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Math Help - exponential graphs

  1. #1
    Member i_zz_y_ill's Avatar
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    Question exponential graphs

    part one: given y=0.5e^-t -0.2e^-2t +0.1sint -0.3cos t

    and another curve

    y=Ae^-t + Be^-2t + 0.1e^t(sint -cost) A B constants

    How do you know that for large values of t that the first one is bounded and the second one is unbounded? And what does bounded mean?

    part two: Given x = Ae^-0.04t +Be^-0.02t +1875

    and y = -Ae^-0.04t + Be^-0.02t +625

    How do you know that they grow exponentially like a normal exponential. I would have thought that they would be reflected in the y or x axis(corrosponding to each) one. And the y curve has a dip in it, how are you supposed to know that.
    Tanswers to these questions would be greatly aprecciated I have an exam soon and this is quite hard. Thanks!!!!
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by i_zz_y_ill View Post
    part one: given y=0.5e^-t -0.2e^-2t +0.1sint -0.3cos t

    and another curve

    y=Ae^-t + Be^-2t + 0.1e^t(sint -cost) A B constants

    How do you know that for large values of t that the first one is bounded and the second one is unbounded? And what does bounded mean?

    part two: Given x = Ae^-0.04t +Be^-0.02t +1875

    and y = -Ae^-0.04t + Be^-0.02t +625

    How do you know that they grow exponentially like a normal exponential. I would have thought that they would be reflected in the y or x axis(corrosponding to each) one. And the y curve has a dip in it, how are you supposed to know that.
    Tanswers to these questions would be greatly aprecciated I have an exam soon and this is quite hard. Thanks!!!!
    Bounded means that for sufficiently large numbers a function is always smaller or larger than another function

    For example

    \sin(x) is bounded above by 2 and below by -2\

    Does that make senes?
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  3. #3
    Member i_zz_y_ill's Avatar
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    Not really. I don't think it means in relation to each function. The mark scheme says that the first one(x) is bounded oscillations and the second one (y) is unbounded oscillations. I can't figure out what it means by large t. I know this would make the first part of the functios zero but then for the sin and cos parts it doesn' really make sense. Is it referring to the fact that for th y function the line connecting its msximum points as t tends to large is a decsyng eponential in which case it is 'not bounded' and for the second one in theory it would be a straight line making it bounded? From an anayltical standpoint however putting a large value of t into the sin and cos values doesnt really get you anywhere,,,,,,is this correct what i'm thinking thnx
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  4. #4
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    Quote Originally Posted by i_zz_y_ill View Post

    given  y= 0.5 e^{-t } -0.2e^{-2t} +0.1\sin t -0.3\cos t

    and

    y=Ae^{-t} + Be^{-2t} + 0.1e^{t}(\sin t -\cos t)

    How do you know that for large values of t that the first one is bounded and the second one is unbounded? And what does bounded mean?
    For large values of t you may neglect the value of e^{-at} (where a is a positive constant) as it becomes very close to zero.

    So for large values of t the first function approaches  y= 0.1\sin t -0.3\cos t to simplify this you need to use an "R" formula. are you familiar with how to write 0.1\sin t -0.3\cos t as R \sin ( t - \theta ) ? but it should be clear that is the a simple sine curve and cannot exceed a particular value.


    for the second one you apply a similar argument so the function approaches y= 0.1e^{t}(\sin t -\cos t) then simplify \sin t -\cos t into the form of R \sin ( t - \alpha) so your function becomes y = 0.1R e^{t}\sin( t - \alpha).

    as -1 \leq \sin ( t - \alpha) \leq 1 then by multiplying by  0.1R e^{t} you get  -0.1R e^{t} \leq 0.1R e^{t}\sin( t - \alpha) \leq 0.1R e^{t} . What can you deduce form this inequality ?

    Bobak
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  5. #5
    Member i_zz_y_ill's Avatar
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    Oh ri thanks. So for the first one the max value is R. I still don't see how that makes it bounded erm and the second one if its limits are between |0.1Re^t| then how is it unbounded. I would have thought in theory this would decay and have limits(max and min) according to its exponential curve. Sill a bi confused sorry!
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  6. #6
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    Quote Originally Posted by i_zz_y_ill View Post
    Oh ri thanks. So for the first one the max value is R. I still don't see how that makes it bounded
    It is bounded because the curves lines between the lines y = R and y = -R

    I attached a graph to illustrate this.


    the second one if its limits are between |0.1Re^t| then how is it unbounded.
    because the limits the seconds ones lies between increase without limits for large t. have you tired sketching the curve e^{x} \sin x ? I have attach a image of that to help you get the general idea.

    Attached Thumbnails Attached Thumbnails exponential graphs-picture-16.png  
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  7. #7
    Member i_zz_y_ill's Avatar
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    oh right yeah!!! I guess this means it is unbounded by having an exponential boundary. Ok so ill use the form Rsin(t+/- alpha) since this is easier to analyse for larger t I suppose. K thanks alot I think I understand now if im correct in presuming this???
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  8. #8
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    Quote Originally Posted by i_zz_y_ill View Post
    oh right yeah!!! I guess this means it is unbounded by having an exponential boundary. Ok so ill use the form Rsin(t+/- alpha) since this is easier to analyse for larger t I suppose. K thanks alot I think I understand now if im correct in presuming this???
    Yes if the bounding function is exponentially increasing then it is unbounded. also your not using Rsin(t+/- alpha) because it is easier to analyse for large t, you use it because the values of R makes your maximum and minimum values apparent.

    Bobak
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  9. #9
    Member i_zz_y_ill's Avatar
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    Oh right yeah!!!! I understand, i appreciate ur time/effort thanks alot!
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  10. #10
    Member i_zz_y_ill's Avatar
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    Oh and you wouldn't happen to know the second part to my original question would you? part 2: I can't draw those graphs
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  11. #11
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    Quote Originally Posted by i_zz_y_ill View Post
    Oh and you wouldn't happen to know the second part to my original question would you? part 2: I can't draw those graphs
    I am unsure about what your question is for that part. but it look easier than the first question. Just remember the e^{-at} terms vanish for large values of t. so you should be able to easily determine what values both x and y are approaching for large values of t.

    Bobak
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