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

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ash2326
 4 years ago
Best ResponseYou've already chosen the best response.1if 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)=1t/5+t^2/50 substitute this in the equation 49t+245(1t/5+t^2/50)545=0 49t+24549t+4.9t^2545=0 combining like terms 4.9t^2=300 \[t=\sqrt{\frac{300}{4.9}}\] \[t=7.8246\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0cool man, series expansion, thanks

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.1phi we don't know about that, so I just assumed it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it would be if the premise is that t is small to begin with, in that case t/5 would be smaller

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but it is that assumption

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks unless you have a belligerent way to do it phi

phi
 4 years ago
Best ResponseYou've already chosen the best response.1did you plug your answer back into the original equation?

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.1phi you're right it won't work, we'll have to include higher order terms

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I just didn't think of doing the expansion in the first place

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I've got to roll, but thanks for the help guys

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.1Yeah daomowon , now you include the cubic power, find t and substitute back, if it doesn't then include higher power

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.0Do you mean no computers and no wolfram? A problem like this without calculators would take a while!

phi
 4 years ago
Best ResponseYou've already chosen the best response.1It'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

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, 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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