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March 29, 2024, 08:17:32 pm

Author Topic: VCE Methods Question Thread!  (Read 4803319 times)  Share 

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pugs

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Re: VCE Methods Question Thread!
« Reply #18030 on: July 24, 2019, 09:48:13 am »
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how do you find the period of 2sin(2x) + 3cos(3x), and hence any 'addition of trig function' (with sin & cos) graph?

thanks!!


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LachlanKarslake

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Re: VCE Methods Question Thread!
« Reply #18031 on: July 24, 2019, 12:03:31 pm »
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how do you find the period of 2sin(2x) + 3cos(3x), and hence any 'addition of trig function' (with sin & cos) graph?

thanks!!


To find the period of a function containing more than one trig term, this is one option:

1. Find the period of each term,
ie. 2sin(2x) is π and 3cos(3x) is 2π/3 (from the period formula 2π/n, sin(nx) or cos(nx))

Converting into degrees will make the lowest common denominator (LCD) calculation easier
2. Convert each result to degrees,
ie. 180° and 120° respectively (from the radian to degree formula 180x/π where x is the radian angle)

Step 3 and 4 is finding the lowest common denominator between the degree angles
3. Decompose each degree angle into its prime parts,
ie.
180 --> 18*10 --> 9*2*5*2 --> 2*2*3*3*5
120 --> 12 * 10 --> 4*3*5*2--> 2*2*2*3*5

4. Group these primes together, removing sequences that contain less of a given number.
ie. 2*2 vs. 2*2*2 (choose 2*2*2 as it contains more 2's)
3 vs. 3*3 (choose 3*3 as it contains more 3's)

The end result will be the group 2*2*2*3*3*5, which equals 360°

Optional: If you are required to have the degrees in radians
5. Convert the degree result back into radians, (using the degree to radian formula πx/180, where x is the degree angle)
ie. 360° becomes 2π radians

6. This result, 2π or 360° is your period of the function: 2sin(2x) + 3cos(3x)

For some intuition, you can think of it like this:
If I count from 1 to 3 and simutanously a buddy counts 4-5. The pattern will repeat (the period) after the 6th count (2*3 from the LCD process)
1 2 3 1 2 3 1
4 5 4 5 4 5 4
This is essentially the same concept.

Hope this helps  ;D

AlphaZero

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Re: VCE Methods Question Thread!
« Reply #18032 on: July 24, 2019, 02:52:15 pm »
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how do you find the period of 2sin(2x) + 3cos(3x), and hence any 'addition of trig function' (with sin & cos) graph?

thanks!!

I believe you don't have to do this without a graph. So, just sketch it in your CAS and simply observe the period.

For your own interest though, to do this by hand, the period of a sum of sinusoidal functions is essentially the lowest common multiple of the periods of the individual functions, provided the resulting function is periodic. For example the function,  \(g(x)=\sin(x/3)-\cos(2x)\)  has period  \(\text{lcm}(6\pi,\,\pi)=6\pi\).

Beware though that it is possible to add sinusoidal functions so that the resulting function is not periodic. For example, take  \(f(x)=\sin(\sqrt{2}\,x)+\sin(x)\),  where taking a lowest common multiple really makes no sense since that would imply that \(\sqrt{2}\in\mathbb{Q}\), which is obviously absurd.
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Srd2000

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Re: VCE Methods Question Thread!
« Reply #18033 on: July 27, 2019, 09:38:45 pm »
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Hey, hope everyone's studies are going well!

Quick transformations question; "a function, \[f(x)=e^{x}\]  undergoes a series of transformations such that it is now equal to \[f(x)=e^{-\frac{1}{2}(x-8)}-1\] . List these transformations".

This is my attempt, please explain if and where it is wrong.
Translated 8 units right
Translated 1 unit down
Reflected about y-axis
Dilated by 1/2 in the x-axis

Thank you very much!!!

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Remy33

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Re: VCE Methods Question Thread!
« Reply #18034 on: July 27, 2019, 11:08:19 pm »
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Hey, hope everyone's studies are going well!

