My Favorite One-Liners: Part 88

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

In the first few weeks of my calculus class, after introducing the definition of a derivative,

\displaystyle \frac{dy}{dx} = y' = f'(x) = \lim_{h \to 0} \displaystyle \frac{f(x+h) - f(x)}{h},

I’ll use the following steps to guide my students to find the derivatives of polynomials.

  1. If f(x) = c, a constant, then \displaystyle \frac{d}{dx} (c) = 0.
  2. If f(x) and g(x) are both differentiable, then (f+g)'(x) = f'(x) + g'(x).
  3.  If f(x) is differentiable and c is a constant, then (cf)'(x) = c f'(x).
  4. If f(x) = x^n, where n is a nonnegative integer, then f'(x) = n x^{n-1}.
  5. If f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 is a polynomial, then f'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + a_1.

After doing a few examples to help these concepts sink in, I’ll show the following two examples with about 3-4 minutes left in class.

Example 1. Let A(r) = \pi r^2. Notice I’ve changed the variable from x to r, but that’s OK. Does this remind you of anything? (Students answer: the area of a circle.)

What’s the derivative? Remember, \pi is just a constant. So A'(r) = \pi \cdot 2r = 2\pi r.

Does this remind you of anything? (Students answer: Whoa… the circumference of a circle.)

Generally, students start waking up even though it’s near the end of class. I continue:

Example 2. Now let’s try V(r) = \displaystyle \frac{4}{3} \pi r^3. Does this remind you of anything? (Students answer: the volume of a sphere.)

What’s the derivative? Again, \displaystyle \frac{4}{3} \pi is just a constant. So V'(r) = \displaystyle \frac{4}{3} \pi \cdot 3r^2 = 4\pi r^2.

Does this remind you of anything? (Students answer: Whoa… the surface area of a sphere.)

By now, I’ve really got my students’ attention with this unexpected connection between these formulas from high school geometry. If I’ve timed things right, I’ll say the following with about 30-60 seconds left in class:

Hmmm. That’s interesting. The derivative of the area of a circle is the circumference of the circle, and the derivative of the area of a sphere is the surface area of the sphere. I wonder why this works. Any ideas? (Students: stunned silence.)

This is what’s known as a cliff-hanger, and I’ll give you the answer at the start of class tomorrow. (Students groan, as they really want to know the answer immediately.) Class is dismissed.

If you’d like to see the answer, see my previous post on this topic.

My Favorite One-Liners: Part 87

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

When teaching Calculus I, I use the following mantra throughout the semester. I heard this from my calculus instructor back in 1984, and I repeat it for my own students:

There are two themes of calculus: approximating curved things by straight things, and passing to limits.

For example, to find a derivative, we approximate a curved function by a straight tangent line and then pass to a limit. Later in the semester, to find a definite integral, we approximate the area under a curve by the sum of a bunch of straight rectangles and then pass to a limit.

For further reading, I’ll refer to this series of posts on what I typically do on the first day of my calculus class.

 

 

 

My Favorite One-Liners: Part 74

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

After presenting the Fundamental Theorem of Calculus to my calculus students, I make a point of doing the following example in class:

\displaystyle \int_0^4 \frac{1}{4} x^2 \, dx

Hopefully my students are able to produce the correct answer:

\displaystyle \int_0^4 \frac{1}{4} x^2 \, dx = \displaystyle \left[ \frac{x^3}{12} \right]^4_0

= \displaystyle \frac{(4)^3}{12} - \frac{(0)^3}{12}

= \displaystyle \frac{64}{12}

= \displaystyle \frac{16}{3}

Then I tell my students that they’ve probably known the solution of this one since they were kids… and I show them the classic video “Unpack Your Adjectives” from Schoolhouse Rock. They’ll watch this video with no small amount of confusion (“How is this possibly connected to calculus?”)… until I reach the 1:15 mark of the video below, when I’ll pause and discuss this children’s cartoon. This never fails to get an enthusiastic response from my students.

If you have no idea what I’m talking about, be sure to watch the first 75 seconds of the video below. I think you’ll be amused.

