Epsilon-delta definition

[Spartacus]

Don't feel bad, so do I.

now that was good

 

[samoth]

now that was good

 

[Yarg!]

hey samoth or courtney can u guys solve this for me please??
two ships leave port at the same time. one sails south at 15mi/h and the other sails east at 20m/h. find a function that models the distance D between ships in terms of the time t (in hrs) elasped since their departure. thanks.

 

[courtneybcca]

Okay, first of all proof will require to show that your δ that you arrive at works when put back in the equation.

You found a number δ such that

|(-x³+1)+7|<ε whenever 0<|x-2|< δ

Since |(-x³+1)+7|=|(-x³+8)|= |(x-2)||(x²+2x+4)|,

You want |(x-2)||(x²+2x+4)|<ε when 0<|x-2|< δ

This gives you |x-2|< ε/|(x²+2x+4)| when 0<|x-2|< δ

So you propose that δ= ε/|(x²+2x+4)|

For the proof, show that this specific value of δ works.
Given ε>0, choose δ= ε/|(x²+2x+4)|

Then if 0<|x-2|< δ,

|(-x³+1)+7|=|(-x³+8)|= |(x-2)||(x²+2x+4)| < |(x²+2x+4)| δ

=(|(x²+2x+4)|) (ε/|(x²+2x+4)|) = ε


SO

|(-x³+1)+7|<ε whenever 0<|x-2|< δ

Thus by the definition of the limit,
Lim (-x³+1) = -7
x→2

Q.E.D.



Lemme know if I made a mistake or somethin'. I only used numerical methods and algebreic manipulations, I didn't graph anything. But I don't think answering it should be dependent on that, since the proof won't be.

Okay so wow I totally understand the way you did that... But that is way more advanced then we have done in are class yet... Why would are teacher give us something like this that we haven't done yet.. That pisses me off any way though thank you so much!!! Do you happen to have msn or yahoo messenger... If so could you please pm me with the address....
Thank you so much

 

[Spartacus]

are should be our
go fuck yourself

 

[samoth]

Okay so wow I totally understand the way you did that... But that is way more advanced then we have done in are class yet... Why would are teacher give us something like this that we haven't done yet.. That pisses me off any way though thank you so much!!! Do you happen to have msn or yahoo messenger... If so could you please pm me with the address....
Thank you so much

It happens quite often in more areas than just calc. Sometimes things are "beyone the scope of the class", which I've been hearin' for the past 4 years, lol, and getting too deeply into something just confuses *most* people or raises too many questions. Or the prof/TA feels that a certain way of progression through the material would be better.

Also, some profs and some texts just plain suck. That's why I never go by just one source of course material anymore.

For the proof I wrote, I used the exact equation given in Stewart's Calculus Early Transcnedentals 4E (section 2.4 pg.115). It was only a matter of plug and chug. You might want to check out some supplementary materials for the course. Or you guys may just do things differently, and your prof/TA doesn't think some things are relevant or would just bog most people down.

I've never used MSN or Yahoo Messenger in my life, but I can be reached at Purdue#NoSpam.Hushmail.com if you would like. I'm sure you can figure the addy out. If not, PM me.

 

[Yarg!]

guys plz help me out anytime youre ready is fine. i fucking hate these word problems. im glad im not at ur guys lvl though.

 

[samoth]

guys plz help me out anytime youre ready is fine. i fucking hate these word problems. im glad im not at ur guys lvl though.

You have speed and direction: you're dealing with vectors. Draw the problem out. Draw the metric between the heads of the two vectors -- this is your distance. It varies as a function of time with respect to the two ships.

Key: Think Pathagorem's triangle. D(t)=sqrt(v1(t)^2 + v2(t)^2). Remember to label your axes correctly, as you're working in the cartesian coordiate system.

 

[Spartacus]

cut it out samoth
get an erection
rub it out
that's my formula