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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: shinny on October 13, 2008, 12:27:47 pm

Title: Differentiating circular functions with modulus
Post by: shinny on October 13, 2008, 12:27:47 pm: The Kilbaha 2006 Exam 2 has a question where you need to differentiate the position vector (I'll simplify things to get to my point);
$r(t)=2e^{-\frac{7}{2}t}|cos(\frac{17\pi t}{10})|$
and to find the velocity vector, they simply used product rule and had the differential of $|cos(\frac{17\pi t}{10})|$ be $-\frac{17\pi}{10}\times|sin(\frac{17\pi t}{10})|$
Isn't that wrong? Because if you graph just say $|cos(x)|$ and try drawing the derivative of that, it definitely isn't just $-|sin(x)|$ . Looks more like $-sin(x)$ with the graph reflected across the x-axis every interval of $(\frac{\pi}{2}+2k\pi, \frac{3\pi}{2}+2k\pi)$ i.e. When the original function is below 0; which logically makes sense.
So firstly, are they wrong/am I right; and secondly, how the hell would you differentiate something like that to get a position vector...hybrid functions with conditions involving intervals separated by $2k\pi$ ? Or was this just a stupid question by Kilbaha that I shouldn't worry about...
Title: Re: Differentiating circular functions with modulus
Post by: Mao on October 13, 2008, 12:36:22 pm: $\frac{d}{dx} \left|\cos (kt)\right| = -k\sin (kt) \cdot \frac{\cos (kt)}{|\cos (kt)|}$

looking at the quadrants, in Q1 cos is positive, sin is positive, hence our derivative is also negative
in Q2 cos is negative, sin is positive, hence our derivative is also positive
in Q3 cos is negative, sin is negative, hence derivative is negative
in Q4 cos is positive, sin is negative, hence derivative is positive

which is really a sine function with the second and third quadrant inverted (which you have found and understand already)

if thats what Kilbaha said, they are definitely wrong here.

and in a MM thread, I think we've come across this for differentiating a modulus function:

$|x|=\sqrt{x^2}$

$\frac{d}{dx} |x| = \frac{d}{dx}\left(x^2\right)^{1/2} = \frac{1}{2}\cdot 2x \cdot \left(x^2\right)^{-1/2} = \frac{x}{|x|}$

$\frac{d}{dx} |f(x)| = f'(x)\cdot \frac{f(x)}{|f(x)|}$
Title: Re: Differentiating circular functions with modulus
Post by: shinny on October 13, 2008, 12:42:07 pm: aaaand...how am I meant to derive that? o_O

EDIT: oh there we go, dw. So is that a standard rule I could just chuck in straight away or would I need to derive that? Never seen it before...
Title: Re: Differentiating circular functions with modulus
Post by: Mao on October 13, 2008, 12:52:10 pm: I doubt you'll have to use that, but yeah, it's fairly standard.

You might have seen it expressed with the "sgn" (sign) function, which is the same. read up on wikipedia =]
Title: Re: Differentiating circular functions with modulus
Post by: shinny on October 13, 2008, 12:53:28 pm: Yeh I know how sign works, but its just that I've never seen much modulus theory in spesh nor methods in school or in textbooks...there isn't really much emphasis on it at this level.
Title: Re: Differentiating circular functions with modulus
Post by: Mao on October 13, 2008, 12:56:57 pm: Quote from: shinjitsuzx on October 13, 2008, 12:53:28 pm
Yeh I know how sign works, but its just that I've never seen much modulus theory in spesh nor methods in school or in textbooks...there isn't really much emphasis on it at this level.

me neither, but such is kilbaha.