SRL Archive - Mathematics

Hiya!
Working on some stupid calculus test and I'm making a very stupid mistake. Unfortunately, I can't find the error (I do know the right way which is to chose the cosine as u but I want to know why my method is wrong as I'll never be smart enough to actually use that).
Let be u, then So the integral becomes which turns out to be zero, yay! Real answer: 4/3.
My error most likely is at the part where I pick u, but I don't know WHY it is wrong :p Anyone who can take a minute to explain please?

sin isn't capitalised. Chicks love the capitalised Sin ;)

I managed 2/3. Give me a minute and I'll have another shot.

EDIT: Got it. I'll upload a pic of my working.

I don't know calculus but I do know that 1-cos²=sin². Maybe that helps?

found this http://answers.yahoo.com/question/index?qid=20080306021501AANSlj0

you should also be able to convert coordinate systems aswell

Cheeky bit of round 1 underneath as well

My attempt. Also stranded at 0. I don't understand what you did Rich. Will look into it tomorrow.

Yours all makes sense until you actually integrate your (-1+a^2)a'. You'd have to use some method of integration to do that other than by inspection/recognition because you have two functions of a multiplied together, whereas you've integrated those functions separately then multiplied them.

From: Rich
Yours all makes sense until you actually integrate your (-1+a^2)a'. You'd have to use some method of integration to do that other than by inspection/recognition because you have two functions of a multiplied together, whereas you've integrated those functions separately then multiplied them.

found it: a' is the same as da

so it would just disappear. Also it ain't 2/3 but 1/3.

#SillyMistakes #IAmNotGoodInMath #IDontEvenUseTwitter

From: masterBB
found it: a' is the same as da

so it would just disappear. Also it ain't 2/3 but 1/3.

#SillyMistakes #IAmNotGoodInMath #IDontEvenUseTwitter

Doesn't a'= -sin(x), or have I misunderstood?

From: Rich
Doesn't a'= -sin(x), or have I misunderstood?

Final calculation:


e:
Except for the last 0, that has to be 4/3

Ah fair enough, we've obviously been taught different ways to handle these sorts of questions.

From: Rich
Ah fair enough, we've obviously been taught different ways to handle these sorts of questions.

I am self-taught.

how is this called:

From: masterBB
I am self-taught.

how is this called:

Oh wow. Impressive. And that's just a forumla we've been taught to use when integrating by parts. Check this out if it's new to you: http://www.mathsrevision.net/alevel/pages.php?page=19

yea, if you got the function h(x) = f(x)*g(x), then H(x) = F(x)*g(x) - ?F(x)*g'(x) dx. +c.

The lame thing is, if I actually do the deriviation it does make sense:

So my question remains: why is picking a = (1-cos²x) wrong?

From: Markus

So my question remains: why is picking a = (1-cos²x) wrong?

Because the derivative of 1-cos²(x) is 2cos(x)sin(x), which doesn't help you at all when substituting. You have to integrate by parts in this question.

YES! That's the mistake I was looking for. Thanks!

From: Markus
YES! That's the mistake I was looking for. Thanks!

I also was able to do it with a default integral:

:(i see that i am too late

Next one!
Find the work done by the force field F = (x+y)i + (x-z)j + (z-y)k in moving an object from (-1, 0, 1) to (0, -2, 3) along any smooth curve.
Hint: the correct answer is 19/2!

My (wrong) approach:

That one's beyond me. I haven't done anything to do with force fields and smooth curves, and since they're not on my syllabus I won't be doing them any time soon. Any idea @masterBB?