Calculus 2, Fall 2016

From Oakley's *A Mind for Numbers* (p. 55, p. 71)

One of the first steps toward gaining expertise in math and science is to create conceptual chunks–mental leaps that unite separate bits of information through meaning…Once you chunk an idea or concept, you don't need to remember all the little underlying details; you've got the main idea–the chunk–and that's enough.

Practicing math and science problems and concepts in a variety of situations helps you build chunks–solid neural patterns with deep, contextual richness.

**In the same amount of time,***by simply practicing and recalling the material*, students learned far more and at a much deeper level than they did using any other approach.

From Oakley's *A Mind for Numbers* (p. 57+)

The first step in chunking, then, is to simply

*focus your attention*on the information you want to chunk.The second step in chunking is to

*understand*the basic idea you're trying to chunk.The third step to chunking is gaining context so you see not just how, but also

*when*to use this chunk.

There is a **bottom-up chunking process** where practice and repetition can help you both build and strengthen each chunk, so you can easily gain access to it when needed. And there is a **top-down "big picture" process** that allows you to see where what you are learning fits in. *Both processes are vital in gaining mastery over the material.* Context is where bottom-up and top-down learning meet.

Is it possible to use triangles and trigonometry to solve interesting calculus integrals?

Given an integral \(\displaystyle \int f(x) \, dx\), is it possible to simplify it by substituting a trig function in for \(x\), that is, \(x = trig(\theta)\)?

When will substituting a trig function in place of \(x\) be helpful?

Three types of trig substitutions:

\(x = a \sin(\theta)\)

\(x = a \tan(\theta)\)

\(x = a \sec(\theta)\)

This right triangle with angle \(\theta\) in the lower left corner:

When \(\sqrt{a^2-x^2}\), \(\sqrt{a^2+x^2}\), or \(\sqrt{x^2-a^2}\), or a power of one of them appears in the integrand, use the appropriate trig substitution.

Sine | Tangent | Secant |
---|---|---|

\(x = a \sin(\theta)\) | \(x = a \tan(\theta)\) | \(x = a \sec(\theta)\) |

\(dx = a \cos(\theta) \, d\theta\) | \(dx = a \sec^2(\theta) \, d\theta\) | \(dx = a \sec(\theta) \tan(\theta) \, d\theta\) |

\(\sin(2\theta) = 2 \sin(\theta) \cos(\theta)\)

\(\displaystyle \cos^2(\theta) = \frac{1+\cos(2\theta)}{2}\)

\(\displaystyle \sin^2(\theta) = \frac{1-\cos(2\theta)}{2}\)

\(\cos^2(\theta) + \sin^2(\theta) = 1\)

\(1 + \tan^2(\theta) = \sec^2(\theta)\)

Evaluate \(\displaystyle \int_{-7}^{7} \sqrt{49 - x^2} \, dx\) in two ways:

by identifying it as the area of a known geometric figure, and

by a trigonometric substitution.

Find the area of an ellipse with major diameter \(6\) along the \(x\)-axis and minor diameter \(2\) along the \(y\)-axis.

Evaluate \(\displaystyle \int \frac{dx}{ \sqrt{16-x^2} }\) for \(|x| < 4\) in two ways:

by factoring and using a known derivative formula, and

by a trigonometric substitution.

\(\displaystyle \int \frac{dx}{ x^2 \sqrt{x^2-9} }\) for \(|x| > 3\)

\(\displaystyle \int \frac{x^2}{ \sqrt{9-x^2} } \, dx\) for \(|x| < 3\)

\(\displaystyle \int \frac{1}{\sqrt{4+x^2}} \, dx\)

Buck the dog is tethered to a sled by a 5 foot rope. The sled starts 5 feet East of Buck, who travels due North. The path of the sled is a curve called a **tractrix** and has the property that the rope is always tangent to the curve. Set up a differential equation for the curve and solve it.

Find the slope of the tangent line \(\frac{dy}{dx}\) using the triangle.

The triangle indicates trig substitution with \(x = 5 \sin(\theta)\).

Since \(y(5) = 0\), \[ y = 5 \ln \left| \frac{ 5 + \sqrt{25-x^2} }{ x } \right| - \sqrt{25 - x^2}.\]