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anonymous

  • 4 years ago

49t+245e^(-t/5)-545=0, by some other means than a calculator or wolfram?

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  1. ash2326
    • 4 years ago
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    if x is very small than 1, then e^x can be expanded as 1+x+x^2/2! and higher order terms can be neglected let's assume -t/5 << 1 then e^(-t/5)=1-t/5+t^2/50 substitute this in the equation 49t+245(1-t/5+t^2/50)-545=0 49t+245-49t+4.9t^2-545=0 combining like terms 4.9t^2=300 \[t=\sqrt{\frac{300}{4.9}}\] \[t=7.8246\]

  2. anonymous
    • 4 years ago
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    cool man, series expansion, thanks

  3. phi
    • 4 years ago
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    but t/5 is not << 1

  4. ash2326
    • 4 years ago
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    phi we don't know about that, so I just assumed it

  5. anonymous
    • 4 years ago
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    it would be if the premise is that t is small to begin with, in that case t/5 would be smaller

  6. anonymous
    • 4 years ago
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    but it is that assumption

  7. anonymous
    • 4 years ago
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    thanks unless you have a belligerent way to do it phi

  8. phi
    • 4 years ago
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    did you plug your answer back into the original equation?

  9. ash2326
    • 4 years ago
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    phi you're right it won't work, we'll have to include higher order terms

  10. anonymous
    • 4 years ago
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    not yet, I just wanted to get an idea of where to go before I worked on it. It seems reasonable if I do an longer expansion

  11. anonymous
    • 4 years ago
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    I just didn't think of doing the expansion in the first place

  12. anonymous
    • 4 years ago
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    rather couldn't

  13. anonymous
    • 4 years ago
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    I've got to roll, but thanks for the help guys

  14. ash2326
    • 4 years ago
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    Yeah daomowon , now you include the cubic power, find t and substitute back, if it doesn't then include higher power

  15. mathmate
    • 4 years ago
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    Do you mean no computers and no wolfram? A problem like this without calculators would take a while!

  16. phi
    • 4 years ago
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    It's not clear what the constraints are on how to do this problem. But generally, you would solve it using numerical techniques. such as http://en.wikipedia.org/wiki/Newton's_method see attached pdf for result

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  17. mathmate
    • 4 years ago
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    I would approach this using numerical methods, like Newton's method. We should note that the middle (e^(-t/5) term is insignificant when t>0, which gives the first approximation as x=545/49=11.1. It converge to10.5116 in a few more iterations. When t<0, both terms come into play, but e^(-t/5) has more influence. It should do the same towards -6.

  18. anonymous
    • 4 years ago
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    yeah, I figure if I can get at these types of things without computers and stuff, I can get at wider selection of material better in the long run. thanks for all the insight though, I couldn't make the leap I needed in the beginning, but this all helps a whole lot

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