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MAE 106 (2018) – Assignment 1

7 September 2018 in MAE 106: Discrete Mathematics with Probability Fall 2018 | No comments

Quiz on 12 September 2018.

  1. Memorize the following definitions:
    • Proposition (Definition 1.2.1)
    • Prime Proposition (Definition 1.2.3)
    • Compound Proposition (Definition 1.2.5)
    • Paradox (Definition 1.2.7)
  2. Know the truth tables for the following:
    • Negation \neg
    • Conjunction \wedge
    • Disjunction \vee
    • Exclusive Disjunction \veebar
    • Implication \rightarrow
  3. Complete the following exercises in the textbook (Kohar).
    • Section 1.2—1.2.3, 1.2.4, 1.2.5
    • Section 1.3—1.3.1, 1.3.2, 1.3.5, 1.3.6, 1.3.10, 1.3.11

Calculus Review Videos

22 April 2018 in MAE 113: Calculus for the Liberal Arts | No comments

Here are some quick review videos for MAE 113.

Derivative Rules

Applications of the Derivative

Solutions

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Assignment 8

5 April 2018 in MAE 113: Calculus for the Liberal Arts | No comments

I have posted assignment 8 here. The solutions for Section 11.3 are here. The solutions for Section 11.4 are here.

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Integrals, integrals, integrals

29 March 2018 in MAE 113: Calculus for the Liberal Arts | No comments

List of indefinite integrals
  • \int x^n \, \d x = \dfrac{1}{n + 1} x^{n + 1} + C (n \neq 1)
  • \int e^x \, \d x = e^x + C and the related \int e^{k x} \, \d x = \dfrac{1}{k} e^{k x} + C (k \neq 0)
  • \displaystyle\int \dfrac{1}{x} = \ln |x| + C
  • \int b^x \, \d x = \dfrac{b^x}{\ln b} + C (b \neq 1)

Any of the above forumale can be verified by differentiating the right hand side.  For example \int x^n \, \d x = \dfrac{1}{n + 1} x^{n + 1} + C (n \neq 1) because \dfrac{ \d \ }{\d x} \left( \dfrac{1}{n + 1} x^{n+1} + C \right) = x^n.  This gives us a way to check our answers.

We can apply the rules from the list of indefinite integrals above to more complicated functions.

Two Useful Properties of Integrals
  1. \int c f(x) \, \d x = c \int f(x) \, \d x

The integral of a constant times a function is equal to the constant times the integral of the function.  Basically, this means that we can pull the constant out in front of the integral.  This is similar to our constant multiple rule with derivatives.

2. \int \big( f(x) + g(x) \big) \, \d x = \int f(x) \, \d x + \int g(x) \, \d x

The integral of a sum of two functions is equal to the integral of the first function plus the integral of the second function. Basically, this means that the integral of the sum of two functions can be broken into two separate integrals for each function.

Examples.

  1. \int 3 x^2 \, \d x = 3 \int x^2 \, \d x by property 1.
  2. \int x + 3x^2 \, \d x = \int x \, \d x + \int 3x^2 \, \d x by property 2.
Examples Illustrating the Use of the Two Properties and the List of Indefinite Integrals
  1. Find \int (2 x^4 - 9x^3 - 2x + \pi) \, \d x.

    \begin{align*} \int (2 x^4 - 9x^3 - 2x + \pi) \, \d x &= \int 2x^4 \, \d x - \int 9x^3 \, \d x - \int 2x \, \d x + \int \pi \, \d x & \text{By Property 2}\\ &= 2 \int x^4 \, \d x - 9 \int x^3 \, \d x - \int 2x \, \d x + \int \pi \, \d x & \text{By Property 1}\\ &= \frac25 x^5 - \frac94 x^4 - x^2 + \pi x + C \end{align*}

2. Evaluate \displaystyle\int_1^4 \frac{1}{\sqrt{x}} \, \d x.

    \begin{align*} \int_1^4 \frac{1}{\sqrt{x}} \, \d x &= \int_1^4 x^{-\frac12} \, \d x\\ &= \left.\frac{1}{-\frac12 + 1} x^{-\frac12 + 1}\right|_1^4\\ &= \left.\frac{1}{\frac12} x^{\frac12}\right|_1^4\\ &= \left. 2x^{\frac12}\right|_1^4\\ &= \left. 2\sqrt{x}\right|_1^4\\ &= 2 \sqrt{4} - 2 \sqrt{1}\\ &= 2 \cdot 2 - 2\\ &= 4 - 2\\ &= 2 \end{align*}

3. Find \displaystyle\int_1^4 \frac{t^3 + \sqrt{t} - 4}{t} \, \d t.

    \begin{align*} \int_1^4 \frac{t^3 + \sqrt{t} - 4}{t} \, \d t &= \int_1^4 \left(t^2 + \frac{1}{\sqrt{t}} - \frac{4}{t}\right) \, \d t\\ &= \int_1^4 \left( t^2 + t^{-\frac12} - \frac{4}{t} \right) \, \d t\\ &= \left.\frac13 t^3 + 2 t^{\frac12} - 4 \ln t \right|_1^4\\ &= \frac13 (4)^3 + 2(4)^{\frac12} - 4 \ln 4 - \left[\frac13 (1)^3 + \underbrace{2(1)^{\frac12}}_2 - 4 \underbrace{\ln 1}_0 \right]\\ &= \frac{64}{3} + 4 - 4 \ln 2^2 - \frac13 - 2\\ &= \frac{64}{3} + 2 - 4 \ln 2^2 - \frac13 \\ &= \frac{63}{3} + \frac{6}{3} - 8 \ln 2\\ &= \frac{69}{3} - 8 \ln 2\\ &= 23 - 8 \ln 2 \end{align*}

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What is the derivative of b^x?

