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Author Topic: VCE Methods Question Thread!  (Read 4802615 times)  Share 

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The Cat In The Hat

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Re: VCE Methods Question Thread!
« Reply #18675 on: August 10, 2020, 09:42:58 pm »
0
Can someone pls confirm this: Dom of f(g(x)) = dom of g(x)
Off the top of a head none too sure about these things, I'd say that isn't right. Off the top of a head wiser than my own, that depends. If you've restricted the domain of f(x), then the domain of g(x) might not exist within that domain (if that makes sense).
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Re: VCE Methods Question Thread!
« Reply #18676 on: August 10, 2020, 09:59:37 pm »
0
Can someone pls confirm this: Dom of f(g(x)) = dom of g(x)
And how would i find the range of f(g(x)) with tech free? Would it just be the intersection of the ranges of both functions?

Yes, this is correct. The domain of f(g(x)) is the domain of g(x). However, you need to check that f(g(x)) exists using a domain and range table and seeing if the range of g(x) is a subset of the domain of f(x).

Re finding the range, I would probably use the endpoints and find turning points to distinguish where the graph is highest and lowest. I haven't seen this asked too much on tech-free but it could come up.
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schoolstudent115

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Re: VCE Methods Question Thread!
« Reply #18677 on: August 10, 2020, 10:06:04 pm »
+2
Can someone pls confirm this: Dom of f(g(x)) = dom of g(x)
And how would i find the range of f(g(x)) with tech free? Would it just be the intersection of the ranges of both functions?
That is not correct. There are two conditions for the domain:
1. x is in the domain of g(x)
2. g(x) is in the domain of f(x) (Because g(x) is the input of f(x) in f(g(x)), so g(x) must be a valid input).

So in a precise manner,
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schoolstudent115

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Re: VCE Methods Question Thread!
« Reply #18678 on: August 10, 2020, 10:34:55 pm »
+2
Can someone pls confirm this: Dom of f(g(x)) = dom of g(x)
And how would i find the range of f(g(x)) with tech free? Would it just be the intersection of the ranges of both functions?
As for the range, just think of it as the range of f(x), where the domain is the domain of f(g(x)). It is really just a restricted domain problem. Generally, your domain will not get any worse than a single set. E.g. [0,inf).

The range of f(g(x)) is a subset of the range of f(x). Reason being, they are in essence the same function, just that the domain of f(g(x)) might be restricted, so f(g(x)) will only pump out a certain portion of the range of f(x).

E.g. f(x)=x^3, g(x)=x^2
dom(f) = R
dom(g) = R
ran(g) = [0,inf)
f(g(x)) = (x^2)^3 = x^6

The range of f(g(x)) is simply the range of f(x), constrained to the fact that the input x is from 0 to infinity (ran g).
Well then the range is just [0,inf)   {x^3 is greater than or equal to 0 for x>=0}
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Coolgalbornin03Lo

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Re: VCE Methods Question Thread!
« Reply #18679 on: August 11, 2020, 05:17:09 pm »
+1
Why won’t my cas let me find the integral of sin squared x ? It keeps saying “(“ missing as in THERES a bracket missing even tho there isn’t!
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Chocolatemilkshake

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Re: VCE Methods Question Thread!
« Reply #18680 on: August 11, 2020, 05:27:47 pm »
+3
Why won’t my cas let me find the integral of sin squared x ? It keeps saying “(“ missing as in THERES a bracket missing even tho there isn’t!
You can't put the squared sign right before the sin (like you write it). Instead you need to put the squared on the outside like this...

(sin(x))^2
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james.358

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Re: VCE Methods Question Thread!
« Reply #18681 on: August 11, 2020, 05:29:08 pm »
+3
Hi El

When you want to square or cube a trig function, you have to type the bracket on the outside. For example, would be error, but would be fine.

Hope this helps!
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Edit: Beaten by chocolatemilkshake
« Last Edit: August 11, 2020, 05:31:07 pm by james.lhr »
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Re: VCE Methods Question Thread!
« Reply #18682 on: August 11, 2020, 07:31:24 pm »
0
Hey Corey!

