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March 29, 2024, 04:37:38 am

Author Topic: 3U Maths Question Thread  (Read 1230331 times)  Share 

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annabeljxde

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Re: 3U Maths Question Thread
« Reply #4095 on: June 08, 2019, 05:10:01 pm »
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Hey there,

You haven't attached question 13! But as for question 19C there are a few approaches. The easiest way I can think of is to try expressing the displacement in terms of a single trigonometric function, and here it's going to be sine or cosine. Then, differentiate the function with respect to t, and the coefficient will be your maximum velocity since the range of sine or cosine is -1<=f(x)<=1. Give it a go! If you don't understand or haven't used the method before, ask again :)

Hope this helps :)
 
Answer
On inspection, the max velocity should be (but tell me if it's wrong!! might be wrong :( )


Thanks for answer! And yes, it is the right answer!
Ah, it seems like I accidentally attached the same question twice  :o. I'll re-post that question with this post.

If you don't mind, could you explain that qu. 13(b) and a new question attached, 21(c) for Simple Harmonic Motion? 

Thank you once again!
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RuiAce

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Re: 3U Maths Question Thread
« Reply #4096 on: June 08, 2019, 05:15:21 pm »
+3
Thanks for answer! And yes, it is the right answer!
Ah, it seems like I accidentally attached the same question twice  :o. I'll re-post that question with this post.

If you don't mind, could you explain that qu. 13(b) and a new question attached, 21(c) for Simple Harmonic Motion? 

Thank you once again!

21c is the same as 19c but with different values. The answer will just be \(n \times \sqrt{2^2+1^2} \) copying what he said.

Hints: With 13c, you know that the centre of motion is at the origin \(x=0\), so you should consider using \( \boxed{v^2 = n^2(a^2-x^2)} \). The amplitude can be found by using your answer from the previous part (i.e. the endpoints), and the value for \(n^2\) is already provided because \( \frac{d^2x}{dt^2}=-n^2x \) for SHM.

annabeljxde

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Re: 3U Maths Question Thread
« Reply #4097 on: June 08, 2019, 05:39:54 pm »
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21c is the same as 19c but with different values. The answer will just be \(n \times \sqrt{2^2+1^2} \) copying what he said.

Hints: With 13c, you know that the centre of motion is at the origin \(x=0\), so you should consider using \( \boxed{v^2 = n^2(a^2-x^2)} \). The amplitude can be found by using your answer from the previous part (i.e. the endpoints), and the value for \(n^2\) is already provided because \( \frac{d^2x}{dt^2}=-n^2x \) for SHM.

ahh.. I get it now, thanks heaps! I realised that I was also using the wrong formula to get velocity from the given acceleration (in qu. 13) and was just simply integrating instead of using d/dx(1/2 v^2), completely ignoring the fact that acceleration is in terms of x, not t!

Thank you!!
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Shaynell01

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Re: 3U Maths Question Thread
« Reply #4098 on: June 10, 2019, 05:26:45 pm »
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Can someone please help me put this into a suitable form to integrate?

richardzzz

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Re: 3U Maths Question Thread
« Reply #4099 on: June 10, 2019, 05:35:46 pm »
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what on earth is that - is that 4 unit ? sorry I can't help

richardzzz

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Re: 3U Maths Question Thread
« Reply #4100 on: June 10, 2019, 05:36:51 pm »
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do u maybe make it pi^2 sinx

RuiAce

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Re: 3U Maths Question Thread
« Reply #4101 on: June 10, 2019, 06:03:34 pm »
+1
Can someone please help me put this into a suitable form to integrate?

If you're intending to say \( \int \cos^{-1}\pi x \sin \pi x\,dx \), i.e. inverse cos of (pi*x) multiplied to sin(pi*x), that is not doable.
Hey guys! I have 3 binomial proof questions that I'm having trouble with. Any help at all would be appreciated!!
Considering you've posted multiple, potentially considerably long questions, can you please show any attempts/progress you have made with any of them? Or pick one for us to focus on first?
« Last Edit: June 10, 2019, 06:31:08 pm by RuiAce »

kaustubh.patel

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Re: 3U Maths Question Thread
« Reply #4102 on: June 11, 2019, 01:47:26 am »
+1
Hey people, some help with a few hard trigonometry questions.
1. simplify
sin(A+B)cosB - cos(A+B)sinA

When i first saw this que, i thought it would be simple cause I'd just need to work back and write it as sin(A+B-A) but the cosB and sinA really troubles this up for me. I tried expanding it, but only lead to a serious mess. Plz help


RuiAce

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Re: 3U Maths Question Thread
« Reply #4103 on: June 11, 2019, 10:29:39 pm »
+1
Hey people, some help with a few hard trigonometry questions.
1. simplify
sin(A+B)cosB - cos(A+B)sinA

When i first saw this que, i thought it would be simple cause I'd just need to work back and write it as sin(A+B-A) but the cosB and sinA really troubles this up for me. I tried expanding it, but only lead to a serious mess. Plz help


It doesn't seem like any product-to-sum formulas get us anywhere with this either to be honest.

