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Latest Posts
An Interesting Problem for First Year Calculus
Submitted by Jeremy Lane on Tue, 02/09/2010 - 19:22The following is a derivative that looks simple on the surface to most, but for a first year math student it can actually be quite puzzling. If I were to ever include an extra-credit problem on a calculus I midterm, this would be it:\[\frac{d}{dx} (-1)^x\]Ok, so some of you might be looking at this problem thinking "thats easy." Half of you are doing this because you know Euler's formula, and the other half haven't tried solving the problem. The first thing we notice is the familiar form:\[\frac{d}{dx} a^x\,=\,(a)^{x}ln(a)\]So it would be logical to conclude that we can follow this method in the following way:\[\frac{d}{dx} (-1)^x\,=\,(-1)^{x}ln(-1)\]But wait, now we have ln(-1). This is definitely an unsatisfactory result at the Calculus I level. Enter Euler's Formula:\[e^{ix} \,=\, cos(x)+(i) sin(x) \]Where $i$ is the familiar $\sqrt{-1}$ . Letting $x \,=\,\pi$ we get:\[e^{i\pi} \,=\, -1+0 \]So from here we can solve the problem like this:\[\frac{d}{dx} (-1)^x\] \[\frac{d}{dx} (e^{i\pi})^x\,=\,(e^{i\pi})^{x}ln(e^{i\pi})\]
5 Helpful Videos about Instantaneous Rates of Change and Derivatives
Submitted by Jeremy Lane on Wed, 08/12/2009 - 04:26While reading this keep in mind that "derivative", "Instantaneous Slope", and "Instantaneous Rate of Change" all mean the same thing: The slope of a function at a single point.
Integration by Trigonometric Substitution
Submitted by Jeremy Lane on Mon, 05/11/2009 - 00:48Here is a PDF on integration by trigonometric substitution.
Some Notes on Trigonometry
Submitted by Jeremy Lane on Thu, 05/07/2009 - 00:25Although you may not have heard of radians before, they are simply another way to measure angles (they relate the arc length to the radius of a circle but thats not important here). \[ \pi radians =180^{\circ}\] If you wanted to express $97^{\circ}$ in radians you would do this: \[ 97^{\circ}*\frac{\pi\,rad}{180^{\circ}} = 1.69 rad\] Trig conversions from radians to degrees and vis versa are important, so is being able to express the answer to a trig function of a simple angle such as $tan(\frac{\pi}{3})$. Here is a chart from Math for Morons Like Us that shows the conversions from degrees to radians for the full 360 degrees. 
This is referred to as a "unit circle" because the radius of the circle is 1 (or one unit). It is important because it means for any right angle triangle drawn from the origin to a point on the circle the length of the hypotenuse will always be 1.
Units
Submitted by Jeremy Lane on Tue, 04/28/2009 - 17:49Here is a brief on using metric units for physics classes. This is pretty basic and very important to know. The physics and engineering classes at the University of Regina mainly focus on SI or International System units (metric).
Common Units
- m
- Meters, unit of length.
- s
- Seconds, unit of time.
- kg
- kilogram, unit of mass. (not Weight)
Most other units are a combination of these units. Units that you will not need for Physics 109 are not listed here.
- $\frac{m}{s^2}$
- Linear Acceleration. Meters per second squared.
- N
- Newtons, unit of force. \[\frac{kgm}{s^2}\]
- J
- Joule, Unit of energy. Refers to work (a form of energy). \[\frac{kgm^2}{s^2}\]
- Hz
- Hertz, unit of frequency (how often something happens per second)$\frac{1}{T}$.
- $\frac{m}{s}$
- Linear Velocity. Meters per second.
Prefixes
Wikipedia's longer list of prefixes Prefixes to units.
Linear Algebra (Math 122)
Submitted by Jeremy Lane on Mon, 04/27/2009 - 22:25This will be the linear algebra book
Integration by Parts: Example 1
Submitted by Jeremy Lane on Mon, 04/27/2009 - 14:48Here is an example integration by parts problem. Remember that the first step is always identifying what integration technique to use. For this example we will start with a relatively easy indefinite integral.
\[ \int\; x\; ln(x)\;dx = ? \]
Introduction to Integration by Parts
Submitted by Jeremy Lane on Thu, 04/23/2009 - 14:21Integration by parts can be a confusing concept to learn because the formula may look intimidating and counter-intuitive at first glance:
\[\int \text{f(x)g'(x)}dx = \text{f(x)g(x)} - \int \text{ f'(x)g(x)} dx \]