz\,\zhat = \left(\yhat + \Partial{f}{y}\,\zhat\right) dy \end{eqnarray} so that \begin{… text/html 2015-08-28T14:12:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/hisym?rev=1440796320 Two of the most fundamental examples in electromagnetism are the magnetic field around a wire and the electric field of a point charge. We consider each in turn. The magnetic field along an infinitely long straight wire along the $z$ axis, carrying uniform current $I$, is given by \begin{equation} \BB = \frac{\mu_0I}{2\pi} \frac{\phat}{r} = \frac{\mu_0I}{2\pi} \frac{x\,\yhat-y\,\xhat}{x^2+y^2} \end{equation} where $\mu_0$ and $I$ are constant. Note that the first expression clearly indicate… text/html 2015-08-28T14:12:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/hisymmath?rev=1440796320 The electric field of a point charge $q$ at the origin is given by \begin{equation} \EE = \frac{q}{4\pi\epsilon_0} \frac{\rhat}{r^2} = \frac{q}{4\pi\epsilon_0} \frac{x\,\xhat+y\,\yhat+z\,\zhat}{(x^2+y^2+z^2)^{3/2}} \end{equation} where $\rhat$ is the unit vector in the radial direction in spherical coordinates. Note that the first expression clearly indicates both the spherical symmetry of $\EE$ and its $\frac{1}{r^2}$ fall-off behavior, while the second expression does neither. text/html 2010-06-20T10:38:38-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/integration?rev=1277055518 Integration is about chopping things up, and adding the pieces. This ``chopping and adding'' viewpoint is often helpful in setting up problems involving integration, especially those involving multiple integrals. For example, consider finding the mass of a straight wire, given the (linear) mass density $\lambda$. Using $x$ to measure distance along the wire, chop the wire into small pieces of length $dx$. What is the mass of each piece? Clearly, $\lambda \,dx$, where in general $\lambda$ … text/html 2015-08-27T10:00:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/intreview?rev=1440694800 Consider all of the different ways that you can find the total mass of $1\over8$ of the unit sphere, lying in the first octant, via integration, assuming that the sphere has constant density, $\rho$. \begin{enumerate}\item Remember the formula (no actual integration): \begin{eqnarray} M &=& \int \, dm \\ &=& \int\rho\, dV \nonumber\\ &=& \rho\int dV \nonumber\\ &=& \rho {1\over 8} \left({4\over 3}\pi R^3\right)\nonumber\\ &=& {\pi\rho R^3\over 6}\nonumber \end{eqnarray} The notation in … text/html 2015-08-22T13:38:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/intro?rev=1440275880 text/html 2010-09-05T12:59:50-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/intsummary?rev=1283716790 In physical situations, you may want to add up something which is not constant from place to place. To do this approximately, it is simple to imagine chopping the relevant part of space up into little pieces, small enough that the something is essentially constant on each little piece and then adding up the contribution from each little piece. Mathematicians spent 200 years understanding what it means to chop the space up into pieces that are ``infinitesimally small'' and showing that, in this… text/html 2015-08-21T16:10:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/lagrange?rev=1440198600 \begin{itemize}\item RECALL: $\displaystyle\grad{g} \perp \{g=\hbox{const}\}$ \end{itemize} Thus, in two dimensions, $\{g(x,y)=\hbox{const}\}$ is a curve whose tangent vector is perpendicular to $\grad{g}$, and in three dimensions $\{g(x,y,z)=\hbox{const}\}$ is a surface containing many curves each of whose tangent vector is perpendicular to $\grad{g}$. text/html 2017-01-15T12:09:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/lagrange2?rev=1484510940 As described in (ss)~Lagrange Multipliers, a standard technique for solving constrained optimization problems is to use Lagrange Multipliers. However, there are also several other techniques for solving such problems. We describe several such methods below. text/html 2011-04-23T16:39:37-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/lagranged?rev=1303601977 Suppose first that you want to extremize a function $f$ of two variables subject to the constraint $g=\hbox{constant}$. In two dimensions, level sets are curves, and we can introduce a coordinate $v$ along the level curves of $g$. Then $v$ and $g$ can be used as coordinates, at least in a small region about any point. Thus, \begin{equation} df = \Partial{f}{g}\,dg + \Partial{f}{v}\,dv \end{equation} The condition that $f$ be extremized at a