There is indeed a much simpler way. One problem with Symbol Lab is that it "brute forces" its way to the answer. It is much easier if you do a bit of algebraic manipulation. If you have any questions with my solution don't hesitate to ask!

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Re: VCE Methods Question Thread!
« Reply #18683 on: August 11, 2020, 07:39:57 pm »
0
I'm just copying my past message because I don't want it to get lost in the past messages lol

Hello! I hope you all are doing well :)

I have a question on sinusoidal graphs, it's a small part of a larger question.
What is h if d=0 when h=0 for the equation d= 2 +/- 3sin(2π/15(t-h))

I wrote the 3 as either negative or positive as we only know that the amplitude is 3. I'm unsure whether this is enough information to find h, and I am confused since wouldn't there be an unlimited values of h? The answer states that h = 1.74202.

I would really appreciate some help!  (•◡•) /
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Re: VCE Methods Question Thread!
« Reply #18684 on: August 11, 2020, 07:47:20 pm »
+1
I'm just copying my past message because I don't want it to get lost in the past messages lol

Hello! I hope you all are doing well :)

I have a question on sinusoidal graphs, it's a small part of a larger question.
What is h if d=0 when h=0 for the equation d= 2 +/- 3sin(2π/15(t-h))

I wrote the 3 as either negative or positive as we only know that the amplitude is 3. I'm unsure whether this is enough information to find h, and I am confused since wouldn't there be an unlimited values of h? The answer states that h = 1.74202.

I would really appreciate some help!  (•◡•) /

I mean, if when d=0, h=0, then h=0? I think you've mislabeled something here.

As for whether the 3 is positive or negative, I think we're going to need to know more context. Can you give us the information from earlier parts? Eg, you've said it's +/-3 because you only know the amplitude is 3, but that's information that us on AN don't even know, so we can only trust your word which may have misinterpreted the actual question

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Re: VCE Methods Question Thread!
« Reply #18685 on: August 11, 2020, 08:04:54 pm »
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Hi,

Methods 1/2 student here. For the image that I've attached and the specific part I've highlighted, when I do that on my CAS, I get {0,0} instead of the highlighted answer. As far as I know, I've done all of the above steps the same, so if anyone could pls help me that would be much appreciated.

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Re: VCE Methods Question Thread!
« Reply #18686 on: August 12, 2020, 12:18:12 pm »
0
I mean, if when d=0, h=0, then h=0? I think you've mislabeled something here.

As for whether the 3 is positive or negative, I think we're going to need to know more context. Can you give us the information from earlier parts? Eg, you've said it's +/-3 because you only know the amplitude is 3, but that's information that us on AN don't even know, so we can only trust your word which may have misinterpreted the actual question

Ughhhh, why do I always make mistakes when writing my questions on AN??  :-[
Yep! I meant 'find h if d=0 when t=0', whoops! The actual question is much longer, but I'll include the answers from the previous questions for extra information. All I really know is that d is the distance from the water surface to point P on a water wheel with radius 3, and t is the time is seconds. Also, the wheel rotates at 4 revolutions per minute.

What is h if d=0 when t=0 for the equation d= 2 +bsin(2π/15(t-h)). For this equation, the period is 15 and the amplitude is 3. I put 'b' because we don't know the exact value, right? (It's either 3 or -3. Correct me if I'm wrong :)) I'm just confused on whether the value of h is indefinite, so any explanation on why h = 1.74 would be great  :D Thank you for your help
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keltingmeith

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Re: VCE Methods Question Thread!
« Reply #18687 on: August 12, 2020, 01:02:17 pm »
+6
Ughhhh, why do I always make mistakes when writing my questions on AN??  :-[
Yep! I meant 'find h if d=0 when t=0', whoops! The actual question is much longer, but I'll include the answers from the previous questions for extra information. All I really know is that d is the distance from the water surface to point P on a water wheel with radius 3, and t is the time is seconds. Also, the wheel rotates at 4 revolutions per minute.