29 March 2018 in MAE 113: Calculus for the Liberal Arts | No comments

If f(x) = b^x, then f'(x) = b^x \ln x.

Let f(x) = b^x. Using the definition of the derivative, we have

    \begin{align*} f'(x) &= \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}\\ &= \lim_{\Delta x \to 0} \frac{b^{x + \Delta x} - b^x}{\Delta x}\\ &= \lim_{\Delta x \to 0} \frac{b^x b^{\Delta x} - b^x}{\Delta x}\\ &= \lim_{\Delta x \to 0} \frac{b^x (b^{\Delta x} - 1)}{\Delta x}\\ &= \lim_{\Delta x \to 0} b^x \underbrace{\lim_{\Delta x \to 0} \frac{b^{\Delta x} - 1}{\Delta x}}_{\ln x}\\ &= b^x \ln b. \end{align*}

It would be nice if \displaystyle\lim_{\Delta x \to 0} \frac{b^{\Delta x} - 1}{\Delta x} = 1. Well, this occurs for a special value of b; we let b = e \approx 2.718.

This leads to the following:

If f(x) = e^x, then f'(x) = e^x.

Assignment 6 & 7

28 March 2018 in MAE 113: Calculus for the Liberal Arts | No comments

Assignment 6 has been handed out. You can find a copy of it here.

Also, assignment 7 has been handed out. You can find a copy of it here.

Each of these assignments refer to the handouts that I gave in class. If you do not have the handouts, please come see me.

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Some homework questions on logarithms

12 March 2018 in MAE 113: Calculus for the Liberal Arts | No comments

Do the questions from the large booklet that I handed out to you after the test.

You should have done Exercise 1 (all questions) already as instructed at the end of the test.

Exercise 2 (p. 352-3)

  1. b, d, f, h
  2. b, d, f, h
  3. a, c, e, g, j
  4. a, c, e, g, h

Exercise 3 (p. 354)

  1. a, b, d, g, h, i, j
  2. a, e, f
  3. b, c, d, e, f

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Letters to a Young Scientist

10 March 2018 in Uncategorized | No comments

These are some highlights from a talk that E. O. Wilson gave at TED.

Keep your eyes lifted and your head turning. The search for knowledge is in our genes. It was put there by our distant ancestors who spread across the world, and it’s never going to be quenched. To understand and use it sanely, as a part of the civilization yet to evolve requires a vastly larger population of scientifically trained people like you. In education, medicine, law, diplomacy, government, business and the media that exist today.

Our political leaders need at least a modest degree of scientific literacy, which most badly lack today — no applause, please. It will be better for all if they prepare before entering office rather than learning on the job. Therefore you will do well to act on the side, no matter how far into the laboratory you may go, to serve as teachers during the span of your career.

…

The longer you wait to become at least semi-literate the harder the language of mathematics will be to master, just as again in any verbal language, but it can be done at any age. I speak as an authority on that subject, because I’m an extreme case. I didn’t take algebra until my freshman year at the University of Alabama. They didn’t teach it before then.

I finally got around to calculus as a 32-year-old tenured professor at Harvard, where I sat uncomfortably in classes with undergraduate students, little more than half my age. A couple of them were students in a course I was giving on evolutionary biology. I swallowed my pride, and I learned calculus.

I found out that in science and all its applications, what is crucial is not that technical ability, but it is imagination in all of its applications. The ability to form concepts with images of entities and processes pictured by intuition. I found out that advances in science rarely come upstream from an ability to stand at a blackboard and conjure images from unfolding mathematical propositions and equations. They are instead the products of downstream imagination leading to hard work, during which mathematical reasoning may or may not prove to be relevant. Ideas emerge when a part of the real or imagined world is studied for its own sake.

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Assignment 5

5 March 2018 in MAE 113: Calculus for the Liberal Arts | No comments

Assignment 5 is posted here. Solutions were handed out in class.

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Test 1

11 February 2018 in MAE 113: Calculus for the Liberal Arts | No comments

The first test of MAE 113 will be on 13 February 2018.

No calculators, notes, or textbooks will be allowed.

You are responsible for all material

  • covered in class lectures
  • Assignments 1, 2, 3, and 4
  • will be asked to prove one of the following differentiation rules:

Tips:

  1. Review your two marked quizzes and identify your mistakes. You do not want to make these same mistakes again on the test.
  2. Practice. You should have completed all the assignments to feel confident in abilities.
  3. Drop-In Math Help Centre. It is open from 7 pm to 10 pm in G327. Tutors are available to help you. Take advantage of this opportunity!

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