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December 08, 2019, 10:19:49 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 1140729 times)  Share 

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AlphaZero

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9510 on: July 09, 2019, 02:38:53 pm »
+4
I used the following syntax to find \(V\). \begin{align*} &\texttt{Define }\ \mathbf{r}(t)=\begin{bmatrix}v\cdot t\cdot \cos(a) & v\cdot t\cdot\sin(a)-\dfrac{49}{10}\cdot t^2\end{bmatrix}\\
&\texttt{solve}\left(\mathbf{r}(t)=\begin{bmatrix}\dfrac{9}{2} & \dfrac{5}{4}\end{bmatrix},v\right)\mid a=60^\circ\ \text{and}\ v>0 \end{align*}
In fact, this gives the exact answers for \(t\) and \(V\), and so I don't believe the CAS has an issue with the complexity of the solution.

However, when I change the second input to contain  \(\{v,t\}\)  or  \(\{t,v\}\),  it returns \(\texttt{false}\),  which suggests that the different arguments change the way the CAS handles solving simultaneous equations.

Thank you for finding this S_R_K and Otter. I'm going to look into this.

I'm actually quite surprised I didn't come across this when doing the exam myself.
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undefined

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9511 on: July 09, 2019, 03:05:15 pm »
+2
For comparison, I plugged in S_R_K's exact equations into Wolfram Mathematica and it gave me correct values (regardless of which variable I entered first) attached below.

The graphing approach would probably take quite a bit of time since it takes so long to move the cursor on the nspire not to mention the 2 frames per second screen (exaggerating but still).
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suskieanna

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9512 on: July 15, 2019, 12:03:19 pm »
0
Can anyone help me with this kinematics question?
I don't know why the answer to question a is like this.

if the gradient is negative, isn't it showing deceleration? But the question is saying that the object is accelerating.  :-\
Anna

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9513 on: July 15, 2019, 12:34:23 pm »
+1
Can anyone help me with this kinematics question?
I don't know why the answer to question a is like this.

if the gradient is negative, isn't it showing deceleration? But the question is saying that the object is accelerating.  :-\
(Image removed from quote.)

The question states that the object is accelerating in the opposite direction. This is probably the reason why.

AlphaZero

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9514 on: July 18, 2019, 12:26:07 am »
+2
I've attached a screenshot of the worked solutions of a textbook differential equation.

The question provided the condition that dx/dt=4 when x=0 which I think is used to find the value of 'c' when you antidiff from the second derivative to the first.

However, when you antidiff the second derivative, you only have 't' and no 'x'. The worked solution subbed x=0 into the t. Is this allowed? I've never seen this before.

It's a typo. It should say  "\(t=0\)".
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persistent_insomniac

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9515 on: July 19, 2019, 03:25:44 pm »
0
Hi!
As constant acceleration formulae are no longer part of the specialist 3/4 course, will be asked in the exam that will require us to do them using this rather than anti-differentiation? Because I'm looking at some kinematics questions from last year and I don't know how to do them using anti-differentiation?

AlphaZero

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9516 on: July 19, 2019, 04:41:26 pm »
+1
Hi!
As constant acceleration formulae are no longer part of the specialist 3/4 course, will be asked in the exam that will require us to do them using this rather than anti-differentiation? Because I'm looking at some kinematics questions from last year and I don't know how to do them using anti-differentiation?

You're right. Constant acceleration formulae are not required, but no one is stopping you if you want to use them.

Regardless of whether acceleration is constant or variable, you can always use one of these \[a=\frac{dv}{dt}=v\frac{dv}{dx}=\frac{d}{dx}\left[\frac{1}{2}v^2\right],\]and just solve the problem as a separable differential equation.
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Tau

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9517 on: August 07, 2019, 10:16:52 pm »
0
(Cambridge) 15B Question 6

Quote
Suppose that the weights of people are normally distributed with a mean of 82 kg and a standard deviation of 9 kg. What is the maximum number of people who can get into an elevator which has a weight limit of 680 kg, if we want to be at least 99% sure that the elevator does not exceed capacity?

So we know that , and we are looking for a multiple, say \(n\), of W such that \(\Pr(nW<680)=0.99\). Where \(nW\sim \mathcal N (\mu=82n, \sigma^2 = 81n^2)\). My trouble is that, on my CAS (TI Nspire CX), I can't figure out a way to numerically solve using a NormCDF function for the requisite value of \(n\). The Cambridge solutions use a guess and check method and, unless I am mistaken, their values of sigma for their guesses don't seem accurate.

I was considering using the formula for the normal distribution and using a numerical approximation to solve, but I'm not sure if that would work. Anyone have any suggestions?
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AlphaZero

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9518 on: August 08, 2019, 10:16:27 am »
+3
(Cambridge) 15B Question 6

So we know that , and we are looking for a multiple, say \(n\), of W such that \(\Pr(nW<680)=0.99\). Where \(nW\sim \mathcal N (\mu=82n, \sigma^2 = 81n^2)\). My trouble is that, on my CAS (TI Nspire CX), I can't figure out a way to numerically solve using a NormCDF function for the requisite value of \(n\). The Cambridge solutions use a guess and check method and, unless I am mistaken, their values of sigma for their guesses don't seem accurate.

