NYU MATH-UA 122.006 Calculus II Spring 2012

A harmonograph uses two pendulums swinging in perpendicular planes to create a graphic representation of harmonies. This device uses the same relationships found in musical harmonies, but at the much lower frequencies of pendulums. The design for this harmonograph came from "100 Amazing Award Winning Science Fair Projects" by Glen Vecchione, with minor alterations during the build and test phases. Another interesting book on this topic is Anthony Ashton's "Harmonograph: A Visual Guide to the Mathematics of Music".

The curve traced out by a point on the rim of a circle rolling along a straight line is called a cycloid. Let the radius of the circle be . Allowing the tracing point to be either within or without the circle at a distance from the center generates "curtate" or "prolate" cycloids, respectively.

Here is the story of the Archimedes Palimpsest, a lost text of his found underneath a recycled prayerbook. Some of what he wrote about involves centers of mass computed with infinitesimals. I.e., Calc II!

Lots of helpful links and PDFs up on Prof. Tsishchanka's site.

Please read and act accordingly.

Here is another good video by Khan Academy explaining integration by substitution for those of you who may need help. There are also other videos on pretty much any topic in calculus which I find useful for reviewing and practicing. Khan Academy is on a mission to provide a free world-class education to anyone anywhere. With over 2,600 videos covering everything from arithmetic to physics, finance, and history and 200 practice exercises, we're helping students learn whatever they want, whenever they want, at their own pace.

Students have a lot of problems with changing the limits when doing u-substitution with definite integrals. But it's easy to remember: if the variable integrated over is u, the limits have to be values of u. You start with limits which are values of x. How do you convert x to u? We have already decided that u=cos(x). So the limits x=π/2 and x=π become cos(π/2) = 0 and cos(π) = -1. He's being a bit sloppy by saying this computes a negative area. Area is always positive. The integral *results* in a negative number because the region we are finding the area of is below the x-axis. So the integral computes the negative of the area.

Slides from my Calc I Lecture

My lecture on substitution from Calc I

Integration using U-Substitution of Indefinite Integrals.

Ilustrations of several u-substitution problems

This applet is designed for the exploration of Riemann sums. The user can overlay the graph of the function with the area represented by the chosen method of sums, the piecewise linear approximation to the antiderivative, and the graph of a guess at the antiderivative.