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Velocity and what not

  1. Feb 26, 2005 #1
    i had this problem on my calculus test the other day that was actually challenging. the whole class tried to get it taken off of the test, because barely any of them did it. :sad: i was wanting to know if i approached it right and yielded the correct answer.

    given [itex]v(t)=13e^{-.02t}sin(t)[/itex]:
    a) find a(t)
    b) find a position equation assuming s(0)=0
    c) find the average velocity along 0<t<pi

    a) [tex]a(t)=\frac{dv}{dt}=13e^{-.02t}cos(t)-.26e^{-.02t}sin(t)][/tex]
    a is one of th easy ones. i had trouble on b, but i think i have the correct answer.
    b)[tex]s(t)=\int13e^{-.02t}sin(t)dt[/tex]
    i took out the 13 and got:
    [tex]s(t)=13 \int e^{-.02t}sin(t)dt[/tex]
    now, i use integration by parts:
    [tex]u=e^{-.02t}~~~du=-.02e^{-.02t}dt~~~dv=sin(t)dt~~~v=-cos(t)[/tex]
    [tex]-e^{-.02t}cos(t)-\frac{1}{50} \int e^{-.02t}cos(t)dt[/tex]
    now, i have to integrate by parts again:
    [tex]u=e^{-.02t}~~~du=-.02e^{-.02t}dt~~~dv=cos(t)dt~~~v=sin(t)[/tex]
    [tex]-e^{-.02t}cos(t)-\frac{1}{50}[e^{-.02t}sin(t)+\frac{1}{50} \int e^{-.02t}sin(t)dt][/tex]
    so:
    [tex]-e^{-.02t}cos(t)-\frac{e^{-.02t}sin(t)}{50}+\frac{1}{2500} \int e^{-.02t}sin(t)dt[/tex]
    now, i can set the given integral to the one i have and solve for the first one:
    [tex]\inte^{-.02t}sin(t)dt=-e^{-.02t}cos(t)-\frac{e^{-.02t}sin(t)}{50}+\frac{1}{2500} \int e^{-.02t}sin(t)dt[/tex]
    so:
    [tex]\frac{2501}{2500} \int e^{-.02t}sin(t)dt = -e^{-.02t}cos(t)-\frac{e^{-.02t}sin(t)}{50}[/tex]
    so:
    [tex]\int e^{-.02t}sin(t)dt=\frac{-2500e^{-.02t}cos(t)}{2501}-\frac{2500e^{-.02t}sin(t)}{125050}[/tex]
    if you multiply both sides by the 13 that was taken out at the beginning, you get:
    [tex]13\int e^{-.02t}sin(t)dt=-12.998e^{-.02t}cos(t)-.2599e^{-.02t}sin(t)+C[/tex]
    now, i can assume that t=0 and s(0)=0 and solve for c:
    [tex]0=-12.998e^{-.02*0}cos(0)-.2599e^{-.02*0}sin(0)+C[/tex]
    [tex]0=-12.998+C[/tex]
    [tex]C=12.998[/tex]
    so:
    [tex]s(t)=-12.998e^{-.02t}cos(t)-.2599e^{-.02t}sin(t)+12.998[/tex]?
    i think thats right, but im not sure.
    c) [tex]\overline{v}(t)=\frac{1}{\pi-0}\int_0^\pi v(t)dt=\frac{s(\pi)}{\pi}\approx 8[/tex]

    thanks in advance for any verification!! :!!)
     
  2. jcsd
  3. Feb 26, 2005 #2

    dextercioby

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    Use this formula in your case
    [tex] \int e^{ax}\sin bx \ dx=\frac{a\sin bx-b\cos bx}{a^{2}+b^{2}}e^{ax}+C [/tex]

    .It can be proven by 2 times part integration.

    Daniel.
     
    Last edited: Feb 27, 2005
  4. Feb 27, 2005 #3
    we had to use integration by parts, so we couldnt use the formula that you provided. but could you tell me if my work looks right? thats all im wondering. id appreciate it. :smile:
     
  5. Feb 27, 2005 #4

    dextercioby

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    Your calculations are almost impossible to follow.The purpose of this HW section is not to check arithmetics.The formula that i've provided is the one you should use and,as i said,but you probably didn't read,it can be proven by 2 times integration by parts.

    I'm sorry,that's all i can do.I doubt anyone can do more.

    Daniel.
     
  6. Feb 27, 2005 #5
    well, when i did the proof, i didnt get what you had. i got:
    [tex] \int e^{ax}\sin bx \ dx=\frac{a\sin bx-b\cos bx}{a^{2}+b^{2}}e^{ax}+C [/tex]

    you have [itex]-bcosax[/itex], and i have [itex]-bcosbx[/itex]. i probably missed something, but oh well im ok now. my calculator has the same answer i have. :smile:
     
  7. Feb 27, 2005 #6

    dextercioby

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    Sorry,it was a typo.I misscopied from the paper.It was almost midnight and i was probably tired...

    Daniel.
     
  8. Feb 27, 2005 #7
    does this look better?

    [tex]13\int e^{-.02t}sin(t)dt=13(\frac{-.02\sin(t)-cos(t)}{1.0004})e^{-.02}+C[/tex]

    [tex]s(t)=e^{-.02t}(-12.998\cos(t)-.2599\sin(t))+C[/tex]

    if we were to assume that s(0)=0, then we could solve for C:

    [tex]0=e^{-.02*0}(-12.998\cos(0)-.2599\sin(0))[/tex]

    [tex]C=12.998[/tex]

    [tex]s(t)=e^{-.02t}(-12.998\cos(t)-.2599\sin(t))+12.998[/tex]

    hopefully, you can help me now. if not, i give up. :frown:
     
  9. Feb 27, 2005 #8

    dextercioby

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    Yes,it looks okay.If you applied the correct formula,then it's correct.

    The point c) is just a formality.

    Daniel.
     
  10. Feb 27, 2005 #9
    thanks, dextercioby, you are of great help! just one more question: to find the average velocity of a function is it
    [tex]\frac{1}{b-a}\int_a^b v(t)dt[/tex]
    that shows the displacement and dividing it by the time should give average velocity? i rarely pay attention in class and never look at my book, because i taught myself this stuff 2 years ago, and im all memory. :frown:
     
  11. Feb 27, 2005 #10

    dextercioby

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    Yes,it is correct.Standard definition for the average of a function on a domain.

    Daniel.
     
  12. Feb 27, 2005 #11
    merci tres beaucoup :smile: i have another question that im going to post as a new thread. tackle it if you please.
     
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