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

Question

ash2326 · Answer

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
$$t=\sqrt{\frac{300}{4.9}}$$
$$t=7.8246$$

anonymous · Answer

cool man, series expansion, thanks

phi · Answer

but t/5 is not << 1

ash2326 · Answer

phi we don't know about that, so I just assumed it

anonymous · Answer

it would be if the premise is that t is small to begin with, in that case t/5 would be smaller

anonymous · Answer

but it is that assumption

anonymous · Answer

thanks unless you have a belligerent way to do it phi

phi · Answer

did you plug your answer back into the original equation?

ash2326 · Answer

phi you're right it won't work, we'll have to include higher order terms

anonymous · Answer

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

anonymous · Answer

I just didn't think of doing the expansion in the first place

anonymous · Answer

rather couldn't

anonymous · Answer

I've got to roll, but thanks for the help guys

ash2326 · Answer

Yeah daomowon , now you include the cubic power, find t and substitute back, if it doesn't then include higher power

mathmate · Answer

Do you mean no computers and no wolfram?
A problem like this without calculators would take a while!

phi · Answer

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

mathmate · Answer

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.

anonymous · Answer

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