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July 21, 2025, 09:08:40 pm

Author Topic: Maths Methods 3/4 Help Thread 2011  (Read 126563 times)  Share 

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b^3

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Re: Maths Methods 3/4 Help Thread
« Reply #210 on: June 21, 2011, 05:34:06 pm »
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ech_93

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Re: Maths Methods 3/4 Help Thread
« Reply #211 on: June 21, 2011, 06:29:53 pm »
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Thanks for that b^3, it helps a bit.

Though, would I be correct in saying how looking at a basic sine graph, for example, you can find its derivative?

Or should i talk more about formulas, etc.
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Re: Maths Methods 3/4 Help Thread
« Reply #212 on: June 21, 2011, 06:57:14 pm »
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i'd say more the formulas/algebraic method because it is reasoned and logical, and well you can't agrue with it or find a flaw in it.
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ech_93

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Re: Maths Methods 3/4 Help Thread
« Reply #213 on: June 21, 2011, 07:50:25 pm »
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Why, according to the unit circle, is tan(x) undefined at pi/2 and 3pi/2? I understand that there are asymptots at these points, but how does the unit circle show that it is undefined?
???
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Re: Maths Methods 3/4 Help Thread
« Reply #214 on: June 21, 2011, 07:54:26 pm »
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tan(x)=sin(x)/cos(x) and at pi/2 and 3pi/2 the value of cos is 0. so anything divided by 0 is undefined, hence tan(x)=sin(x)/0 = undefined (at x=pi/2 and 3pi/2)
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Re: Maths Methods 3/4 Help Thread
« Reply #215 on: June 21, 2011, 07:57:17 pm »
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tan(theta)=y/x, this means that tan(theta) actually calculates the gradient of the line from origion to a point on the unit circle. If theta=pi/2, the line is straight up (x=0), thus gradient can not be defined as we can't write the line x=0 in y=mx+c format
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ech_93

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Re: Maths Methods 3/4 Help Thread
« Reply #216 on: June 21, 2011, 08:01:59 pm »
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Thankyou for all your replies b^3 and thanks xZero. One last question; what is the domain of tan(x)? As there are jumps in the graph due to asymptotes, i was unsure whether or not you could just say R. Or could you?
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Re: Maths Methods 3/4 Help Thread
« Reply #217 on: June 21, 2011, 08:10:19 pm »
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domain of tan(kx)=R\{} where k is a member of z (integers) this means it is defined fow all values except those with asymptotes

you could also look at it that the domain is R\{}
« Last Edit: June 21, 2011, 08:13:28 pm by b^3 »
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Re: Maths Methods 3/4 Help Thread
« Reply #218 on: June 22, 2011, 04:33:45 pm »
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Hey there,

I was just looking at this : http://math.furman.edu/~dcs/book/c3pdf/sec35.pdf
and was wondering how sin(x + h) =sin(x) cos(h) + sin(h) cos(x)...

Should I know this from methods?
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Re: Maths Methods 3/4 Help Thread
« Reply #219 on: June 22, 2011, 04:52:29 pm »
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it's just the compound angle formula, you do it in specialist but i don't think you learn it in methods. its handy though so I suggest that you learn it as it can make things easiers sometimes.
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Re: Maths Methods 3/4 Help Thread
« Reply #220 on: June 22, 2011, 04:56:57 pm »
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Thanks, I just realised that I recongnised that formula as the compound angle formula.

« Last Edit: July 09, 2011, 09:46:47 pm by ech_93 »
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exams worry me
« Reply #221 on: July 06, 2011, 10:18:32 pm »
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Worried about the exams. up if u are feeling the same! :-\
« Last Edit: July 06, 2011, 10:31:04 pm by TheArabTank »

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Re: Maths Methods 3/4 Help Thread
« Reply #222 on: July 09, 2011, 09:14:19 pm »
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Ok guys and girls, I have a multiple choice question off the 2010 VCAA exam for you. Q20.
Let f be a differentiable function defined for all real x, where
If , then is equal to
A. 2a+6
B. 10a+6
C. 20a
D. 40a
E. 50a

The answer is D. 40a
now vcaa gives this now I get the part with the 3 which gives 30a but I don't understand how the first part with the function gives 10a? I have tried to visualise it with a dilation but cannot get it to look right, getting a instead of 5a, Can someone please explain why or draw it out for me?
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Re: Maths Methods 3/4 Help Thread
« Reply #223 on: July 09, 2011, 09:26:41 pm »
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Ok guys and girls, I have a multiple choice question off the 2010 VCAA exam for you. Q20.
Let f be a differentiable function defined for all real x, where
If , then is equal to
A. 2a+6
B. 10a+6
C. 20a
D. 40a
E. 50a

The answer is D. 40a
now vcaa gives this now I get the part with the 3 which gives 30a but I don't understand how the first part with the function gives 10a? I have tried to visualise it with a dilation but cannot get it to look right, getting a instead of 5a, Can someone please explain why or draw it out for me?
I honestly am not sure about this, but wouldn't this situation only occur if f(x)=0
because if
then it is equal to
so
« Last Edit: July 09, 2011, 09:29:33 pm by tony3272 »
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Re: Maths Methods 3/4 Help Thread
« Reply #224 on: July 09, 2011, 09:28:50 pm »
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I'm sorry I don't understand what you mean. I thought it should be a aswell, then times it by 2, but they do 2*5a =10a so I'm lost.
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