ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: hifer on November 02, 2007, 07:12:27 pm
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Hi there, do we always have to put the absolute sign when antidiffing to get logs? wat if the specified limits are positive?
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i dont think in methods...its in specialist
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No if the terminals are positive you don't have to put it. However, it doesn't take anymore time, neither is it wrong if you do.
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It's just useful for things such as evaluating areas under the x-axis, when you don't know the area will come out without the absolute value signs as negatives.
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i remember there was one VCAA exam 1 paper that u had to put modulus after anti into a log to be able to draw the graph
its prob in 2006 sample exam1
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Wtf are you guys talking about *is intimidated* I don't know this stuff ><;
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Wtf are you guys talking about *is intimidated* I don't know this stuff ><;
Don't worry yourself too much.
Basically when you integrate a function with yields a log, (ie. dy/dx (1/x) --> y= log_e x), you put the brackets around the (in my example) x as modulus signs.
Ie. dy/dx = ( 1/x ) --> y= log_e |x|
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Wtf are you guys talking about *is intimidated* I don't know this stuff ><;
Don't worry yourself too much.
Basically when you integrate a function with yields a log, (ie. dy/dx (1/x) --> y= log_e x), you put the brackets around the (in my example) x as modulus signs.
Ie. dy/dx = ( 1/x ) --> y= log_e |x|
Whyyy? Lol ....I'm so screwed for exams ><;
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Poor Tankia :( there there, i'm sure you'll be fine =]
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Whyyy? Lol ....I'm so screwed for exams ><;
dy/dx = 1/x, find y: for R\{0}
Since 1/x is defined for all values except 0, the antiderivative (and the derivative) should also exist for those values.
Notice that ln(x) is only defined for x>0. To correct for this, we place: ln|x| so that it is defined for R\{0}. That's a floppy argument, and what I'm about to suggest is still floppy, but if you look at ln|x| for x is less than zero, you'll notice that it is decreasing. Look at 1/x, it is the derivative of ln|x| and for values less than zero, it has negative numbers! It makes sense.
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Whyyy? Lol ....I'm so screwed for exams ><;
dy/dx = 1/x, find y: for R\{0}
Since 1/x is defined for all values except 0, the antiderivative (and the derivative) should also exist for those values.
Notice that ln(x) is only defined for x>0. To correct for this, we place: ln|x| so that it is defined for R\{0}. That's a floppy argument, and what I'm about to suggest is still floppy, but if you look at ln|x| for x is less than zero, you'll notice that it is decreasing. Look at 1/x, it is the derivative of ln|x| and for values less than zero, it has negative numbers! It makes sense.
_.... You lost me at dy/dx >_>;
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Whyyy? Lol ....I'm so screwed for exams ><;
dy/dx = 1/x, find y: for R\{0}
Since 1/x is defined for all values except 0, the antiderivative (and the derivative) should also exist for those values.
Notice that ln(x) is only defined for x>0. To correct for this, we place: ln|x| so that it is defined for R\{0}. That's a floppy argument, and what I'm about to suggest is still floppy, but if you look at ln|x| for x is less than zero, you'll notice that it is decreasing. Look at 1/x, it is the derivative of ln|x| and for values less than zero, it has negative numbers! It makes sense.
_.... You lost me at dy/dx >_>;
Tanika I'll explain it to you on msn =]
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Okay :D
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It seriously doesn't matter too much :p
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It seriously doesn't matter too much :p
LOL. I was doing my FIRST practice exam (one that i actually finished...Thank the lord for solutions :D) and i finally got what you guys were talking about...kinda...some what..maybe...maybe not...Okay...Yea..I'll go back to procrastinating :D
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When you antiderive 1/x, loge|X| has a modulus sign because :
loge(x) = y
e^y = x
theres no way that x can be negative because a positive constant (e) to the power of anything will never equal a negative.
eg. x^(1/50) = 50Rtx ... x^50... no matter which way you go it doesnt hit 0 or lower
not sure how to express what im trying to say.
e^x=y... as x --> 0, y --> 1.... and as x --> infinity, y also does
At least, thats why i thought we put a modulus in... why we only do it after antideriving im not sure, it could be something to do with the gradient of the graph going in a certain direction.
(edit: I just read what coblin wrote, lol ... i think ive just dumbed down what he said)