blueollie

How To Succeed in Calculus Class and Why it is Worth It

This post is intended to help the student who is willing to put time and effort into succeeding in a college calculus class.

Part One: How to Study

The first thing to remember is that most students will have to study outside of class in order to learn the material. There are those who pick things up right away, but these students tend to be the rare exception.

Think of it this way: suppose you want to learn to play the piano. A teacher can help show you how to play it and provide a practice schedule. But you won’t be any good if you don’t practice.

Suppose you want to run a marathon. A coach can help you with running form, provide workout schedules and provide feedback. But if you don’t run those workouts, you won’t build up the necessary speed and endurance for success.

The same principle applies for college mathematics classes; you really learn the material when you study it and do the homework exercises.

Here are some specific tips on how to study:

1. It is optimal if you can spend a few minutes scanning the text for the upcoming lesson. If you do this, you’ll be alert for the new concepts as they are presented and the concepts might sink in quicker.

2. There is some research that indicates:
a. It is better to have several shorter study sessions rather than one long one and
b. There is an optimal time delay between study sessions and the associated lecture.

Look at it this way: if you wait too long after the lesson to study it, you would have forgotten much of what was presented. If you study right away, then you really have, in essence, a longer class room session. It is probably best to hit the material right when the initial memory starts to fade; this time interval will vary from individual to individual. For more on this and for more on learning for long term recall, see this article.

3. Learn the basic derivative formulas inside and out; that is, know what the derivatives of functions like sin(x), cos(x), tan(x), sec(x), arctan(x), arcsin(x), exp(x), ln(x) are on sight; you shouldn’t have to think about them. The same goes for the basic trig identities such as \sin ^{2}(x)+\cos ^{2}(x)=1 and \tan^{2}(x)+1 = \sec^{2}(x)

Why is this? The reason is that much of calculus (though not all!) boils down to pattern recognition.

For example, suppose you need to calculate:

\int \dfrac{(\arctan (x))^{5}}{1+x^{2}}dx=

If you don’t know your differentiation formulas, this problem is all but impossible. On the other hand, if you do know your differentiation formulas, then you’ll immediately recognize the arctan(x) and it’s derivative \dfrac{1}{1+x^{2}} and you’ll see that this problem is really the very easy problem \int u^{5}du .

But this all starts with having “automatic” knowledge of the derivative formulas.

Note: this learning is something your professor or TA cannot do for you!

4. Be sure to do some study problems with your notes and your book closed. If you keep flipping to your notes and book to do the homework problems, you won’t be ready for the exams. You have to kick up the training wheels.
Try this; the difference will surprise you.

5. When reviewing for an exam, study the problems in mixed fashion. For example, get some note cards and write problems from the various sections on them (say, some from 3.1, some from 3.2, some from 3.3, and so on), mix the cards, then try the problems. If you just review section by section, you’ll go into each problem knowing what technique to use each time right from the start. Many times, half of the battle is knowing which technique to use with each problem; that is part of the course! Do the problems in mixed order.

If you find yourself whining complaining “I don’t know where to start” it means that you don’t know the material well enough. Remember that a trained monkey can repeat specific actions; you have to be a bit better than that!

6. Read the book, S L O W L Y, with pen and paper nearby. Make sure that you work through the examples in the text and that you understand the reasons for each step.

7. For the “more theoretical” topics, know some specific examples for specific theorems. Here is what I am talking about:

a. Intermediate value theorem: recall that if f(x)=\frac{1}{x} , then f(-1)=-1,f(1)=1 but there is no x such that f(x) = 0 . Why does this not violate the intermediate value theorem?

b. Mean value theorem: note also that there is no c such that f'(c) = \frac{f(1)-f(-1)}{2} = 0 . Why does this NOT violate the Mean Value Theorem?

c. Series: it is useful to know basic series such as those for exp(x), sin(x), cos(x) . It is also good to know some basic examples such as the geometric series, the divergent harmonic series \sum \frac{1}{k} and the conditionally convergent series \sum (-1)^{k}\frac{1}{k} .

d. Limit definition of derivative: be able to work a few basic examples of the derivative via the limit definition: f(x) = x^{n}, f(x) = \frac{1}{x}, f(x)=\sqrt{x} and know why the derivative of f(x) = |x| and f(x) = x^{1/3} do not exist at x = 0 .

Having some “template” example can help you master a theoretical concept.

Part II: Attitude
Your attitude will be very important.