point $P$ on a given level curve of $g$ is precisely… text/html 2012-10-30T08:50:47-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/laplacian?rev=1351612247 One second derivative, the divergence of the gradient, occurs so often it has its own name and notation. It is called the Laplacian of the function $V$, and is written in any of the forms \begin{eqnarray*} \triangle V = \nabla^2 V = \grad\cdot\grad V \end{eqnarray*} In rectangular coordinates, it is easy to compute \begin{eqnarray*} \triangle V = \grad\cdot\grad V = \PARTIAL{V}{x} + \PARTIAL{V}{y} + \PARTIAL{V}{z} \end{eqnarray*} text/html 2015-08-16T15:27:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/law?rev=1439764020 Figure 2: The Law of Cosines is just the definition of the dot product. The definition of the dot product can be used to prove several familiar formulas. For example, consider Figure 2, in which $\CC=\BB-\AA$. Then \begin{eqnarray} \CC\cdot\CC &=& (\BB -\AA) \cdot (\BB -\AA) \nonumber\\ &=& \AA\cdot\AA + \BB\cdot\BB - 2 \,\AA\cdot\BB \end{eqnarray} or equivalently \begin{equation} \,|\CC|^2 = |\AA|^2 + |\BB|^2 - 2|\AA||\BB|\cos\theta \end{equation} which is just the Law of Cosines. text/html 2015-08-22T08:48:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/level?rev=1440258480 It is important to distinguish between two quite different situations: Some surfaces describe actual regions in space, while others are merely the graphs of functions of two variables. One can talk about the ``volume'' under the surface in either case, but only in the first case is it really a volume, with the correct units. text/html 2015-08-26T21:52:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/limits?rev=1440651120 If the limits are not constant, double integrals can still be evaluated in either order; the answer is still the same, although the limits and actual integrals evaluated may not be, as we now demonstrate. Figure 2:Chopping the region under a parabola vertically. text/html 2012-10-31T17:48:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/lineint?rev=1351730880 If you want to add up something along a curve, you need to compute a line integral. Common examples are determining the length of a curve, the mass of a wire, or how much work is done when moving a mass along a particular path. Consider the problem of trying to find the length of a quarter of a circle. What do you know? In polar coordinates, a circle is given by $r=\hbox{constant}$, so that $dr=0$. Inserting this fact into the expression~(2) of (ss)~{Other Coordinate Systems} for arclength … text/html 2012-09-15T09:44:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/lineintp?rev=1347727440 Suppose you want to evaluate the line integral \begin{equation} W = \Lint \FF\cdot d\rr \end{equation} and you have an explicit parameterization $\rr=\rr(u)$ of the curve. You can differentiate this expression in order to determine \begin{equation} d\rr = {d\rr\over du}\,du \label{drparam} \end{equation} thus turning your line integral into an ordinary integral with respect to $u$. text/html 2015-08-19T14:50:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/lineints?rev=1440021000 What if you want to determine the mass of a wire in the shape of the curve $C$ if you know the density $\lambda$? The same procedure still works; chop and add. In this case, the length of a small piece of the wire is $ds=|d\rr|$, so its mass is $\lambda\,ds$, and the integral becomes \begin{equation} m = \Lint \lambda \, ds \end{equation} which can also be written as \begin{equation} m = \Lint \lambda(\rr) \, |d\rr| \end{equation} which emphasizes both that $\lambda$ is not constant, and that … text/html 2012-09-15T09:50:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/lineintv?rev=1347727800 Consider now the problem of finding the work $W$ done by a force $\FF$ in moving a particle along a curve~$C$. We begin with the relationship \begin{equation} \hbox{work} = \hbox{force} \times \hbox{distance} \end{equation} Suppose you take a small step $d\rr$ along the curve. How much work was done? Since only the component along the curve matters, we need to take the dot product of $\FF$ with $d\rr$. Adding this up along the curve yields \begin{equation} W = \Lint \FF\cdot d\rr \end{equatio… text/html 2015-08-28T14:11:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/losym?rev=1440796260 The reader may have the feeling that two quite different languages are being spoken here. The tilted plane in (ss)~{Flux} was treated in essentially the traditional manner found in calculus