ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: 2007vce on November 15, 2007, 10:12:29 pm
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HOW DO U DO THESE QUESTIONSS??? ITS HARDDD
(http://img91.imageshack.us/img91/8589/mav07ay2.jpg)
(http://img91.imageshack.us/img91/4087/tsfxup4.jpg)
ALSO!!!!!!!!!!!
when do u use R=MA --> (x)i+(y)j = ma (the vector crap)
and when do u just resolve, resolve horizontally, resolve vertically
???? I DUNNO, WHEN U USE WHICH ONE???? ANY???
THANKSSSSSSSS
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A point of inflection is when a double derivative crosses the x-axis. In other words, when double derivative is zero and dy/dx is not zero (so that there is no turning point, and this ensures the double derivative crosses the x-axis).
This is painful to the eye, and I'm on the phone. I'll get back to this later.
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f'' = second derivative
f' = first derivative
where f'' is positive it is said that the curve is concave up
where f'' is negative it is said that the curve is concave down
An inflection point is a point on the curve where the curvature (concavity) changes.
Therefore, at the inflection point, f'' changes sign
If f'' changes sign (from either +ve to -ve or vice versa), and if f'' is continuous, then by the intermediate value theorem, you know that f'' crosses the x axis. Let x = a be where f'' crosses the x-axis. Thus f''(a) = 0. Therefore, f' does not change
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f'' = second derivative
f' = first derivative
where f'' is positive it is said that the curve is concave up
where f'' is negative it is said that the curve is concave down
An inflection point is a point on the curve where the curvature (concavity) changes.
Therefore, at the inflection point, f'' changes sign
If f'' changes sign (from either +ve to -ve or vice versa), and if f is continuous, then by the intermediate value theorem, you know that f'' crosses the x axis. Let x = a be where f'' crosses the x-axis. Thus f''(a) = 0. Therefore, f' changes sign at x = a
Don't confuse her with your university first semester Maths knowledge! My explanation does not go into the gory details, but it's all you need to know for Specialist (and any further probably is too hard to understand). Not good when your exam is tomorrow, haha.
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The more general case for a point of inflection is when the double derivative changes sign. Brendan highlighted (on MSN) the counter-example to what I said originally with: y = x^(1/3)
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If f'' changes sign (from either +ve to -ve or vice versa), and if f is continuous, then by the intermediate value theorem, you know that f'' crosses the x axis. Let x = a be where f'' crosses the x-axis. Thus f''(a) = 0. Therefore, f' does not change
I find this a bit weird. You need f'' to be continuous not f to apply the intermediate value theorem, by virtue of the fact that f'' exists f is continuous.
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f''[0] = 0, but clearly f'
if f'' is zero, then that implies that at that point f' is not changing.
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f''[0] = 0, but clearly f'
if f'' is zero, then that implies that at that point f' is not changing.
Yeah but you're checking for sign changes on the sides of f'.
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POI discussion has come up a few times on here already. I've already written about it lol. :x
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Heh, it's a maths-debate. :lol: (Had to say it)
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f''[0] = 0, but clearly f'
if f'' is zero, then that implies that at that point f' is not changing.
Yeah but you're checking for sign changes on the sides of f'.
but you don't need f' to change sign for a point of inflection.
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Heh, it's a maths-debate. :lol: (Had to say it)
Healthy discussion, not debate!
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Heh, it's a maths-debate. :lol: (Had to say it)
Healthy discussion, not debate!
Wanking is healthy.
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I thought you were trying to show that dy/dx does not change sign.
Anyway here is how I'd do it:
f''[a] = 0 and f'' changes sign at a. Therefore f'[a] is either a minimum or maximum and therefore does not change sign.
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I thought you were trying to show that dy/dx does not change sign.
Anyway here is how I'd do it:
f''[a] = 0 and f'' changes sign at a. Therefore f'[a] is either a minimum or maximum and therefore does not change sign.
what about the function f(x) = x^(1/3)
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What about it?
Edit: Thinking about it now, I guess it depends on the definition you take as a POI.
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Q21 the projectile is moving up so resistance and weight act down on the object.
Will this exam be difficult and long? I'm scared what happen to methods ppl will happen to us. Who is doing multiple choice first?
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f'' to be continuous not f to apply the intermediate value theorem, by virtue of the fact that f'' exists f is continuous.
yeah i wud concur with that.
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you should really know all this.
That ski question is a slope, of course its covered in spesh. mass parallel to plane is mgsintheta, and the normal reaction is mgcostheta.
in 22, you take direction as negative because its ascending so you need to change teh value to keep the formulas working.
and its negative g resistance because they say in hte question that its velocity upwards, therefore accelleration is negative.
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Yeah, the theorem is proven by observing that f([a,b]) is connected because the image of a connected set under a continuous function is connected, where f([a,b]) denotes the image of the interval [a,b] under the function f. C is between f(a) and f(b) so its in connected set.