Mahuta ♥ Posted February 12, 2009 Report Share Posted February 12, 2009 alright sure:sin(root of x).derivative of sinx = cos xI used the chain ruley=sinu u=rootx= x^(1/2)y'= cos u u'= (1/2)x^(-1/2)y'= cosu * (1/2)x^(-1/2) =(1/2)cosx^(-1/2) =[cosx^(-1/2)] /2 Reply Link to post Share on other sites More sharing options...
Hyperbole Posted February 12, 2009 Report Share Posted February 12, 2009 But aren't you missing the extra [x^(-1/2)] there? And shouldn't the "u" thing remain the same in the derivative as it is in the original function (ie. positive)? I'm not entirely sure (calculus is hard! ) but I would say it should becos[x^(1/2)]/[2x^(1/2)]and I keep getting it to that every time I redo the calculation. Just like the OP.As for the integration of cos^(2) x, I think that might have something to do with tanx since (tanx)'=1/cos^(2) x. But I'm definitely not sure on that one!And it's too late for me to even try to think about the third question. Woes, calculus! (But we do love it, really...) Reply Link to post Share on other sites More sharing options...
moneyfaery Posted February 12, 2009 Report Share Posted February 12, 2009 (edited) is correct.What's that complicated thing, Ash?Double angle formula:Edit: LaTex looks so nice. I'll do #3 when I get back.... Edited February 13, 2009 by Irene Reply Link to post Share on other sites More sharing options...
Ashika Posted February 13, 2009 Report Share Posted February 13, 2009 I don't know where you're getting that first bit from.Or is that some sort of proof? Reply Link to post Share on other sites More sharing options...
moneyfaery Posted February 13, 2009 Report Share Posted February 13, 2009 That's the double angle formula! Silly Ash. When's your calc exam? Reply Link to post Share on other sites More sharing options...
Ashika Posted February 13, 2009 Report Share Posted February 13, 2009 Tomorrow... and no wonder, we aren't required to know the double angle formula for this course haha. Reply Link to post Share on other sites More sharing options...
moneyfaery Posted February 13, 2009 Report Share Posted February 13, 2009 (edited) Ok, time for #3!From x=0 to x=pi/2, f(x) = cosx is the upperbound and g(x) = sinx is the lowerboundWrong answer: see next post for correct ans Spoiler - Click me! [Close]If there's a mistake, that's because 1) LaTex is a pain to use and 2) first time touching this stuff in almost a year. Going to fail math exam... Edited February 14, 2009 by Irene Reply Link to post Share on other sites More sharing options...
Abu Posted February 13, 2009 Report Share Posted February 13, 2009 Ok, time for #3!From x=0 to x=pi/2, f(x) = cosx is the upperbound and g(x) = sinx is the lowerboundIf there's a mistake, that's because 1) LaTex is a pain to use and 2) first time touching this stuff in almost a year. Going to fail math exam...How did you get from the second step to the third? I'm confoosed. Reply Link to post Share on other sites More sharing options...
Mahuta ♥ Posted February 13, 2009 Report Share Posted February 13, 2009 WOW! I seriously suck!!!This makes no sense to me, lol. My Math mock is after tomorrow.and irene, its good for someone who hasnt done that in a year. lol. You rock Reply Link to post Share on other sites More sharing options...
moneyfaery Posted February 13, 2009 Report Share Posted February 13, 2009 (edited) How did you get from the second step to the third? I'm confoosed.That's because I made a ****load of errors. In fact, f(x)=cos(x) isn't even the upperbound because it's only the upperbound from x=0 to x=pi/4. WHOOPS.Although, I guess you could find the volume from x=0 to x=pi/4 then multiply that by 2 to find the total volume. Anyway, that whole thing up there is wrong so ignore it. Answer looks nice so more probable. Aboo or someone else check it please. Edited February 13, 2009 by Irene Reply Link to post Share on other sites More sharing options...
SharkSpider Posted February 14, 2009 Report Share Posted February 14, 2009 You happen to be rather correct, Irene, though you could just have moved everything in to 2*cos(x)^2 - 1 before starting. Reply Link to post Share on other sites More sharing options...
moneyfaery Posted February 14, 2009 Report Share Posted February 14, 2009 (edited) You happen to be rather correct, Irene, though you could just have moved everything in to 2*cos(x)^2 - 1 before starting.LaTex is a pain to type! So I cut that part out. That and I wanted to show separately where the integrals of cos2(x) and sin2(x) came from.Double angle formulas if anyone's really interested. Edited February 14, 2009 by Irene Reply Link to post Share on other sites More sharing options...