textbooks, using rectangular coordinates. While the ``use what you know'' strategy may be somewhat unfamiliar, the basic idea should not be. On the other hand, the examples in (ss)~{Highly Symmetric Surfaces} will be quite unfamiliar to most mathematicians, due to their use of adapted basis vectors such … text/html 2012-12-31T11:44:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/mapleintro?rev=1356983040 A short worksheet to help you get started using Maple arithmetic and graphics can be found at: $\qquad\qquad\qquad\qquad$ [First Maple Worksheet] A more extended worksheet, including basic calculus, linear algebra, and complex numbers can be found at: $\qquad\qquad\qquad\qquad$ [Extended First Maple Worksheet] text/html 2015-08-18T20:36:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/mass?rev=1439955360 Integrals in curvilinear coordinates are just like any other integrals: chop and add. The most important difference is that you chop along curves for which one of the coordinates is constant; these curves are no longer straight lines. It is useful to know how to convert between various coordinate systems, but in practice one often constructs the integral directly in the appropriate coordinate system. Round problems should be done in round coordinates! text/html 2015-08-28T13:45:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/maxmin?rev=1440794700 Before considering functions of several variables, let us first review how to find maxima and minima for functions of one variable. Recall that a local max/min can only occur at a critical point, where the derivative either vanishes or is undefined. The second derivative test can help to determine whether a critical point is a max or a min: If the second derivative is positive or negative, then the graph is concave up or down, respectively, and the critical point is a local min or a local max,… text/html 2015-08-28T12:20:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/maxmin2?rev=1440789600 Why does the second derivative test work? Recall that first derivatives tell you how to find the linear approximation to a function. For functions of two variables, we have \begin{equation} f(x+\Delta x,y+\Delta y) \approx f(x,y) + \Partial{f}{x}\Delta x + \Partial{f}{y}\Delta y \end{equation} which we can also write as \begin{equation} \Delta f \approx \Partial{f}{x}\Delta x + \Partial{f}{y}\Delta y \end{equation} At a local max/min, however, both partial derivatives vanish, and the linear a… text/html 2015-08-28T14:11:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/mmm?rev=1440796260 We describe here a variation of the usual procedure for determining whether a vector field is conservative and, if it is, for finding a potential function. Figure 1a:Symbolic ``tree diagram'' for computing mixed partial derivatives with two variables. text/html 2010-10-23T09:56:38-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/motion?rev=1287852998 The vector version of a parametric curve is given by interpreting $\rr=\rr(u)$ as the position vector of an object moving along the curve. The derivatives of position are velocity $\vv$ and acceleration $\aa$: \begin{eqnarray} \vv &=& {d\rr\over du}\\ \aa &=& {d\vv\over du} = {d^2\rr\over du^2} \end{eqnarray} and speed is the magnitude of velocity: \begin{equation} v = |\vv| = \left| {d\rr\over du} \right| \end{equation} text/html 2015-08-28T14:11:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/msurfaces?rev=1440796260 There are many ways to describe a surface. How many can you think of? Take a moment to make a list before reading further. Consider the following descriptions of a surface: \begin{itemize}\item the upper half of the unit sphere; \item $x^2+y^2+z^2=1$ with $z\ge0$; \item the graph of $z=+\sqrt{1-x^2-y^2}$; \item $x=\sin\theta\cos\phi$, $y=\sin\theta\sin\phi$, $z=\cos\theta$, with $\theta\in[0,\frac{\pi}{2}]$ and $\phi\in[0,2\pi)$; \end{itemize} And there are more, involving spherical coordinat… text/html 2015-08-22T13:35:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/notation?rev=1440275700 Notational conventions are just that: conventions. Even closely related disciplines tend to have slightly different ways of saying the same things. We summarize here several instances where our notation differs from standard mathematical usage as found in typical calculus texts, along with a brief discussion of our reasons for adopting an alternative. text/html 2013-10-20T13:40:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/notdiff?rev=1382301600 Differentials are a wonderful tool for