My Favorite One-Liners: Part 60

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

I’m a big believer using scaffolded lesson plans, starting from elementary ideas and gradually building up to complicated ideas. For example, when teaching calculus, I’ll use the following sequence of problems to introduce students to finding the volume of a solid of revolution using disks, washers, and shells:

  • Find the volume of a cone with height h and base radius r.
  • Find the volume of the solid generated by revolving  the region bounded by y=2, y=2\sin x for 0 \le x \le \pi/2, and the y-axis about the line y=2.
  • Find the volume of the solid generated by revolving the region bounded by y=2, y=\sqrt{x}, and the y-axis about the line y=2 .
  • Find the volume of the solid generated by revolving the region bounded by y=2, y=\sqrt{x}, and the y-axis about the y-axis .
  • Find the volume of the solid generated by revolving the region bounded by x=\sqrt{2y}/(y+1), y=1, and the y-axis about the y-axis .
  • Find the volume of the solid generated by revolving  the region bounded by the parabola x=y^2+1 and the line x=3 about the line x=3.
  • Water is poured into a spherical tank of radius R to a depth h. How much water is in the tank?
  • Find the volume of the solid generated by revolving the region bounded by y= x^2, the x-axis, and x=4 about the x-axis.
  • Find the volume of the solid generated by revolving the region bounded by y= x^2, the x-axis, and x=4 about the y-axis.
  • Repeat the previous problem using cylindrical shells.

In this sequence of problems, I slowly get my students accustomed to the ideas of horizontal and vertical slices, integrating with respect to either x and y, the creation of disks and washers and (eventually) cylindrical shells.

As the problems increase in difficulty, I enjoy using the following punch line:

To quote the great philosopher Emeril Lagasse, “Let’s kick it up a notch.”

My Favorite One-Liners: Part 25

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Consider the integral

\displaystyle \int_0^2 2x(1-x^2)^3 \, dx

The standard technique — other than multiplying it out — is using the substitution u = 1-x^2. With this substitution du = -2x \, dx. Also, x = 0 corresponds to u = 1, while x = 2 corresponds to u = -3. Therefore,

\displaystyle\int_0^2 2x(1-x^2)^3 \, dx = - \displaystyle\int_0^2 (-2x)(1-x^2)^3 \, dx = -\displaystyle\int_1^{-3} u^3 \, du.

My one-liner at this point is telling my students, “At this point, about 10,000 volts of electricity should be going down your spine.” I’ll use this line when a very unexpected result happens — like a “left” endpoint that’s greater than the “right” endpoint. Naturally, for this problem, the next step — though not logically necessary, it’s psychologically reassuring — is to absorb the negative sign by flipping the endpoints:

\displaystyle\int_0^2 2x(1-x^2)^3 \, dx =  -\displaystyle\int_1^{-3} u^3 \, du = \displaystyle\int_{-3}^1 u^3 \, du,

and then the calculation can continue.

My Favorite One-Liners: Part 24

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Here’s a problem that could appear in my class in probability or statistics:

Let f(x) = 3x^2 be a probability density function for 0 \le x \le 1. Find F(x) = P(X \le x), the cumulative distribution function of X.

A student’s first reaction might be to set up the integral as

\displaystyle \int_0^x 3x^2 \, dx

The problem with this set-up, of course, is that the letter x has already been reserved as the right endpoint for this definite integral. Therefore, inside the integral, we should choose any other letter — just not x — as the dummy variable.

Which sets up my one-liner: “In the words of the great philosopher Jean-Luc Picard: Plenty of letters left in the alphabet.”

We then write the integral as something like

\displaystyle \int_0^x 3t^2 \, dt

and then get on with the business of finding F(x).

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 5

Check out this lovely integral, dubbed the Sophomore’s Dream, found by Johann Bernoulli in 1697 (Gamma, page 44):

\displaystyle \int_0^1 \frac{dx}{x^x} = \displaystyle \frac{1}{1^1} + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{4^4} + \dots.

I’ll refer to either Wikipedia or Mathworld for the derivation.

green line

When I researching for my series of posts on conditional convergence, especially examples related to the constant \gamma, the reference Gamma: Exploring Euler’s Constant by Julian Havil kept popping up. Finally, I decided to splurge for the book, expecting a decent popular account of this number. After all, I’m a professional mathematician, and I took a graduate level class in analytic number theory. In short, I don’t expect to learn a whole lot when reading a popular science book other than perhaps some new pedagogical insights.

Boy, was I wrong. As I turned every page, it seemed I hit a new factoid that I had not known before.

