Results 1 to 6 of 6

Math Help - FACULTY OF MECH ENGINEERING exam #2

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    3

    FACULTY OF MECH ENGINEERING exam #2

    Hello, im new to the forum and this is my debut thread/problem. I am currently a student enrolling in mechanical engineering at the University of Ljubljana, Slovenia. Anyways we had an exam a few days ago and it was suprising to me that only 2 people managed to get above 50% out of 100 students. The differential equation that gave me problems was this:

    f(x)=ln(tg(2x))

    Prove that: f "(x)+4ctg(4x)*f '(x)=0

    *=product
    ln=natural logarithm
    tg=tangent
    ctg=cotangent

    if you have trouble understanding my post please reply. thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,234
    Thanks
    27
     f(x)=\ln(\tan(x))

    Using the chain rule find f'(x) and f''(x)

    and check them in the equation you have.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2010
    Posts
    3
    yeah i did that but its very complicated.
    i got f '(x)= 2/(cos^2 (2x) * tg(2x))
    and f ''(x)= -4 sin(2x)*tg(2x)-2*(cos^2 (2x) - 1/cos^2 (2x))/ (cos^4 (2x)*tg^2 (2x))
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by spanky489 View Post

    f(x)=ln(tg(2x))

    Prove that: f "(x)+4ctg(4x)*f '(x)=0

    *=product
    ln=natural logarithm
    tg=tangent
    ctg=cotangent
    f'(x) = \frac{2\sec^2(2x)}{\tan(2x)} = \frac{2}{\cos^2(2x) \tan(2x)} = \frac{2}{\sin(2x)\cos(2x)} = \frac{4}{2\sin(2x)\cos(2x)} = 4\csc(4x)

    \textcolor{red}{f'(x) = 4\csc(4x)}

    \textcolor{blue}{f''(x) = -16\csc(4x)\cot(4x)}

    \textcolor{blue}{-16\csc(4x)\cot(4x)} + 4\cot(4x) \cdot \textcolor{red}{4\csc(4x)} = 0
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Aug 2008
    Posts
    903
    y''+4\cot(4x)y'=0

    \frac{dy'}{y'}=-4\cot(4x)dx

    \log(y')=-\log(\sin(4x))+C

    y'=\frac{k}{\sin(4x)}

    y=\frac{k}{4}\log(tan(2x))+c_2

    so y(x)=\log(tan(2x) is a particular solution.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Feb 2010
    Posts
    3
    Quote Originally Posted by shawsend View Post
    y''+4\cot(4x)y'=0

    \frac{dy'}{y'}=-4\cot(4x)dx

    \log(y')=-\log(\sin(4x))+C

    y'=\frac{k}{\sin(4x)}

    y=\frac{k}{4}\log(tan(2x))+c_2

    so y(x)=\log(tan(2x) is a particular solution.
    hehe thanks! but i think this is too complex for 1st semester math. anyways thx for help guys im off to bed gotta wake up early tomorrow and protest my exam score. thx again!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Engineering economics
    Posted in the Business Math Forum
    Replies: 0
    Last Post: March 14th 2011, 11:57 AM
  2. Replies: 2
    Last Post: February 14th 2011, 10:28 AM
  3. static friction, need help for mech final!
    Posted in the Advanced Applied Math Forum
    Replies: 0
    Last Post: May 12th 2009, 03:54 AM
  4. little engineering/little calc
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: March 7th 2009, 11:17 PM
  5. Reactions (engineering)
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: January 28th 2009, 10:51 AM

Search Tags


/mathhelpforum @mathhelpforum