manipulating derivatives. However, it is important to remember that differentials themselves always refer to the total change in a quantity. Ratios of differentials can often be interpreted as ordinary derivatives, but not as partial derivatives. Put differently, correct statements about differentials can be obtained by pulling apart an ordinary derivative, but never by pulling apart a partial derivative. text/html 2015-08-26T17:45:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/order?rev=1440636300 What if the limits of integration are not constant? Figure 1:A triangular region. Let's start with a simple example, and find the area of the triangle between the graph of the line $2x+3y=6$ and the axes, as shown in Figure~1. First of all, the area of a triangle is half of its base times its height, so we know that the area must be $(2)(3)/2=3$ square units. Let's check this via integration. text/html 2015-08-27T22:31:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/orthogonal?rev=1440739860 The basis vectors adapted to a particular coordinate system are perpendicular to each other at every point. In particular, \begin{eqnarray} \rhat\cdot\phat &= \phat\cdot\zhat &= \zhat\cdot\rhat = 0 \qquad\hbox{(cylindrical)} \\ \rhat\cdot\that &= \that\cdot\phat &= \phat\cdot\rhat = 0 \qquad\hbox{(spherical)} \\ \end{eqnarray} Figure 2 of (ss)~{Curvilinear Basis Vectors} shows this orthogonality in the case of polar basis vectors. These basis vectors are also normalized; they are unit vec… text/html 2015-08-28T14:10:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/parametric?rev=1440796200 The traditional approach to curves and surfaces involves parameterization, which we have deliberately saved for last. Recall that a parametric curve can be written \begin{equation} \rr = \rr(u) = x(u)\,\xhat + y(u)\,\yhat + z(u)\,\zhat \end{equation} together with an appropriate domain for the parameter $u$. A parametric surface is similar, except there are now two parameters $u$,$v$ (and an appropriate domain): \begin{equation} \rr = \rr(u,v) = x(u,v)\,\xhat + y(u,v)\,\yhat + z(u,v)\,\zhat \en… text/html 2019-01-07T14:15:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/partial?rev=1546899300 The tangent line to the graph of $y=f(x)$ at the point ($x_0$,$y_0$) is given by \begin{equation} y-y_0 = m \left( x-x_0 \right) \end{equation} where the slope $m$ is of course just the derivative $\frac{df}{dx} \big|_{x=x_0}$. It is tempting to rewrite the equation of the tangent line as text/html 2010-06-20T10:38:38-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/parts?rev=1277055518 For ordinary functions of one variable, the rule for integration by parts follows immediately from integrating the product rule \begin{eqnarray*} \frac{d}{dx}(fg) &=& \frac{df}{dx}g+f\frac{dg}{dx}\\ \int_a^b\frac{d}{dx}(fg)\, dx &=& \int_a^b\frac{df}{dx}g\, dx + \int_a^b f\frac{dg}{dx}\, dx\\ \left. fg \right\vert_a^b &=& \int_a^b\frac{df}{dx}g\, dx + \int_a^b f\frac{dg}{dx}\, dx \end{eqnarray*} Rearranging, we obtain $$\int_a^b\frac{df}{dx}g\, dx = \left. fg \right\vert_a^b - \int_a^b f\… text/html 2015-08-27T18:00:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/path?rev=1440723600 RECALL: $\qquad\displaystyle\Int_a^b{df\over dx}\,dx = f \Big|_a^b$ Figure 1:Two different paths between the same two points. This is the Fundamental Theorem of Calculus, which just says that the integral of a derivative is the function you started with. We could also write this simply as $$\int df = f$$ But recall the master formula~(2) of (ss)~{Gradient}, which says $$df = \grad{f} \cdot d\rr$$ Putting this all together, we get the fundamental theorem for line integrals, which says that … text/html 2015-08-28T14:10:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/pcurves?rev=1440796200 A parametric curve consists of the functions $$\{x(u),y(u),z(u)\}$$ together with an appropriate domain for the parameter $u$. The parameter $u$ determines both the direction and the speed that a given curve is traced out; a parametric curve is not merely a set of points. Consider for example a circle. You can go around it in either direction, and you can go around just once or many times. All of these are different parametric curves (can you write them down?)