In this series, I’d like to compile some of my favorites — while giving the book a very high recommendation.

The antiderivative of 1/(x^4+1): Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on the computation of

\displaystyle \int \frac{dx}{x^4+1}

Part 1: Introduction.

Part 2: Factoring the denominator using De Moivre’s Theorem.

Part 3: Factoring the denominator using the difference of two squares.

Part 4: The partial fractions decomposition of the integrand.

Part 5: Partial evaluation of the resulting integrals.

Part 6: Evaluation of the remaining integrals.

Part 7: An apparent simplification using a trigonometric identity.

Part 8: Discussion of the angles for which the identity holds.

Part 9: Proof of the angles for which the identity holds.

Part 10: Implications for using this identity when computing definite integrals.

 

 

How I Impressed My Wife: Index

Some husbands try to impress their wives by lifting extremely heavy objects or other extraordinary feats of physical prowess.

That will never happen in the Quintanilla household in a million years.

But she was impressed that I broke an impasse in her research and resolved a discrepancy between Mathematica 4 and Mathematica 8 by finding the following integral by hand in less than an hour:

\displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

Yes, I married well indeed.

In this post, I collect the posts that I wrote last summer regarding various ways of computing this integral.

Part 1: Introduction
Part 2a, 2b, 2c, 2d, 2e, 2f: Changing the endpoints of integration, multiplying top and bottom by \sec^2 x, and the substitution u = \tan x.
Part 3a, 3b, 3c, 3d, 3e, 3f, 3g, 3h, 3i: Double-angle trig identity, combination into a single trig function, changing the endpoints of integration, and the magic substitution u = \tan \theta/2.
Part 4a, 4b, 4c, 4d, 4e, 4f, 4g, 4h: Double-angle trig identity, combination into a single trig function, changing the endpoints of integration, and contour integration using the unit circle
Part 5a, 5b, 5c, 5d, 5e, 5f, 5g, 5h, 5i, 5j: Independence of the parameter a, the magic substitution u = \tan \theta/2, and partial fractions.
Part 6a, 6b, 6c, 6d, 6e, 6f, 6g:Independence of the parameter a, the magic substitution u = \tan \theta/2, and contour integration using the real line and an expanding semicircle.
Part 7: Concluding thoughts… and ways that should work that I haven’t completely figured out yet.

Thoughts on Infinity (Part 3f)

In recent posts, we’ve seen the curious phenomenon that the commutative and associative laws do not apply to a conditionally convergent series or infinite product: while

\displaystyle 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - ... = \ln 2,

a rearranged series can be something completely different:

\displaystyle 1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \frac{1}{9} + \frac{1}{11} - \frac{1}{6} ... = \displaystyle \frac{3}{2} \ln 2.

This very counterintuitive result can be confirmed using commonly used technology — in particular, Microsoft Excel. In the spreadsheet below, I typed:

  • =IF(MOD(ROW(A1),3)=0,ROW(A1)*2/3,IF(MOD(ROW(A1),3)=1,4*(ROW(A1)-1)/3+1,4*(ROW(A1)-2)/3+3)) in cell A1
  • =POWER(-1,A1-1)/A1 in cell B1
  • =B1 in cell C1
  • I copied cell A1 into cell A2
  • =POWER(-1,A2-1)/A2 in cell B2
  • =C1+B2 in cell C2

The unusual command for cell A1 was necessary to get the correct rearrangement of the series.

Then I used the FILL DOWN command to fill in the remaining rows. Using these commands cell C9 shows the sum of all the entries in cells B1 through B9, so that

\displaystyle 1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \frac{1}{9} + \frac{1}{11} - \frac{1}{6} \approx 0.961544012

log2series6

Filling down to additional rows demonstrates that the sum converges to \displaystyle \frac{3}{2}\ln 2 and not to \ln 2. Here’s the sum up to 10,000 terms… the entry in column E is the first few digits in the decimal expansion of \displaystyle \frac{3}{2} \ln 2. log2series7

Clearly the partial sums are not approaching \ln 2 \approx 0.693, and there’s good visual evidence to think that the answer is \displaystyle \frac{3}{2} \ln 2 instead. (Incidentally, the 10,000th partial sum is very close to the limiting value because 10,000 is one more than a multiple of 3.)