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Midnight Tutor - Videos To Help You Learn Calculus

  1. Nov 28, 2007 #1
    http://www.midnighttutor.com/

    I searched and it looked like no one had posted this before, so here it is. The Midnight Tutor is an awesome site in which some guys make videos of themselves in front of boards and teach calculus, be it a concept or how to solve a certain kind of problem. It is modeled on the AP Calculus program (including test specific approaches and questions), and seems to me one of the most awesome things I have seen. They also post their videos on YouTube, too. And if you have a question, you can e-mail them and they will (eventually) do a video on how to solve it! It's great! ^.^
     
  2. jcsd
  3. Nov 29, 2007 #2

    Kurdt

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    The guy knows his stuff and they are certainly useful. One criticism is that he should have planned the videos a bit more carefully as he tends to stop and start quite a lot. Never the less I'm sure lots of students would find the videos exceptionally useful.

    P.S. Not too sure about his policy of solving peoples homework questions. I know he explains them thoroughly but examples would be better so the student can work through themselves.
     
  4. Nov 29, 2007 #3

    Mk

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    That's an awesome site.
     
  5. Nov 30, 2007 #4
    I am working on something similar, but based on general Calculus lessons.
     
  6. Dec 8, 2007 #5
    thanks for sharing!
     
  7. Dec 9, 2007 #6
    Awesome!.. This site is what I am looking for. There are many problems that I could not solve. Thanks for your sharing.
     
  8. Dec 31, 2007 #7
    Very nice. Even with my bachelors in Physics from 7 years ago I still need a refresher.
     
  9. Jan 8, 2008 #8
    thanks for sharing.
     
  10. Jan 11, 2008 #9
    pretty interesting
    thanks for sharing the link
     
  11. Jan 13, 2008 #10
    I found tutorial with mistake http://www.midnighttutor.com/integral_1_over_xsqrt(4x+1).html
    Type 1/(x Sqrt[4x+1]) into http://integrals.wolfram.com/index.jsp
    and you will see diferent
    1/(1-x^2) integral in integral tables I can't found at all. only in one book found that it's equal 0.5ln|(1+x)/(1-x)|+C. But even if repair mistake wolfram reasearch still giving ln((4x+1)^0.5-1)-ln((4x+1)^0.5+1) instead -[ln(1+(4x+1)^0.5)-ln(1-(4x+1)^0.5)] if use this formula 0.5ln|(1+x)/(1-x)|+C by trying end integration calculation corectly.
    btw wolfram 1/(1-x^2) integrating as 0.5(ln(x+1)-ln(x-1))=0.5ln|(x+1)/(x-1)|. So then final answer by continuing guy work must be if using wolfram formula:
    -[ln((4x+1)^0.5+1)-ln((4x+1)^0.5-1)]= ln((4x+1)^0.5-1)-ln((4x+1)^0.5+1).
     
    Last edited: Jan 13, 2008
  12. Jan 13, 2008 #11

    Kurdt

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    Thats the same answer.

    [tex] \tanh^{-1}(x) = \frac{1}{2} \ln (x+1) - \ln (x-1) [/tex]
     
    Last edited: Jan 13, 2008
  13. Jan 14, 2008 #12
    I don't think so.
    arctg(1.1)=0.83298
    0.5ln[(1.1+1)/(1.1-1)]=0.5ln(21)=1.52226

    Final his integral
    [tex]2 \int \frac{du}{u^2-1} [/tex] possible integrate by taking integral from table (which don't exist in all integral tables) [tex]\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\ln|\frac{x-a}{x+a}|+C [/tex] and then
    [tex]2 \int \frac{du}{u^2-1}=\ln|\frac{u-1}{u+1}|+C [/tex]
     
    Last edited: Jan 14, 2008
  14. Jan 14, 2008 #13

    Kurdt

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    Its a well known identity. Besides you have calculated arctan(x) not arctanh(x). Arctanh(1.1) does not exist.

    From the table you have:

    [tex] \ln\left|\frac{u-1}{u+1}\right| +C = \ln |u-1| -\ln |u+1| +C = -2\tanh^{-1} (u) +C[/tex]
     
    Last edited: Jan 14, 2008
  15. Jan 15, 2008 #14
    Just type 1/(x sqrt(4x+1)) into http://integrals.wolfram.com/index.jsp and answer will be
    [tex]\int \frac{dx}{x\sqrt{4x+1}}=\ln(\sqrt{4x+1}-1)-\ln(\sqrt{4x+1}+1)+C[/tex]
    And hyperbolic arctangent http://mathworld.wolfram.com/InverseHyperbolicTangent.html is
    arctanh(z)=0.5[ln(1+z)-ln(1-z)] and (arctanh(z))'=1/(1-z^2)
    So this guy looks like integrate correctly, but then it means that wolfram integrator integrate wrong? Becouse according to this guy final answer will be
    [tex]-2\int \frac{du}{1-u^2} =\ln(1-u)-\ln(1+u)+C=\ln(1-\sqrt{4x+1})-\ln(1+\sqrt{4x+1})[/tex]
    so answers are different, computer vs human, who lieing?
    BTW wolfram integrator showing that 1/(1-x^2) is
    [tex]\int\frac{dx}{1-x^2}=\frac{1}{2}(\ln(x+1)-\ln(x-1))[/tex]
    Looks like I found bug in wolfram integrator, becouse everywhere on internet [tex] (\tanh^{-1} (x))'=\frac{1}{1-x^2}[/tex]

    And in this lecture http://www.midnighttutor.com/int_x^2sinxdx.html he missed 2 in final answer.
     
    Last edited: Jan 15, 2008
  16. Jan 15, 2008 #15

    Kurdt

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    Sorry yes in post 11 I gave the identity for arccoth (x) by mistake. I think That wolfram may be taking the absolute value in which case it doesn't matter which way round you write them but you have to be aware that arctanh has a particular domain. So in that respect perhaps its not so good.
     
    Last edited: Jan 15, 2008
  17. Jan 19, 2008 #16
    Great link, thanks!
     
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