Gloamy's Thread of Kweshchuns

[Martoman]

yeah this one is pretty simple with simple manipulation.



=

Now for that annoying sin term.




Now that will get rid of our sin term and we can take the -2 factor outside.



=

=

so

edit: +c will be the end of me. *sigh* I should learn to add water.

[Gloamglozer]

Ah, thank you /0 and Martoman.  Thanks for the two alternatives.  I can't believe I forgot that .  Silly me, basic Methods and I've failed. 

I've got another question.  This one is pretty general though so no working, just explanation:

With partial fractions, just say I had:



As we cannot factorise the denominator fully, does that mean we have to complete the square to become:



?  Then it will become:



right?

[Ilovemathsmeth]

I got the same answer as Martoman but I used a different method. Though I think his is easier

For your next one Gloamglozer, I did that. You're supposed to change the numerator so that you have a multiple of the derivative in the denominator. So you can write it as:

(3x - 6 + 1)/(x^2 - 4x + 5)

And then go (3x - 6)/(x^2 - 4x + 5) + 1/(x^2 - 4x + 5)

Then you can say u = x^2 - 4x + 5, and du/dx = 2x - 4, and 3x - 6 = 3/2 * du/dx. Use substitution from there. And for the second part of your fraction, complete the square in the denominator to give 1/[(x - 2)^2 + 1] which when integrated gives you arctan ( x - 2 ). Hope that helps

[superflya]

I got the same answer as Martoman but I used a different method. Though I think his is easier

For your next one Gloamglozer, I did that. You're supposed to change the numerator so that you have a multiple of the derivative in the denominator. So you can write it as:

(3x - 6 + 1)/(x^2 - 4x + 5)

And then go (3x - 6)/(x^2 - 4x + 5) + 1/(x^2 - 4x + 5)

Then you can say u = x^2 - 4x + 5, and du/dx = 2x - 4, and 3x - 6 = 3/2 * du/dx. Use substitution from there. And for the second part of your fraction, complete the square in the denominator to give 1/[(x - 2)^2 + 1] which when integrated gives you arctan ( x - 2 ). Hope that helps

ahaha rage

[Martoman]

I got the same answer as Martoman but I used a different method. Though I think his is easier

For your next one Gloamglozer, I did that. You're supposed to change the numerator so that you have a multiple of the derivative in the denominator. So you can write it as:

(3x - 6 + 1)/(x^2 - 4x + 5)

And then go (3x - 6)/(x^2 - 4x + 5) + 1/(x^2 - 4x + 5)

Then you can say u = x^2 - 4x + 5, and du/dx = 2x - 4, and 3x - 6 = 3/2 * du/dx. Use substitution from there. And for the second part of your fraction, complete the square in the denominator to give 1/[(x - 2)^2 + 1] which when integrated gives you arctan ( x - 2 ). Hope that helps

"her"  :smitten:

With the partial fractions question with an irreducible quadratic, then there is a rule for that which tells you what to put on A and B, google is your friend here.

[Ilovemathsmeth]

Sorry Martoman - just your name, I subconsciously thought it was a guy

Really, what rule is that??

[Martoman]

Most of which can be found here.

http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/partialfracdirectory/PartialFrac.html

Go down to the partial fractions which are irreducible. Basically, if the denominator is then numerator has

[dcc]

I asked my tutor about it and he said that UoM doesn't like matty's method.  Huh?

protip from uom - in second year, they don't care if you write statements like or something else which doesn't make sense.  But that's because you're supposed to know how to do it the 'hard way' by second year anyway.

edit:
oh and just a personal thing - I hate the questions they give in VCE (indefinite integration - seriously?).  Something like is infinitely more useful.  p.s. peter taylor rules

[/0]

they even encourage it, separation of variables lol

[Gloamglozer]

Can I please get some clarification if I'm doing things right here?  Just clarification so far:

1.  Consider these three curves:







Find the volume of the solid generated by rotating this region with about the y-axis.

So far, I have:





To find volume:









Or is there an easier way?

[Martoman]

This seems like the most suitable way. Take out the pi tho

[Ilovemathsmeth]

I don't get why the 4 wasn't squared too? I think I'm a bit lost, why didn't he do 4^2 - ((10-6)^0.5 + 4)^2

[Gloamglozer]

This seems like the most suitable way. Take out the pi tho

Yeah thanks anyway, I did that in the end, after realising that it would be easier a few hours later after the last post lol.  Fail, eh?

I don't get why the 4 wasn't squared too? I think I'm a bit lost, why didn't he do 4^2 - ((10-6)^0.5 + 4)^2

Do you mean the 4 in:



?

[Ilovemathsmeth]

nope i mean why didnt you use washers? i used washers :S. i thought the "solid" required was that formed outside the cone shape in the middle. the q was ambiguous.

[Martoman]

oooo you do it the intuitive way. *high five* What Gloamy is doing is exactly that. Formally.