Quick transformations question; "a function, \[f(x)=e^{x}\]  undergoes a series of transformations such that it is now equal to \[f(x)=e^{-\frac{1}{2}(x-8)}-1\] . List these transformations".

This is my attempt, please explain if and where it is wrong.
Translated 8 units right
Translated 1 unit down
Reflected about y-axis
Dilated by 1/2 in the x-axis

Thank you very much!!!

I thought you have to follow the DRT order? This is what I got:

1 - Dilate by a factor of 1/2 in the x-axis
2 - Reflected in the y-axis
3 - Translated 4 units units in the negative direction of the x-axis
4 - Translated 1 unit in the positive direction of the y-axis

Not sure if I'm correct but I'd love someone to explain the right way to go about this question too, thanks!
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redpanda83

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Re: VCE Methods Question Thread!
« Reply #18035 on: July 28, 2019, 12:27:40 am »
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I thought you have to follow the DRT order? This is what I got:

1 - Dilate by a factor of 1/2 in the x-axis
2 - Reflected in the y-axis
3 - Translated 4 units units in the negative direction of the x-axis
4 - Translated 1 unit in the positive direction of the y-axis

Not sure if I'm correct but I'd love someone to explain the right way to go about this question too, thanks!
Yes as you said transformations must be applied in DRT order to achieve the desired result.
1* - Dialation of factor of 2 parallel to x-axis or from the y-axis. (factor of 1/n)
3* - translation of  8 units in the positive direction of x axis. (flip the sign)
4* - translation of 1 unit in the negative direction of y-axis. (u dont flip the sign here)
for your visual check the assignment. As you can see the effect of transformation on the e^x.  ;D
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Srd2000

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Re: VCE Methods Question Thread!
« Reply #18036 on: July 28, 2019, 12:32:25 pm »
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Thanks for that! That is making more sense. But what about the reflection in the y-axis?
Shouldn't it be:
Dilation of 2 from y-axis
Reflection in y-axis
Translated 8 units right
Translated 1 unit down

Thank you!
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Re: VCE Methods Question Thread!
« Reply #18037 on: July 28, 2019, 12:42:24 pm »
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Thanks for that! That is making more sense. But what about the reflection in the y-axis?
Shouldn't it be:
Dilation of 2 from y-axis
Reflection in y-axis
Translated 8 units right
Translated 1 unit down

Thank you!

Yes you are right. I am pretty sure, using the DRT method for graphs it should be stated as the following.

1. Dilated by a factor of 2. (As redpanda mentioned, a factor of 1/n so 1 divided 1/2 will give 2)
2. Reflected in the y-axis, since the exponent is now the opposite  sign.
3. Translated 8 units to the from left to right (or the positive direction).
4. Translated 1 unit down.

The important thing is, do you understand how we transformed the graph?(like why it is translated 1 unit down, why it is reflected in the y-axis etc.?)

redpanda83

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Re: VCE Methods Question Thread!
« Reply #18038 on: July 28, 2019, 03:12:50 pm »
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But what about the reflection in the y-axis?

ye reflection is there, i just fixed the ones that were wrong. hence star next number. :3

The important thing is, do you understand how we transformed the graph?(like why it is translated 1 unit down, why it is reflected in the y-axis etc.?)

I agree, always think back to what is actually happening to the graph. It will start to make much more sense when you do. It doesnt matter what the function look, once you see it (can sketch/visualise it), every question is the same, doesnt matter how hard it is.
Try playing around with your calculator, or even use desmos (online graphing website, pretty good) to see different functions and affect of the transformations.
You can even try to see f(x) = g(x) solutions , and how they are affected by transforming both graphs. Always helped me!
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S_R_K

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Re: VCE Methods Question Thread!
« Reply #18039 on: July 28, 2019, 04:17:56 pm »
+1
Perhaps the best method for transformations is to use mapping (ie. write equations relating coordinates of points on the old graph to coordinates of points on the new graph). Note that transformations do not necessarily have to be written in the "DRT order". Having an inflexible approach to transformations is not beneficial in the long run because it hinders understanding.