Ongfufu Posted February 16, 2009 Report Share Posted February 16, 2009 If you want softwares on your computer or calculator... I suggest you forget about that. First, you don't have you computer with you during tests. Second, they will clear your calculator before exams. And third, I find it pretty confusing and time consuming trying to enter equations into computers and calculators. Just use your brain! Reply Link to post Share on other sites More sharing options...
ezex Posted February 17, 2009 Report Share Posted February 17, 2009 This post has been so cool . Nice job explaining everything Irene, I don't think there could be a better way to explain than how you did. One thing though. For number 3 you're right that cosine is only upperbound from 0 to pi over four. Unfortunately you cant just multiply that area by two to get the second part. You actually have to integrate, from 0 to pi over four, cosine squared minus sine squared PLUS integration, from pi over four to pi over two, of sine squared - cosine squared.And by the way, now that we're using really cool formulas and proofs, cosine squared minus sine squared is cos(2x) so really the integration is just sin(2x)/2 No need to go and integrate individually And sine squared minus cosine squared is -cos(2x)...oh wow, you CAN multiply the volumes by two, oh well that was just lucky . Oher than that great explanation. Reply Link to post Share on other sites More sharing options...
moneyfaery Posted February 17, 2009 Report Share Posted February 17, 2009 It wasn't entirely luck haha. If you look at a sine/cosine graph, you see that the area bound by the functions from x=0 to x=pi/4 equals the area from x=pi/4 to x=pi/2. If these areas are then rotated about the x-axis, then it makes sense that the volumes of revolution are equal.Spoiler - Click me! [Close]I just thought it'd be easier to multiply by 2 since LaTex is a pain to use! Whoaaaaa, I found a sine/cosine dragon. Spoiler - Click me! [Close] Reply Link to post Share on other sites More sharing options...
luk3tm Posted February 26, 2009 Report Share Posted February 26, 2009 I have merged all the topics with questions on MATH SL into this.-MahaHi im doing math SL and recently we have covered the second derivitaive in finding stationary points after already being able to do the same thing without the second derivativeMore specifically, we found whether the points where maximum or minimum or points of inflection by plugging in numbers (ie if both sides are negative then it is an inflection, if one side is positive and the next negative then it is a maximum)After this we learnt that if the second derivative < 0 then the point is a maximum and >0 it is a minimum. I was just wondering, what is the point in learning this if we can just the first method without the second derivative to find maximums, minimums and points of inflection it just seems like a bit of a waste of time.... can anyone explain? Is it just to check or something? Reply Link to post Share on other sites More sharing options...
Mahuta ♥ Posted February 26, 2009 Report Share Posted February 26, 2009 Second derivative, when finding stationary points are much more straight foreward and you have almost no chance in going wrong.In the IBquestion, you are usually asked to find f''(x) (second derivative) then asked to find a stationary point, it was delibertly put in that order so that you can use the2nd derivative to immediatly come up with the answer.I personally find it easier, as i can just find the second derivative put the values x and see if its negative or positive, or zero.According to my teacher, your first way is just a preparation for the second. And if you have time in the exam, yuo can use it as a checking.I have done alot of quizzes on this and my mock, and always found that second derivative usage is much easier. In fact, you dont need to use anything for paper 2, just use your GDC. Reply Link to post Share on other sites More sharing options...
Hyperbole Posted February 26, 2009 Report Share Posted February 26, 2009 There is also the thing where you may end up putting in values that are too far from the stationary points when you do it with the first derivative and the slopes. I did that once: we had a narrow parabola in a polynomial, but no illustration of it, and I just took the x value for the stationary point plus and minus one. It didn't work...I sort of fail at explaining it right now (yay mocks), but maybe you get the picture? As long as the second derivative is either negative or positive, you can be certain that you have the right result. Reply Link to post Share on other sites More sharing options...
Tarz Posted March 7, 2009 Report Share Posted March 7, 2009 I have a calculus test on Tuesday. I've basically not done anything for the past 1 month. What's the best way to learn integration really fast? The teacher only gave us photocopies of some questions. He never gave us the book because he didn't have enough of them. Is there some software that can test my integration skills or teach me how to do it? Reply Link to post Share on other sites More sharing options...
moneyfaery Posted March 7, 2009 Report Share Posted March 7, 2009 You can differentiate, yes? Then just learning to recognize the integrals and practicing. Do all those questions first and also learn substitution and integration by parts. Reply Link to post Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.