View Full Version : Spot the error
Markus
04-14-2013, 08:38 PM
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).
http://www.texify.com/img/%5Cnormalsize%5C%21%5Cint_%7B0%7D%5E%5Cpi%20%5Csin %5E3%28%5Cphi%29%20d%5Cphi.gif
http://www.texify.com/img/%5Cnormalsize%5C%21%5Cint_%7B0%7D%5E%5Cpi%20%281-%5Ccos%5E2%28%5Cphi%29%29%5Csin%28%5Cphi%29%20d%5C phi.gif
Let http://www.texify.com/img/%5Cnormalsize%5C%211-%5Ccos%5E2%28%5Cphi%29.gif be u, then http://www.texify.com/img/%5Cnormalsize%5C%21du%20%3D%202%5Csin%28%5Cphi%29. gif
So the integral becomes http://www.texify.com/img/%5Cnormalsize%5C%21%5Cfrac%7B1%7D%7B4%7D%28%7B1-cos%5E2%28%5Cphi%29%7D%29%5E2%7C%5E%7B%5Cpi%7D_%7B 0%7D.gif 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.
Nebula
04-14-2013, 09:18 PM
I don't know calculus but I do know that 1-cos²=sin². Maybe that helps?
tealc
04-14-2013, 09:23 PM
found this http://answers.yahoo.com/question/index?qid=20080306021501AANSlj0
you should also be able to convert coordinate systems aswell
http://i48.tinypic.com/35a557c.png
Cheeky bit of round 1 underneath as well
masterBB
04-14-2013, 10:34 PM
My attempt. Also stranded at 0. I don't understand what you did Rich. Will look into it tomorrow.
http://i.imgur.com/NTFq7Oa.png
http://i.imgur.com/J1eM3cm.png
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.
masterBB
04-14-2013, 10:48 PM
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 (http://villavu.com/forum/usertag.php?do=list&action=hash&hash=SillyMistakes) #IAmNotGoodInMath (http://villavu.com/forum/usertag.php?do=list&action=hash&hash=IAmNotGoodInMath) #IDontEvenUseTwitter (http://villavu.com/forum/usertag.php?do=list&action=hash&hash=IDontEvenUseTwitter)
found it: a' is the same as da
so it would just disappear. Also it ain't 2/3 but 1/3.
#SillyMistakes (http://villavu.com/forum/usertag.php?do=list&action=hash&hash=SillyMistakes) #IAmNotGoodInMath (http://villavu.com/forum/usertag.php?do=list&action=hash&hash=IAmNotGoodInMath) #IDontEvenUseTwitter (http://villavu.com/forum/usertag.php?do=list&action=hash&hash=IDontEvenUseTwitter)
Doesn't a'= -sin(x), or have I misunderstood?
masterBB
04-14-2013, 11:00 PM
Doesn't a'= -sin(x), or have I misunderstood?
Final calculation:
http://i.imgur.com/hSMEGtQ.png
http://i.imgur.com/YLnlVVE.png
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.
masterBB
04-14-2013, 11:32 PM
Ah fair enough, we've obviously been taught different ways to handle these sorts of questions.
I am self-taught.
how is this called: http://i.imgur.com/Qlkg9FM.png
I am self-taught.
how is this called: http://i.imgur.com/Qlkg9FM.pngOh 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
Markus
04-15-2013, 08:16 AM
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:
http://www.texify.com/img/%5Cnormalsize%5C%21%5Cfrac%7B%5Cfrac%7B1%7D%7B4%7D %281-%5Ccos%5E2%28x%29%29%5E2%7D%7Bdx%7D%20%3D%20%5Cfra c%7B1%7D%7B4%7D%5Ccdot%202%20%281-%5Ccos%5E2%28x%29%29%5Ccdot%202%5Csin%28x%29%20%3D %20%281-%5Ccos%5E2%28x%29%29%5Ccdot%5Csin%28x%29%20%3D%20% 5Csin%5E3%28x%29.gif
So my question remains: why is picking a = (1-cos²x) wrong?
litoris
04-15-2013, 02:34 PM
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.
Markus
04-15-2013, 02:42 PM
YES! That's the mistake I was looking for. Thanks!
masterBB
04-15-2013, 03:11 PM
YES! That's the mistake I was looking for. Thanks!
I also was able to do it with a default integral:
http://i.imgur.com/t72zG2m.png
Enslaved
04-15-2013, 08:48 PM
:(i see that i am too late
Markus
04-16-2013, 05:59 PM
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:
http://i.imgur.com/2t3fdJ8.jpg
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?
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