~although the set of points invo… text/html 2015-08-27T17:47:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/planes?rev=1440722820 Two points $A$ and $B$ determine a line. But there are also other ways to describe a line. Rather than specifying two points, we can specify just one ($A$), then give a vector $\vv$ along the line. Figure 1:The geometry behind the vector description of a line. So let $\AA$ be the vector from the origin to the point $A$, and $\vv$ be a vector from $A$ that points along the line. Then any other point $P$ on the line can be reached by going to $A$ along $\AA$, then going along $\vv$. Thus, th… text/html 2015-08-21T16:51:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/polar?rev=1440201060 Polar coordinates are useful for situations with circular symmetry in the plane. Figure 1:The construction of the polar coordinates ($r$,$\phi$) at an arbitrary point. The polar coordinates ($r$,$\phi$) of a point $P$ give the distance $r$ that from the given origin to $P$, together with the angle $\phi$ from the positive $x$-axis to $P$, as shown in Figure~1. You will also see this angle called $\theta$ instead of $\phi$; we use $\phi$ to agree with our conventions for (cylindrical and) s… text/html 2015-08-26T21:57:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/polarint?rev=1440651420 Suppose you want to find the total mass of a round flat plate? How do you chop a round region? Figure 1:Chopping a circular region. Figure 2:An infinitesimal box in polar coordinates. When working with round regions, it usually helps to use round coordinates! In polar coordinates, one chops a region into pie shaped regions using radial lines and circles, as shown in Figure~1. These lines are orthogonal to each other, so that a small enough piece is nearly rectangular, which means that its… text/html 2015-08-21T17:45:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/powerapprox?rev=1440204300 In the Maple worksheet $\qquad\qquad\qquad\qquad$ [Power Series Worksheet] or the Mathematica notebook $\qquad\qquad\qquad\qquad$ [Power Series Notebook] you have the opportunity to explore what it means to approximate a given function, in this case $f(\theta)=\sin\theta$, by looking at the first few terms of its power series. Using the coefficients that you found in (ss)~{Finding Power Series Coefficients}, you can plot both the function $\sin\theta$ and its various approximations… text/html 2015-08-21T17:25:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/powerapproxhint?rev=1440203100 See Figure 1 below. Figure 1: The function $\sin\theta$ and three approximations formed from truncating the power series after the first, second, and third non-zero terms (i.e.~at the first, third, and fifth order.) Notice that if you truncate a power series, then you are looking at a polynomial whose dominant term (i.e.~highest power) is the highest term that you kept in the power series. Since $\sin\theta$ is bounded for large $\theta$ and a polynomial in $\theta$ is not bounded, the fir… text/html 2012-09-15T16:54:06-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/powermemorize?rev=1347753246 The following power series for common functions are used so often in approximations in physics, that you should make the extra effort to memorize the first few terms of each one. \begin{eqnarray} \sin(z) &=& \displaystyle\sum_{n=0}^{\infty} (-1)^n\frac{z^{2n+1}}{(2n+1)!} \qquad\qquad\qquad\qquad \hbox{valid $\;\forall z$}\\ &=& z-\frac{z^3}{3!}+\frac{z^5}{5!}-\frac{z^7}{7!}+\dots\\ \end{eqnarray} \begin{eqnarray} \cos(z) &=& \displaystyle\sum_{n=0}^{\infty} (-1)^n\frac{z^{2n}}{(2n)!} … text/html 2015-08-21T17:43:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/powerseriescoeff?rev=1440204180 To check your understanding of (ss)~{Power Series}, you should complete the following activity. A discussion of some of the most important observations you may make is contained in (ss)~{ %* Wrap-Up: Finding Power Series Coefficients}. \begin{enumerate}\item Find the first five nonzero coefficients for $\sin(\theta)$ expanded around the origin. \item Write out a series approximation, correct to fourth order, for $\sin(\theta)$ expanded around the origin. $$\sin(\theta) = \qquad\qquad\qqua… text/html 2012-10-28T10:28:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/powerseriescoeffhint?rev=1351445280 (ss)~{Activity: Finding Power Series Coefficients} should have given you practice using Taylor's theorem to find the coefficients for the power series expansion of a known function, in this case $\sin\theta$. Some important observations are given below. \begin{enumerate}\item Pay attention to the name of the independent variable. The equation for the coefficients is given in terms of the variable $z$. What is the independent variable in $\sin\theta$? \item Typically, there are an infinite num… text/html 