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Difficult Calculus Problems

  1. Jun 17, 2005 #1
    Hey I have a course project where I have to find/develop 15 slightly difficult calculus problems and solve them. Any suggestions?
    (I'm not asking for proofs but if you want to share those as well I really really don't object :smile: )

    Oh, and this is first/second year calculus not multivariable.
  2. jcsd
  3. Jun 17, 2005 #2
    If I were in your shoes I would pick 15 problems that involve the trig identities. Those were the hardest for me, because you have to remember the formulas which you can use as tricks to solve the problems. If your taking calc 2, Id do problems with trig substitution, those are kind of hard and will help you out on tests and stuff.
  4. Jun 17, 2005 #3


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    Here's kind of a fun site from Harvey Mudd College with tutorials and quizzes about calculus. You can go into the tutorials for 1st year calculus subjects, go to the harder subjects, and at the end of some of the tutorial pages, they'll have an "Explore" page where you can play with graphing or other stuff. Check out the Taylor Series subject and its Exploration page -- very cool.

  5. Jun 17, 2005 #4
    [tex]\int \sqrt{tan\theta} d\theta[/tex]

    I remember this problem from calc 2 :mad:
  6. Jun 17, 2005 #5
    Prove that

    [tex] \lim_{x\rightarrow 0} \frac{\sin x}{x} = 1 [/tex]
  7. Jun 17, 2005 #6
    Actually that wouldn't be too hard.

    Ignoring the limit and rearranging the equation gives you sinx = x. That's only true when x = 0.

    The only reason the limit exists in the original equation is because you cannot divide by zero.

    p.s. first post, hope i'm not wrong :smile:
  8. Jun 17, 2005 #7
    [tex] \lim_{x\rightarrow 0} \frac{\sin x}{x} = 1 [/tex]

    To prove this I would set up a unit circle and find the area of the triangle with legs sin(x) and cos(x), area of the sector of angle x, and area of triangle with legs 1 and tan(x). Now rearrange and set up and equation with these which will lead you to the fact that [tex]1\leq \lim_{x\to 0}\frac{\sin x}{x}\leq 1[/tex]
  9. Jun 17, 2005 #8
    hello there

    well i remember this from high school, but i know that some of the people i go to uni with, were not able to do it
    [tex]\int \sec\theta d\theta[/tex]
    there would be two ways of doing it, a long and short way, i come to look at this now its pretty simple anyway good luck with solving

    by the way it would be a great idea to list some topics that you have went though in your calculus class?

  10. Jun 18, 2005 #9
    [tex]\int_{0}^{\pi/2} \frac{dx}{1+(tan(x))^{\sqrt2}}[/tex]

    (the answer is pi/4, & it doesn't matter what number goes where the sqrt 2 is)
  11. Jun 18, 2005 #10


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    You can't just ignore the limit or declare it to be 1 since sin(x)=x at x=0. Notice sin(x)=x^(1/3) only at x=0 as well, but [tex]\lim_{x\rightarrow 0} \frac{\sin x}{x^{1/3}} \neq 1[/tex]. You have to treat indeterminant limits like this with care.
  12. Jun 18, 2005 #11
    Well i guess you have to understand why we took the limit in the first place. suppose we just had the equation w/o the limit. Then solving for x you get 0. But in current mathematics, you cannot do that since it gives you division by 0 (in the times befor Sir Wallis, mathematicians like Fermat did just that).

    The reason we're saying sin(x)=x and not anything else is because we don't want to change the original equation.

    In your equation, the limit would be x^(-2/3) which is solved in exactly the same manner.
  13. Jun 18, 2005 #12
    The limit involving sinx/x is prooved by drawing a unit circle. What you told is wrong because what you do is intersecting the x=y line and y=sinx curve. sinx/x has a totally different curve.
    Using the areas in the unit circle, it can be proved.

    Not so difficult but try prooving that as x approaches 0, sinkx/x=k.
  14. Jun 18, 2005 #13
    I'm sorry i still don't see how my method is wrong. If you want to use unit circles, that's fine, but my method works unless it's one of those math coincidences.

    If someone could explain why it's wrong, i would be grateful. :smile:
  15. Jun 18, 2005 #14
    I think shmoe did a good job of that. I guess the better problem as wisredz suggested was find

    [tex] \lim_{x\rightarrow 0} \frac{\sin kx}{x} [/tex]
  16. Jun 18, 2005 #15
    Another good one would be using some tabular integration to find a reduction formula for

    [tex] \int \sin ^n x dx [/tex] for even integers 'n'.
  17. Jun 18, 2005 #16


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    Are you saying [tex]\lim_{x\rightarrow 0} \frac{\sin x}{x^{1/3}}=\frac{1}{x^{2/3}}[/tex]? This wouldn't make any sense at all, could you please clarify?

    By your method if we ignore the limit in the 'equation'

    [tex]\lim_{x\rightarrow 0} \frac{\sin x}{x^{1/3}}=1[/tex]

    and rearrange we get [tex]\sin x=x^{1/3}[/tex] which is true only when x=0 so this limit is correct (of course it's actually wrong).

    Please explain to me how this is different from what you've proposed for sin(x)/x.
  18. Jun 18, 2005 #17
    here's another one:

    maximize [tex]f(x) = \frac{1}{2^{x}} + \frac{1}{2^{1/x}}[/tex] for x>0.
  19. Jun 21, 2005 #18

    wow what a great response thanks so much guys :biggrin:
  20. Jun 21, 2005 #19
    oh and some of the topics we covered in that class are pretty basic:
    improper integrals
    parametric eq
    taylor mclauren series
    length of a curve/along a path
    but i've also done multivariable calc.
  21. Oct 18, 2009 #20
    hello i was trying to solve this problem but i really couldn't could you please show me step by step how to solve it i would really appreciate that, thank you very much..
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