Where did you get the question from? Looks extremely out of place for a 3U question.

spnmox

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Re: 3U Maths Question Thread
« Reply #4104 on: June 11, 2019, 10:41:05 pm »
+1
Hey, I'm having trouble with these general solution questions and I'm not sure why.

sinx+cosx=1

I squared both sides and ended up with sin2x=-1, then I wasn't really sure about my next stops because the reference angle is either not acute or negative. But I guessed and did x=-pi/4. So then I got x=-npi + (-1)^n*pi/4. However there are actually 2 answers. Help please?

(sqr3 cosx)-sinx=1

I did the auxiliary angle method and got sin(x-pi/3) = 1/2, so x-pi/3 = npi + (-1)^n*pi/6, and then I added pi/3 to both sides to get x=4npi/3 + (-1)^n*pi/6. However the answer is 2pin +- pi/6. Not sure how to get that.

RuiAce

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Re: 3U Maths Question Thread
« Reply #4105 on: June 11, 2019, 11:03:43 pm »
+2
Hey, I'm having trouble with these general solution questions and I'm not sure why.

sinx+cosx=1

I squared both sides and ended up with sin2x=-1, then I wasn't really sure about my next stops because the reference angle is either not acute or negative. But I guessed and did x=-pi/4. So then I got x=-npi + (-1)^n*pi/4. However there are actually 2 answers. Help please?

(sqr3 cosx)-sinx=1

I did the auxiliary angle method and got sin(x-pi/3) = 1/2, so x-pi/3 = npi + (-1)^n*pi/6, and then I added pi/3 to both sides to get x=4npi/3 + (-1)^n*pi/6. However the answer is 2pin +- pi/6. Not sure how to get that.
To resolve the problem with the first question firstly, note that the general solution to \( \sin x = \alpha\) is \( x=n\pi + (-1)^n\sin^{-1}\alpha\).

\(\sin^{-1}\alpha\) takes values between \( -\frac\pi2 \leq \sin^{-1}\alpha \leq \frac\pi2\). This is due to the range of the function \(f(x) = \sin^{-1}x\). So therefore when dealing with the general solution for \(\sin\) with negative RHS's, negative angles are valid. Therefore solving \( \sin 2x = -1 \) would give \( \boxed{2x = n\pi + (-1)^n \left( -\frac{\pi}{2} \right)} \).

Which then becomes \(x = \frac{n\pi}{2} + (-1)^{n+1} \frac\pi4 \). Which is (possibly surprisingly) wrong.

The dangerous thing you did was "square both sides". It may seem surprising, but squaring (especially in the context of trigonometric equations) typically introduces new solutions to the equations that we didn't have before. Your general solution now has the values \(x = 0, \pm \frac\pi2, \pm \frac{2\pi}{2}, \pm \frac{3\pi}{2}, \pm \frac{4\pi}{2}, \pm \frac{5\pi}{2}, \pm \frac{6\pi}{2}\) and so on, when the original question only has the solutions \( x=0, \pm \frac\pi2, \frac{4\pi}{2}, \frac{5\pi}{2},\) etc.

It is recommended to use the auxiliary angle method for the first one as well, and deal with \( \sqrt{2} \sin \left( x + \frac\pi4 \right) = 1 \) instead.
____________________

For your second one, you should've obtained \( \sin x - \sqrt{3}\cos x = -1 \). This would've given you \(2\sin \left( x - \frac\pi3 \right) = -1 \), and hence \(x - \frac\pi3 = n\pi + (-1)^n \left( -\frac\pi6\right) \).