2015-08-21T19:18:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/powertheory?rev=1440209880 Why should you care about power series? Because they allow us to approximate functions to any desired accuracy. For instance, consider the function $h(x)=-1-2x+x^2$. Draw its graph. Figure 1: The graph of the polynomial $h(x)=-1-2x+x^2$. (Notice that we have chosen a particularly simple funciton for this example, a polynomial. Its power series, expanded around the origin, is just the algebraic expression for the function itself. Just by looking at it; you can see that it has only three… text/html 2015-08-22T13:38:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/preface?rev=1440275880 text/html 2011-04-23T16:08:31-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/rearrange?rev=1303600111 Using differentials allows algebraic operations to yield information about differentiation. Not only do we know that \begin{equation} df = \Partial{f}{x}\,dx + \Partial{f}{y}\,dy \end{equation} but we can run this argument in reverse. Suppose we know that \begin{equation} du = A\,dv + B\,dw \end{equation} Then we also know that \begin{eqnarray} A &=& \Partial{u}{v} \\ B &=& \Partial{u}{w} \end{eqnarray} Furthermore, we can use algebra to solve for $dv$, obtaining \begin{equation} dv = \frac{1}… text/html 2012-10-10T20:17:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/rules?rev=1349925420 We digress briefly to discuss product rules for vector derivatives, which are discussed in (ss)1.2.6 of Griffiths, and summarized on the inside front cover. All types of derivatives have product rules, all of which take the form \begin{itemize}\item The derivative of a product is the derivative of the first quantity times the second plus the first quantity times the derivative of the second. \end{itemize} For example, the familiar product rule for functions of one variable is $$\frac{d}{dx}(… text/html 2015-08-27T18:03:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/second?rev=1440723780 A conservative vector field is the gradient of some function, for example \begin{eqnarray*} \EE = - \grad V \end{eqnarray*} Figure 1:Two different paths between the same two points. Figure 2:A closed path. But integrals of conservative vector fields are independent of path, so that evaluating the integral along two different paths between the same two points yields the same answer, as illustrated in Figure~1. Combining two such paths into a closed loop changes the orientation of one path,… text/html 2015-08-21T17:30:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/series?rev=1440203400 Most functions can be represented as a power series, whose general form is given by: \begin{eqnarray} f(z) &=& \sum_{n=0}^{\infty} c_n(z-a)^n \nonumber\\ &=& c_0 + c_1(z-a) + c_2(z-a)^2 + c_3(z-a)^3 + \dots \label{Series} \end{eqnarray} where $z$ is the independent variable of the function, $a$ represents the point ``around'' which the function is being expanded, each of the constants $c_n$ is called the coefficient of the $n$th term, and the entire $n$th term, i.e. $c_n(z-a)^n$, is called the … text/html 2012-09-14T19:54:54-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/seriesdimension?rev=1347677694 When we consider a power series expansion of a special function such as \begin{equation} \sin z = z -\frac{1}{3!} z^3 + \frac{1}{5!} z^5 + \dots \end{equation} we can notice an interesting fact. If the variable $z$ were to have any kind dimensions (e.g.~length, $L$) then the power series expansion of that special function would add together terms with different dimensions ($L$, $L^3$, $L^5$, etc.). Since this is impossible, it implies that the argument of such special functions must always be … text/html 2012-10-31T17:43:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/seriesthms?rev=1351730580 The power series of a function, if it exits, is unique, i.e.~there is at most one power series of the form $\sum_{n=0}^{\infty} c_n (z-a)^n$ which converges to a given function within a circle of convergence centered at $a$. We call this a power series ``expanded around $a$''. text/html 2015-08-26T17:24:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/single?rev=1440635040 Theory Integration is about chopping things up, and adding the pieces. This ``chopping and adding'' viewpoint is often helpful in setting up problems involving integration, and will be especially useful later for multiple integrals. Figure~1:Chopping a line into small pieces. For example, consider finding the total amount of chocolate on a straight piece of wafer (like