I was considering using the formula for the normal distribution and using a numerical approximation to solve, but I'm not sure if that would work. Anyone have any suggestions?

Hey Tau,

In this question, we are NOT concerned with the distribution of \(nW\). This does NOT represent the distribution of the combined weight of \(n\) randomly selected persons, and actually represents the distribution of \(n\) times the weight of a single randomly selected person.

First, we define a new random variable \(U\) so that \(U=W_1+\dots+W_n\),  where \(W_i\),  \(i=1,...,n\),  are independent identically distributed (i.i.d) copies of \(W\). Then, we have \begin{align*}\text{E}(U)&=\text{E}(W_1+\dots+W_n)\\ &=\text{E}(W_1)+\dots+\text{E}(W_n)\\ &=n\text{E}(W),\end{align*}\begin{align*}\text{Var}(U)&=\text{Var}(W_1+\dots+W_n)\\ &=\text{Var}(W_1)+\dots+\text{Var}(W_n)\\&=n\text{Var}(W)\end{align*} Notice that for the variance obtained is \(n\text{Var}(W)\) instead of \(n^2\text{Var}(W)\).

Since the sum of i.i.d normal variables is still a normal variable, \[U\sim\mathcal{N}(\mu=82n,\ \sigma^2=81n).\] Thus, where \(Z\sim \mathcal{N}(0,\ 1)\), we solve the following: \begin{align*}&\Pr(U<680)\geq 0.99\\ \implies &\Pr\left(Z<\frac{680-82n}{9\sqrt{n}}\right)\geq 0.99\\ \implies &\frac{680-82n}{9\sqrt{n}}\geq 2.32635...\\ \implies &n\leq 7.58928... \end{align*}That is, maximum is \(n=7\) people.
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Tau

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9519 on: August 08, 2019, 10:28:18 am »
0
Hey Tau,

In this question, we are NOT concerned with the distribution of \(nW\). This does NOT represent the distribution of the combined weight of \(n\) randomly selected persons, and actually represents the distribution of \(n\) times the weight of a single randomly selected person.

First, we define a new random variable \(U\) so that \(U=W_1+\dots+W_n\),  where \(W_i\),  \(i=1,...,n\),  are independent identically distributed (i.i.d) copies of \(W\). Then, we have \begin{align*}\text{E}(U)&=\text{E}(W_1+\dots+W_n)\\ &=\text{E}(W_1)+\dots+\text{E}(W_n)\\ &=n\text{E}(W),\end{align*}\begin{align*}\text{Var}(U)&=\text{Var}(W_1+\dots+W_n)\\ &=\text{Var}(W_1)+\dots+\text{Var}(W_n)\\&=n\text{Var}(W)\end{align*} Notice that for the variance obtained is \(n\text{Var}(W)\) instead of \(n^2\text{Var}(W)\).

Since the sum of i.i.d normal variables is still a normal variable, \[U\sim\mathcal{N}(\mu=82n,\ \sigma^2=81n).\] Thus, where \(Z\sim \mathcal{N}(0,\ 1)\), we solve the following: \begin{align*}&\Pr(U<680)\geq 0.99\\ \implies &\Pr\left(Z<\frac{680-82n}{9\sqrt{n}}\right)\geq 0.99\\ \implies &\frac{680-82n}{9\sqrt{n}}\geq 2.32635...\\ \implies &n\leq 7.58928... \end{align*}That is, maximum is \(n=7\) people.

Thank you so much, I see the flaw in my original logic now. That’s really helpful.
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AnonymooseUser

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9520 on: August 17, 2019, 01:58:12 pm »
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Hello everyone, I feel like I'm missing something completely obvious, but I cannot understand why c=0 here.This is VCAA 2007 Exam 1 Q6a. Would really appreciate it if someone could explain.

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9521 on: August 17, 2019, 03:10:41 pm »
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Hello everyone, I feel like I'm missing something completely obvious, but I cannot understand why c=0 here.This is VCAA 2007 Exam 1 Q6a. Would really appreciate it if someone could explain.



a) To find position vector we integrate the velocity vector:



Notice that \[\mathbf{c}\] is a vector quantity, not a scalar. Substituting in the initial condition we have:

.
« Last Edit: August 17, 2019, 03:14:16 pm by Tau »
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undefined

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9522 on: August 17, 2019, 03:14:37 pm »
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Hello everyone, I feel like I'm missing something completely obvious, but I cannot understand why c=0 here.This is VCAA 2007 Exam 1 Q6a. Would really appreciate it if someone could explain.

so r(t)=2cos(2t)i+3sin(2t)j+c, where c is a vector

r(0)=2cos(0)i+3sin(0)j+c=2i
so 2i+c=2i
so c=0
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AnonymooseUser

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9523 on: August 17, 2019, 09:10:36 pm »
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r(0)=2cos(0)i+3sin(0)j+c=2i
so 2i+c=2i
so c=0



.

I just realised that I was registering cos (0) as pi/2 instead of 1... Thank you both and sorry for the dumb question.

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9524 on: August 23, 2019, 08:10:54 pm »
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Do you guys Recommend Doing Specialist 3/4 without 1/2?