Which of course, does not look like what the answers had. The thing is, your approach (once you fix the sign error) is equally valid. Basically, they chose to use the other auxiliary angle formula instead:
\[ a\cos x + b\sin x = R\cos (x-\alpha)\\ \text{where }R=\sqrt{a^2+b^2}\text{ and }\tan \alpha = \frac{b}{a}. \]
If you try doing it that way instead, you'll find your answer matches their's. In the exam, either they will specify which to use, or both will be marked correct.

spnmox

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Re: 3U Maths Question Thread
« Reply #4106 on: June 11, 2019, 11:27:58 pm »
+1
To resolve the problem with the first question firstly, note that the general solution to \( \sin x = \alpha\) is \( x=n\pi + (-1)^n\sin^{-1}\alpha\).

\(\sin^{-1}\alpha\) takes values between \( -\frac\pi2 \leq \sin^{-1}\alpha \leq \frac\pi2\). This is due to the range of the function \(f(x) = \sin^{-1}x\). So therefore when dealing with the general solution for \(\sin\) with negative RHS's, negative angles are valid. Therefore solving \( \sin 2x = -1 \) would give \( \boxed{2x = n\pi + (-1)^n \left( -\frac{\pi}{2} \right)} \).

Which then becomes \(x = \frac{n\pi}{2} + (-1)^{n+1} \frac\pi4 \). Which is (possibly surprisingly) wrong.

The dangerous thing you did was "square both sides". It may seem surprising, but squaring (especially in the context of trigonometric equations) typically introduces new solutions to the equations that we didn't have before. Your general solution now has the values \(x = 0, \pm \frac\pi2, \pm \frac{2\pi}{2}, \pm \frac{3\pi}{2}, \pm \frac{4\pi}{2}, \pm \frac{5\pi}{2}, \pm \frac{6\pi}{2}\) and so on, when the original question only has the solutions \( x=0, \pm \frac\pi2, \frac{4\pi}{2}, \frac{5\pi}{2},\) etc.

It is recommended to use the auxiliary angle method for the first one as well, and deal with \( \sqrt{2} \sin \left( x + \frac\pi4 \right) = 1 \) instead.
____________________

For your second one, you should've obtained \( \sin x - \sqrt{3}\cos x = -1 \). This would've given you \(2\sin \left( x - \frac\pi3 \right) = -1 \), and hence \(x - \frac\pi3 = n\pi + (-1)^n \left( -\frac\pi6\right) \).

Which of course, does not look like what the answers had. The thing is, your approach (once you fix the sign error) is equally valid. Basically, they chose to use the other auxiliary angle formula instead:
\[ a\cos x + b\sin x = R\cos (x-\alpha)\\ \text{where }R=\sqrt{a^2+b^2}\text{ and }\tan \alpha = \frac{b}{a}. \]
If you try doing it that way instead, you'll find your answer matches their's. In the exam, either they will specify which to use, or both will be marked correct.

Thank you for the answers!!

For the first question: I solved it using auxiliary angle method but got 3npi/4 + (1)^n *pi/4 when the answer is given as x=pi n + (-1)^n * pi/2, 2pin

For the second question, am I not allowed to use rin(x-alpha) ?

Shaynell01

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Re: 3U Maths Question Thread
« Reply #4107 on: June 12, 2019, 02:30:38 pm »
+1
If you're intending to say \( \int \cos^{-1}\pi x \sin \pi x\,dx \), i.e. inverse cos of (pi*x) multiplied to sin(pi*x), that is not doable.Considering you've posted multiple, potentially considerably long questions, can you please show any attempts/progress you have made with any of them? Or pick one for us to focus on first?
Yes Rui, that is what I was trying to say. My teacher says it is possible, and we have to do it for an assessment task. Any idea how to re-arrange it to a integratable expression, I would suppose you have the restrict the domain when you express your final answer?

RuiAce

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Re: 3U Maths Question Thread
« Reply #4108 on: June 12, 2019, 04:01:39 pm »
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Yes Rui, that is what I was trying to say. My teacher says it is possible, and we have to do it for an assessment task. Any idea how to re-arrange it to a integratable expression, I would suppose you have the restrict the domain when you express your final answer?
I will have to disagree as it stands. The function has no elementary anti-derivative. (To be secure about it I double checked on an integration calculator.)

With some rigged boundaries it may be computable as a definite integral. But finding a primitive function is not possible. If your teacher claims it’s doable, it’s either not of an elementary form, or I would be suspecting a mistake if anything.

Youssefh_

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Re: 3U Maths Question Thread
« Reply #4109 on: June 13, 2019, 12:39:58 pm »
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I just need a little assistance, I need an example equation of a Parabola with Two Real and Distinct Roots,