What is h if d=0 when t=0 for the equation d= 2 +bsin(2π/15(t-h)). For this equation, the period is 15 and the amplitude is 3. I put 'b' because we don't know the exact value, right? (It's either 3 or -3. Correct me if I'm wrong :)) I'm just confused on whether the value of h is indefinite, so any explanation on why h = 1.74 would be great  :D Thank you for your help


All good! At least you're honest about it - I've had some students (not from AN) come to me and go, "no!! Of course I copied the question right, maybe you're just a bad tutor?!", and then when I gave them proof that the question they asked couldn't be answered, they'd come back with, "oh, turns out I copied it wrong. It's actually this" without even an apology.

Okay, so now that I've got a better understanding of what's going on, let's work with what we know. d=0 when t=0, this gives us:



Since h is a translation factor, we know that it can have an INFINITE possible number of values - but all of them should be spaced 15 seconds apart. The reason it's 15 seconds apart is because the period is 15 seconds (feel free to confirm this for yourself). So, let's find the first value of h for 0<t<15, so first we solve for h:



Okay, so this is immediately frustrating because we can't use exact values, and don't know if b is negative or positive!!. But, that's not as difficult as you think. Here, consider the two sin waves below:

https://www.desmos.com/calculator/9cvmthhh0u (note that you can turn off graphs by clicking the red or blue circle)

So, there is logic I could use to solve this. Notice that the red curve immediately goes up, while the right curve immediately goes down. That would mean that the sign of b is going to depend on whether that point P initially goes up or down. But that logic will rely on information you don't necessarily know - so instead, let's just be simple, and assume b=+3. If we do that, we get:



Which is right! Note that in the third line, we didn't need to worry about solving for a specific domain, because we only care about the first possible value that h can be in this instance. For example, let's say you were instead interested in a probability question. If I asked you what's the probability of drawing an ace, it's going to be 4/52=1/13. Now, what if I asked yo uwhat's the probability of drawing an ace if you use TWO decks? You might immediately jump on it being 8/104 - but the trick is, the answer is still 1/13. And it's going to be 1/13 no matter how many decks you add, because the amount of cards being added matches the ratio of aces being added. It's the same idea here - because sine waves are periodic, it doesn't matter which h you pick, because they won't affect what the graph looks like at the end. (and if it made sense to you before I used the card analogy, and the card analogy just confused you, simply ignore it and stick with your initial instincts)

But if that doesn't sit well with you, it shouldn't - it doesn't with me, either. So, let's see what happens if we let b=-3:



How weird - we get the exact same number, just a different sign! If you're curious, that's because sin(x) is an odd function (f(-x)=-f(x), I think this definition unfortunately got removed from the methods curriculum, but it's still kinda cool!), and doesn't really have bearing on this question. So, how do our graphs look with the h value added in?

https://www.desmos.com/calculator/sbrggivf0k

Well, the truth is, they both have d=0 when t=0, so as far as I'm concerned - both are valid answers. See your teacher for any reason as to why one of these answers isn't correct - it might be due to some information that you didn't even realise was information!

h is that value since if you shift it, it each minute it begins and ends at y = 0 after 4 rotations.

If you graph the funciton you'll notice.
(Image removed from quote.)
That at 0 seconds and 60 seconds it is at y = 0, I'm assuming that's the reason for the shift is there any part of the quesetion that states that d starts at 0?
If so, t = 0 and d = 0 that is (0, 0). There are technically infinte solutions yes. That's why people write it as a function like cos (45 + 360k) where k is an integer. But they probably chose the closest solution or something. It's basically asking 'for what value of h does the function intersect (0, 0)'

The example cos (45 + 360k) was made up btw.

You're not wrong, the problem is your answer misses the question that was being asked. In your graph, you've chosen there are two points that intersect with the x-axis - one in which the curve then starts going down, and one in which the curve starts going up. We don't know which of these points is making the intersection in Azila's answer, hence why I've explored what happens for both cases.