a stick of Pocky), given the density of chocolate on the wafer. What does ``density'' mean? In this case, the amount of… text/html 2010-06-20T10:38:38-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/spherecyl?rev=1277055518 Surprisingly, it often turns out to be simpler to solve problems involving spheres by working in cylindrical coordinates. We indicate here one of the reasons for this. The equation of a sphere of radius $a$ in cylindrical coordinates is \begin{equation} r^2 + z^2 = a^2 \end{equation} so that \begin{equation} 2r\,dr + 2z\,dz = 0 \label{Spherecyl} \end{equation} Proceeding as for the paraboloid, we take \begin{eqnarray} d\rr_1 &=& r\,d\phi\,\phat \\ d\rr_2 &=& dr\,\rhat + dz\,\zhat ~=~ \left… text/html 2015-08-27T09:46:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/sphereint?rev=1440693960 What is the surface area of a sphere of radius $a$? You surely know the answer: $4\pi a^2$. But do you know why? How do you chop up a sphere? In spherical coordinates, of course. As we have seen in (ss)~{Scalar Surface Elements}, an infinitesimal rectangle on the surface of the sphere has sides $r\,d\theta$ and $r\,\sin\theta\,d\phi$, so the scalar surface element on a sphere is \begin{equation} dA = r^2\sin\theta\,d\theta\,d\phi \end{equation} and we know that $r=a$. The surface area is … text/html 2015-08-22T08:24:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/step?rev=1440257040 The step function $\Theta(x)$, also called the Heaviside function or theta function, is defined to be $0$ if $x<0$ and $1$ if $x>0$. See Figure~1. Figure 1: The step function $\Theta(x)$. Step functions are used to model idealized physical situations where some quantity changes rapidly from one value to another in such a way that the exact details of the change are irrelevant for the solution of the problem, e.g.~edges of materials or a process that switches on abruptly at a particular tim… text/html 2015-08-21T16:38:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/stokes?rev=1440200280 The total circulation of the magnetic field around a small loop is given by \begin{eqnarray*} {\rm circulation} = \sum_{\rm box} \BB \cdot d\rr = (\grad\times\BB) \cdot d\AA \end{eqnarray*} where $d\AA=\nn\,dA$, where $\nn$ is perpendicular to the (filled in) loop. text/html 2015-08-28T14:10:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/surfaceints?rev=1440796200 Consider again the example in (ss)~{Flux}, which involved the part of the plane $x+y+z=1$ which lies in the first quadrant. Suppose you want to find the average height of this triangular region above the $xy$-plane. To do this, chop the surface into small pieces, each at height $z=1-x-y$. In order to compute the average height, we need to find \begin{equation} \hbox{avg height} = \frac{1}{\hbox{area}} \Sint z \,\dS \end{equation} where the total area of the surface can be found either as \be… text/html 2015-08-28T14:09:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/surfaces?rev=1440796140 Figure 1:The graph of a function of 2 variables. There are many ways to describe a surface. Consider the following descriptions: \begin{itemize}\item the unit sphere; \item $x^2+y^2+z^2=1$; \item $r=1$ (where $r$ is the spherical radial coordinate); \item $x=\sin\theta\cos\phi$, $y=\sin\theta\sin\phi$, $z=\cos\theta$; \item $\rr(\theta,\phi) = \sin\theta\cos\phi\,\xhat + \sin\theta\sin\phi\,\yhat + \cos\theta\,\zhat$; \end{itemize} all of which describe the same surface. Here are some more … text/html 2011-04-23T15:52:54-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/thermo?rev=1303599174 COMING SOON text/html 2015-08-29T14:28:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/thick?rev=1440883680 In (ss)~{Review of Single Variable Differentation}, we briefly discussed representing derivatives symbolically (as ratios), graphically (as slopes), and even verbally (as the ratio of small quantities). What about numerically or experimentally? text/html 2015-08-21T19:12:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/tint?rev=1440209520 Integration in three dimensions is still about chopping and adding. For example, to find the total amount of chocolate in a region in space given its mass density $\rho$, first chop the region into small ``boxes'', as indicated symbolically in Figure~1. How big are the pieces? Surely text/html 2015-08-22T08:23:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/triple?rev=1440256980 Just as a parallelogram is the region in the plane spanned by 2 vectors, a parallelepiped is the region in space spanned