For the example given above, y = e^((-1/2)(x – 8 )) is an image of y = e^x under a transformation (x, y) -> (x', y') if and only if y = y' + 1 and x = –(1/2)(x' – 8 ). Rearranging gives y' = y – 1 and x' = –2x + 8 = –2(x – 4 ).

So the sequence of transformations could be written:
Dilation from the y-axis by a factor of 2, followed by a reflection in the y-axis, followed by a translation by 8 units right, followed by a translation 1 unit down.

Or it could be written:
Translation by 4 units left, followed by a dilation from the y-axis by a factor of 2, followed by a reflection in the y-axis, followed by a translation 1 unit down.

JR_StudyEd

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Re: VCE Methods Question Thread!
« Reply #18040 on: July 28, 2019, 09:13:12 pm »
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Can someone explain what a sign diagram is?
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milanander

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Re: VCE Methods Question Thread!
« Reply #18041 on: July 28, 2019, 09:37:11 pm »
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Can someone explain what a sign diagram is?

Pretty sure it's the diagram that you draw where + indicates that particular part of the graph is above the x-axis (i.e. positive) and - means that particular part of the graph is below the x-axis (i.e. negative), so you can get a sense of what the graph would look like even without really graphing it out.
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redpanda83

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Re: VCE Methods Question Thread!
« Reply #18042 on: July 28, 2019, 09:41:13 pm »
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Can someone explain what a sign diagram is?
so its pretty much like a cartesian plane, but instead of drawing the graph you just use + and - signs to show if the graph is above or below the axis.
It can also be used to simplify derivative function, and identify type of stationary points. So like a positive gradient, 0, negative gradient, / -  \ (using linear lines or + - signs what ever you prefer). From the shape it looks like local maximum. inflection point will have -ve  0 -ve or +ve 0 +ve.
In summary simplified version of curve in terms of symbol
Hope that helps
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pugs

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Re: VCE Methods Question Thread!
« Reply #18043 on: July 31, 2019, 09:44:00 pm »
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hi guys, i'm not sure if anyone else has experienced this, but i'm having trouble using the solve function on my calculator with equations involving n & loge(x)

i've been trying to use the solve function on my calculator to find the answer to
(0.8 )n + n x 0.8(n−1) × 0.2 < 0.05
^^ that is supposed to be 0 point 8 btw, pls ignore the gap! idk how to pull up the maths writing tool

ideally, it would give me a decimal answer, but i'm receiving 5n - 5 x (n + 4) x 4n > 0 instead

is there any way to fix this/is there something i'm doing wrong? i really don't want this to happen during my sac/exams haha

thanks!!
« Last Edit: July 31, 2019, 09:47:39 pm by pugs »


2019 vce journal here

redpanda83

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Re: VCE Methods Question Thread!
« Reply #18044 on: July 31, 2019, 09:49:24 pm »
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hi guys, i'm not sure if anyone else has experienced this, but i'm having trouble using the solve function on my calculator with equations involving n & loge(x)

i've been trying to use the solve function on my calculator to find the answer to
(0.8 )n + n x 0.8(n−1) × 0.2 < 0.05
^^ that is supposed to be 0 point 8 btw, pls ignore the gap! idk how to pull up the maths writing tool

ideally, it would give me a decimal answer, but i'm receiving 5n - 5 x (n + 4) x 4n > 0 instead

is there any way to fix this/is there something i'm doing wrong? i really don't want this to happen during my sac/exams haha

thanks!!
wats x and wats multiply sign, i am blind.
Uhh try to put multiply sign in the middle of variables x and n. usually this will solve the problem as calc considers xn as 1 variable sometimes.
solve ((0.8)^n +0.2nx(0.8)^n-1  <0.05,x)      [n times x]
so i got x>(0.2(1.25)^n)/n - 4/n    , n<0
       and x< (0.2(1.25)^n)/n - 4/n   , n>0
« Last Edit: July 31, 2019, 09:55:14 pm by redpanda83 »
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