1. Remember that your effort will be essential! Again, you can’t learn to run a marathon without getting off of the couch and making your muscles sore. Learning mathematics involves some frustration and, yes, at times, some tedium. Learning is fun OVERALL but it isn’t always fun at all times. You will encounter discomfort and unpleasantness at times.

2. Remember that winners look for ways to succeed; losers and whiners look for excuses for failure. You can always find those who will be willing to enable your underachievement; unfortunately, on occasion, some of the enablers will be those few misguided faculty and staff members. Here is such an example.
Hang out with the winners; don’t let the losers infect you with a defeatist, excuse making attitude.

3. Success is NOT guaranteed; that is what makes success rewarding! Think of how good you’ll feel about yourself if you mastered something that seemed impossible to master at first. And yes, anyone who has achieved anything that is remotely difficult has taken some lumps and bruises along the way. You will NOT be spared these.

Put on your “big boy” or “big girl” underpants and buck it up!

Remember that if you duck the calculus challenge, you are, in essence, slamming many doors of opportunity shut right from the get-go.

4. On the other hand, remember that Calculus (the first two semesters anyway) is a Freshman level class; exceptional mathematical talent is not a prerequisite for success. True, calculus is easy for some but that isn’t the point. A reasonably intelligent person can have success, if they are willing to put forth the proper effort in the proper manner.

Just think of how good it will feel to succeed in an area that isn’t your strong suit!

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December 13, 2008 - Posted by | education, mathematics

21 Comments »

  1. My math lecture teach me that tan(x)=sin(x)/cos(x), and the tan(x) was obtained by integrating arctan(x) respect to dx. Further, he gives me a homework how to create the definition of tan(x)=sin(x)/cos(x) from the arctan(x). So, I need your help because I have idea to answer the homework. Thx.

    Comment by Nadya Fermega | December 14, 2008 | Reply

  2. Sorry “because I have no idea to answer the homework”. Thx.

    Comment by Nadya Fermega | December 14, 2008 | Reply

  3. Sorry, but something must be lost in translation. I do know that we frequently use the definition of tan(x) and that the derivative of arctan(x) is obtained by the implicit function theorem and I can guess that one can use the derivative of arctan(x) to find out information about tan(x).

    In short, I don’t understand your question and if I did, I probably couldn’t help you anyway as I don’t know what you have to work with.

    Update: perhaps it is this?

    A Direct Proof of the Integral Formula for Arctangent
    Arnold J. Insel, Illinois State University, Normal, IL
    The College Mathematics Journal, May 1989, Volume 20, Number 3, pages 235–237.

    Try google.

    Comment by blueollie | December 14, 2008 | Reply

  4. Many thanks for the explanation and your information about the the book reference. I’ve just browsed via google, and I’ve read both definition of the tangent and arctangent functions. Best Regards.

    Comment by Nadya Fermega | December 15, 2008 | Reply

  5. This is the question of my homework.
    Create the definition of tan(x)=sin(x)/cos(x) from
    dy/dx=1+y^2, for x(0)=0 and y(0)=0. Maybe my lecture wants
    to know the verifying of the tangent function from his students. Now, I am really confusing. Thank You for your attention.

    Comment by Nadya Fermega | December 15, 2008 | Reply

  6. Again, I don’t know what you are allowed to use.

    Clearly dy/dx = 1 + y^2 with x(0)=0, y(0)=0 has the solution y = tan(x) IF you know that integral(1/1+x^2)dx = arctan(x)

    Then use the inverse function theorem to derive that dy/dx = sec^2(x).

    Then the original equation gives you sec^2(x) = 1 + tan^2(x). But I don’t know what you are allowed to use.

    Comment by blueollie | December 15, 2008 | Reply

  7. How to prove the following definite integral:

    [tex]$\int_{\infty}^{\infty}exp(-x^2) dx=\sqrt{\pi}$[/tex]

    Is in this case we need gamma function? Thank’s for your help

    Comment by Denaya | December 18, 2008 | Reply

  8. No, you don’t need the gamma function although you can use it.

    Hint: first look at the above integral from 0 to infinity then multiply this integral by the the same integral that uses “y” as the dummy variable.

    Then use multi-variable calculus results to obtain the double integral (0 to infinity for both limits) of exp(-(x^2 + y^2))dxdy (which equals integral(exp(-x^2))dx*integral(exp(-y^2)dxdy )
    Now switch to polar coordinates; the “rdrd(theta)” gives the double integral (exp(-r^2)r drd(theta); this is an integral that you can do.