Azila2004

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Re: VCE Methods Question Thread!
« Reply #18688 on: August 12, 2020, 09:59:17 pm »
0
All good! At least you're honest about it - I've had some students (not from AN) come to me and go, "no!! Of course I copied the question right, maybe you're just a bad tutor?!", and then when I gave them proof that the question they asked couldn't be answered, they'd come back with, "oh, turns out I copied it wrong. It's actually this" without even an apology.

Okay, so now that I've got a better understanding of what's going on, let's work with what we know. d=0 when t=0, this gives us:



Since h is a translation factor, we know that it can have an INFINITE possible number of values - but all of them should be spaced 15 seconds apart. The reason it's 15 seconds apart is because the period is 15 seconds (feel free to confirm this for yourself). So, let's find the first value of h for 0<t<15, so first we solve for h:



Okay, so this is immediately frustrating because we can't use exact values, and don't know if b is negative or positive!!. But, that's not as difficult as you think. Here, consider the two sin waves below:

https://www.desmos.com/calculator/9cvmthhh0u (note that you can turn off graphs by clicking the red or blue circle)

So, there is logic I could use to solve this. Notice that the red curve immediately goes up, while the right curve immediately goes down. That would mean that the sign of b is going to depend on whether that point P initially goes up or down. But that logic will rely on information you don't necessarily know - so instead, let's just be simple, and assume b=+3. If we do that, we get:



Which is right! Note that in the third line, we didn't need to worry about solving for a specific domain, because we only care about the first possible value that h can be in this instance. For example, let's say you were instead interested in a probability question. If I asked you what's the probability of drawing an ace, it's going to be 4/52=1/13. Now, what if I asked yo uwhat's the probability of drawing an ace if you use TWO decks? You might immediately jump on it being 8/104 - but the trick is, the answer is still 1/13. And it's going to be 1/13 no matter how many decks you add, because the amount of cards being added matches the ratio of aces being added. It's the same idea here - because sine waves are periodic, it doesn't matter which h you pick, because they won't affect what the graph looks like at the end. (and if it made sense to you before I used the card analogy, and the card analogy just confused you, simply ignore it and stick with your initial instincts)

But if that doesn't sit well with you, it shouldn't - it doesn't with me, either. So, let's see what happens if we let b=-3:



How weird - we get the exact same number, just a different sign! If you're curious, that's because sin(x) is an odd function (f(-x)=-f(x), I think this definition unfortunately got removed from the methods curriculum, but it's still kinda cool!), and doesn't really have bearing on this question. So, how do our graphs look with the h value added in?

https://www.desmos.com/calculator/sbrggivf0k

Well, the truth is, they both have d=0 when t=0, so as far as I'm concerned - both are valid answers. See your teacher for any reason as to why one of these answers isn't correct - it might be due to some information that you didn't even realise was information!

h is that value since if you shift it, it each minute it begins and ends at y = 0 after 4 rotations.

If you graph the funciton you'll notice.
(Image removed from quote.)
That at 0 seconds and 60 seconds it is at y = 0, I'm assuming that's the reason for the shift is there any part of the quesetion that states that d starts at 0?
If so, t = 0 and d = 0 that is (0, 0). There are technically infinte solutions yes. That's why people write it as a function like cos (45 + 360k) where k is an integer. But they probably chose the closest solution or something. It's basically asking 'for what value of h does the function intersect (0, 0)'

The example cos (45 + 360k) was made up btw.

Thank you guys so much!  ;D I really appreciate the effort you put in to help, it's great!
Just someone who likes to learn a lot of questions.

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Corey King

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Re: VCE Methods Question Thread!
« Reply #18689 on: August 13, 2020, 11:28:15 am »
+1
First off, I'd like to thank James for his last response to my question :)
It helped.

Second, I have a new question regarding simultaneous equations.
I've gone wrong somewhere in the process of answering this question on the Elimination method. I can't see where.
If anyone here is willing to help I would much appreciate it :)
Many thanks,
Corey