by 3 vectors. Each side of a parallelepiped is therefore a parallelogram. Figure 1:The triple product gives the volume of a parallelepiped. What is the volume of a parallelepiped? The area of its base times its height. But the base of a parallelepiped is a parallelogram, and the area of a parallelogram is just the magnitude of the cross product of its vector sides. The … text/html 2015-08-28T14:08:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/vectorint?rev=1440796080 Vector fields can also be integrated. This is easiest in rectangular coordinates, since the rectangular basis vectors are constants, and hence pull through the integral. For instance, if \begin{equation} \FF = F_x \,\xhat + F_y \,\yhat \end{equation} then \begin{equation} \int \FF \,dx = \xhat \int F_x \,dx + \yhat \int F_y \,dx \end{equation} text/html 2010-07-28T16:17:59-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/vectorint2?rev=1280359079 In curvilinear coordinates, however, care must be taken to remember that the basis vectors are not constant, and must also be integrated. This is not always easy! One example can be obtained by comparing the expressions for $\rr$ and $d\rr$ in polar coordinates, namely \begin{eqnarray} \rr &=& r \rhat \\ d\rr &=& dr \,\rhat + r\,d\phi \,\phat \end{eqnarray} from which it follows, using the product rule, that \begin{equation} d\rhat = d\phi \,\phat \end{equation} which in turn implies that \beg… text/html 2015-08-27T17:51:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/vectors?rev=1440723060 A vector $\ww$ is an arrow in space, having both a magnitude and a direction. Examples of vectors include the displacement from one point to another and your velocity when moving along some path. An explicit example is shown in Figure 1. Figure 1:The vector $\uu$. The magnitude of $\ww$ is denoted by $|\ww|$, also written $||\ww||$ or sometimes just as $w$, without an arrow on top. The magnitude of $\ww$ is often casually called the ``length'' of $\ww$, but that usage is only correct if … text/html 2015-08-16T09:12:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/vfields?rev=1439741520 Adapted basis vectors form a simple example of the geometric notion of a vector field, which is nothing more than a vector at each point. Each such vector should be thought of as “living” (having its tail at) the point at which it is defined, rather than at the origin. text/html 2015-08-22T08:22:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/visconserv?rev=1440256920 Which of the above vector fields is conservative? It is usually easy to determine that a given vector field is not conservative: Simply find a closed path around which the circulation of the vector field doesn't vanish. But how does one show that a given vector field is conservative? Consider the figures below. It is easy to see that the figure on the right is not conservative. But what about the one on the left? text/html 2015-08-22T08:19:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/visdivcurl?rev=1440256740 The Divergence Theorem says \begin{equation} \Int_{\rm box} \FF \cdot d\SS = \Int_{\rm inside} \grad\cdot\FF \> dV \end{equation} which tells us that \begin{equation} \grad\cdot\FF \approx \frac{\Int \FF \cdot d\SS}{\hbox{volume of box}} = \frac{\rm flux}{\rm unit~volume} \end{equation} so that the divergence measures how much a vector field ``points out'' of a box. Similarly, Stokes' Theorem says \begin{equation} \oint\limits_{\rm loop} \FF \cdot d\rr = \Int_{\rm inside} (\grad\times\FF) \… text/html 2012-09-15T11:34:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/volumeint?rev=1347734040 The basic building block for volume integrals is the infinitesimal volume, obtained by chopping up the volume into small “parallelepipeds”. Our approach for surface integrals can be extended to volume integrals using the triple product. The volume element becomes \begin{equation} \dV = (d\rr_1\times d\rr_2)\cdot d\rr_3 \end{equation} for the $d\rr$'s computed for (any!)~3 non-coplanar families of curves. (The volume element is often written as $dV$, but we prefer $\dV$ to avoid confusio… text/html 2015-08-27T15:05:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/zap?rev=1440713100 Derivatives are instantaneous rates of change, which are in turn the ratios of small changes. There are two traditional notations for derivatives, which you have likely already seen. Newton: In this notation, due to Newton, the primary objects are functions, such as $f(x)=x^2$, and derivatives are written with a prime, as in $f'(x)=2x$.