1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Various Questoins

  1. Jul 6, 2004 #1
    I just took a take home calc test this weekend, turned it in this morning. There were a few questions I couldn’t answer. Will you guys tell me (or even better show me) how to solve these?

    #1 [tex] \int \frac{dx}{3sinx - 4cosx} [/tex]

    #2 Find g’(x) where g is an inverse function of f(x)
    f(x) = 3 + x^2 + sin([pi]x) -0.4 < x < 0.4

    #3 find the exact value of sin[arcsine(1/3) + arcsine(2/3)]
     
  2. jcsd
  3. Jul 6, 2004 #2
    #1. You can rewrite 3sin(x) - 4cos(x) as k * sin(x + v), where k and v are constants. Then you just have integrate 1/k * csc(x + v).

    #2. Hmm, I wonder if you can leave g(x) in the answer?

    #3. Simpify using the addition formula for sine, and find a formula for cos(arcsin(x/y)).
     
    Last edited: Jul 6, 2004
  4. Jul 6, 2004 #3

    Zurtex

    User Avatar
    Science Advisor
    Homework Helper

    Here is always a useful substitution for integrals of that and similar form:

    [tex]t = \tan \frac{x}{2}[/tex]


    [tex]\cos x = \frac{1-t^2}{1+t^2}[/tex]


    [tex]\sin x = \frac{2t}{1+t^2}[/tex]


    [tex]\frac{dx}{dt} = \frac{2}{1+t^2}[/tex]

    (Not 100% sure I have that last one right)
     
    Last edited: Jul 7, 2004
  5. Jul 6, 2004 #4
    #2 - f(x) = 3 + x^2 + sin([pi]x)
    g(x) is the inverse of f(x). Remember, just because it is written as g(x) doesnt mean that you have to make g a function of x. You are not asked to find the inverse of the function. You are rather being asked to find dy/dx of the inverse.

    The inverse of f(x) is:
    x = 3 + y^2 + sin([pi]y)

    now can you find dy/dx? (think of it as implicit differentiation)

    EDIT: nevermind you can't do this. I just realized they don't want dy/dx they instead want g'(x) which is not the same thing.
     
    Last edited: Jul 7, 2004
  6. Jul 6, 2004 #5
    How does this take-home test system work? What is stopping you from asking these same questions during the test period (other than your conscience - but who ever listens to it anyways?)?
     
  7. Jul 7, 2004 #6
    thats about it. It was 25 questions and these are the only 3 that i didn't get right i'm pretty sure.
     
  8. Jul 7, 2004 #7
    [tex]sin(x+y) = ?[/tex]
    [tex]sin(sin^-^1(x))= ?[/tex]
     
  9. Jul 7, 2004 #8
    forget to say it is just another way to solve the problem because the answer given by Muzza can also be applied..
     
  10. Jul 7, 2004 #9
    Seems to me like we were thinking of the same solution. Since you know, sin(x + y) will not only include sin(x) and sin(y), but also cos(x) and cos(y)...
     
  11. Jul 7, 2004 #10
    Yeah, i should have thought about what you wrote more deeper....Nothing big right ? --lol

    Uhmm, the same!
     
  12. Jul 7, 2004 #11
    Thanks for the help guys, those problems were a lot easier then I was making them out to be…

    i have a new question…

    Is the equation [tex] y = \lim_{n \rightarrow \infty} \pm(1 - x^{2n})^{1/2n} [/tex] a square?

    Is [tex] \lim_{n \rightarrow \infty} \int \pm(1 - x^{2n})^{1/2n} dx [/tex] = to 4? i.e. a 2x2 square?
     
    Last edited: Jul 7, 2004
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Various Questoins
Loading...