    Comment by blueollie | December 19, 2008 | Reply

  9. Good explanation,
    I am soon understand why the above definite integral (I) questioned by Denaya gives the result pi^(1/2). The square of I corresponds to 4 times the square of quarter of a circle, or I^2=4*[integral(exp(-r^2)rdr]*integral[d(theta)] with the bound of integrals for dr from 0 to infinity, while for d(theta) are from 0 to pi/2. Hence we can write
    I^2=4*(1/2)(pi/2)=pi
    and finally we can show that :
    I =integral(exp(-x^2))dx = pi^(1/2).
    Again, thank for your guide.

    Comment by Juliani | December 19, 2008 | Reply

  10. I am very glad now, because I can meet in this forum with mrs.Juliani my math lecture. But, I want to ask you, what’s the main reason the integral(exp(-x^2))dx with the bound of integrals from -infinity to +infinity always be transformed into polar coordinate?Thx.

    Comment by Nadya Fermega | December 19, 2008 | Reply

  11. exp(-x^2) does not possess an elementary anti-derivative (that can be proved mathematically)

    For more discussion: see this article (it is written by a department colleague): this deals with exp(-x^2).

    For what can be integrated in elementary terms and what can’t, start here.

    Comment by blueollie | December 19, 2008 | Reply

  12. Hi, last night I search on the internet and I find in this address
    http://math152.wordpress.com/2008/11/09/dunhams-talk-on-euler/
    there is an alternative way in creating Euler formula by using Bernoulli Integral. The author (Mr.Rohedi from Indonesia) also created a new technique of solving first order ODE especially that can be transformed into Bernoulli differential equation. Maybe the paper be useful for us.

    Comment by Juliani | December 20, 2008 | Reply

  13. Whether the derivative of sine and cosine functions also not exist at x=0. If not, how to get each of its derivative.

    Comment by Denaya | December 21, 2008 | Reply

  14. Apologize, my purpose whether the absolute of sine and cosine functions also not exist at x=0. If not, how to get each of its derivative?

    Comment by Denaya | December 21, 2008 | Reply

  15. Hint: try using the limit definition of the derivative; you may or may not need to use the trig identities.

    Note that with the absolute value:

    lim (abs(sin(x))-abs(sin(0))/x = lim abs(sin(x))/x
    Then use left and right handed limits (lim as x goes to zero from the left, and again on the right).

    You can do something similar with cosine except one wouldn’t expect to get a different value for the derivative of abs(cos(x)) since cos(x) stays positive near x = 0.

    Comment by blueollie | December 22, 2008 | Reply

  16. Thank You very much for your hint, I soon try it. I am very glad visit to this blog, because you always help me and all visitors.

    Comment by Denaya Lesa | December 22, 2008 | Reply

  17. How to get the analytic solution of the following chini differential equation : dy/dx + py = qy^3 + r, for all of the three p,q,and r are constants. The ODE is integrable, isn’t, hence actually we can solve the ODE by integration as soon as we perform the separating x and y variables. But, why symbolic softwares such as Matematica, Maple, and Matlab until now do not give the above ODE in analytic form?. Thx for your attention sir.

    Comment by Juliani | December 24, 2008 | Reply

  18. Re: answer to 17 (way late): try the substitution
    z = y^(-2)
    Then z’ = (-2)y^(-3)y’
    which means y’ =-1/2 * y^3* z’
    y = y^3*y^(-2)=z*y^3
    So the differential equation is transformed into:

    z'(y^3) + pzy^3 =qy^3 +r
    z’ + pz = q + r*y^(-3)
    z’ + pz = q + r*z^(3/2)

    And now do a second substitution:

    w = z^(-1/2) which should reduce your differential equation to a linear first order one.

    Comment by blueollie | February 3, 2009 | Reply

  19. Thank’s for your attention @blueollie. But apologise, I’ve tried the second substitution of the above advise, and back into the initial form of the Chini ODE in w of dependent variable, not transforming into a linear first order ODE.
    What do you think? Thx.

    Comment by Juliani | February 5, 2009 | Reply

  20. I’ll have to think about this; I forgot that this differential equation is non-linear; hence superposition of solutions doesn’t work.

    My first substitution would work if that extra constant wasn’t there. 🙂

    Comment by blueollie | February 5, 2009 | Reply

  21. Reblogged this on Think About It, Steph.

    Comment by stephsodopeee | May 5, 2015 | Reply


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