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Math Help - Factorizing -Where to next?

  1. #1
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    Factorizing -Where to next?

    Hi everyone, having a lot of trouble with Calculus and was wondering if anyone can help.

    I have d(t)= -5e^(-5t) x cos(10t)

    u = -5e^(-5t)
    v = cos(10t)

    du/dx = 25e^(-5t)
    dv/dx = -10sin10t

    duv/dx = 25e^(-5t) x cos(10t) + (-10sin10t) x -5e^(-5t)

    = 25e^(-5t)cos(10t) + 50e^(-5t)sin10t
    = 2cos10t+sin10t

    From here I am not sure where to go? Do I have the first bit correct?

    0=(sin+cos=1)
    0=2 x 1 x 10t
    0=20t
    t=1/20 ???

    Or do i have to find out tan = sin(a) / cos (a)

    If anyone could give me some kicks in the right direction it would be much appreciated, and i can stop wasting paper.

    Thanks
    Kayne
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  2. #2
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    Quote Originally Posted by kayne View Post
    Hi everyone, having a lot of trouble with Calculus and was wondering if anyone can help.

    I have d(t)= -5e^(-5t) x cos(10t)

    u = -5e^(-5t)
    v = cos(10t)

    du/dx = 25e^(-5t)
    dv/dx = -10sin10t

    duv/dx = 25e^(-5t) x cos(10t) + (-10sin10t) x -5e^(-5t)

    = 25e^(-5t)cos(10t) + 50e^(-5t)sin10t
    = 2cos10t+sin10t

    From here I am not sure where to go? Do I have the first bit correct?

    0=(sin+cos=1)
    0=2 x 1 x 10t
    0=20t
    t=1/20 ???

    Or do i have to find out tan = sin(a) / cos (a)

    If anyone could give me some kicks in the right direction it would be much appreciated, and i can stop wasting paper.

    Thanks
    Kayne
    The derivative is 25e^{-5t} \left( \cos (10t) + 2 \sin ( {\color{red}1}0t) \right).

    If you're trying to find when the derivative is equal to zero, you need to solve

    \cos (10t) + 2 \sin ({\color{red}1}0t) = 0 \Rightarrow \tan (10t) = - \frac{1}{2}.

    The general solution is 10 t = - \tan^{-1} \left( \frac{1}{2} \right) + n \pi \Rightarrow t = \, .... where n is an integer.

    I assume that:

    1. You have a domain for the solutions ....
    2. You don't want exact solutions.
    Last edited by mr fantastic; October 7th 2008 at 02:52 AM.
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  3. #3
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    Mr Fantastic,

    Sorry for not being very clear, yes you are right about the equation equalling zero. The question is to find the Max and min points for a shock absorber rebound .

    Once the equation = 0, i have to find the second derivative to measuse the rate of change of the curve, and subistute the x values into the original equation.

    Thanks for the help.. If i get stuck again i will post a reply.

    kayne
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  4. #4
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    Mr Fantastic,

    I have been looking over your post today and i am a little lost to how you factorized the equation.

    The deritive that you stated

    25e^(-5t)(cos(10t) + -2sin(20t))

    How did you come to this every time i do this i end up with some thing different so i must be factorizing wrong. I would like to know where the 2 and 20 come from seeing that cos(10t) = -10sin(10t)

    and the next part of the equation where did the 25e^(-5t) go to as you ended up with

    0 = Cos(10t) + 2sin(20t) = tan(10t) = -1/2

    When i do it i end up with

    0 = Cos(10t) + 2sin(10t) = 2tan(10t)


    ?Where have i lost my factorizing?

    I have attached the graph that i am working from to find the max positive rebound and when it occurs.

    Attached Files Attached Files
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  5. #5
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    Quote Originally Posted by kayne View Post
    Mr Fantastic,

    I have been looking over your post today and i am a little lost to how you factorized the equation.

    The deritive that you stated

    25e^(-5t)(cos(10t) + -2sin(20t))

    How did you come to this every time i do this i end up with some thing different so i must be factorizing wrong. Mr F says: I took out the common factor of 25 e^(-5t).

    I would like to know where the 2 and 20 come from seeing that cos(10t) = -10sin(10t) Mr F says: The 20 comes from the fact that I made a typo that unfortunately got carried through when I did a copy and paste. I've edited the reply in red.

    and the next part of the equation where did the 25e^(-5t) go to as you ended up with

    Mr F says: {\color{red}25e^{-5t} \left( \cos (10t) + 2 \sin ( {\color{red}1}0t) \right) = 0} implies that either {\color{red}25e^{-5t} = 0} or {\color{red}\cos (10t) + 2 \sin (10t) = 0}. I thought it would be obvious that {\color{red}25e^{-5t} = 0} has no real solution so I just gave {\color{red}\cos (10t) + 2 \sin (10t) = 0}.

    0 = Cos(10t) + 2sin(20t) = tan(10t) = -1/2

    When i do it i end up with

    0 = Cos(10t) + 2sin(10t) = 2tan(10t) Mr F says: I'm sorry but that's totally wrong. Cos(10t) + 2sin(10t) is not the same as 2 tan(10t). Cos(10t) + 2sin(10t) = 0 => Cos(10t) = -2sin(10t) => 1 = -2 sin(10t)/cos(10t) = -2 tan(10t) => -1/2 = tan(10t).


    ?Where have i lost my factorizing?

    I have attached the graph that i am working from to find the max positive rebound and when it occurs.

    ..
    Last edited by mr fantastic; October 7th 2008 at 03:07 AM.
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  6. #6
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    Thanks again, for the detailed explanation. I can see where i have been going wrong now. i will work on this now and finish the problem.

    kayne
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