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January 24, 2020, 08:25:49 pm

### AuthorTopic: VCE Specialist 3/4 Question Thread!  (Read 1171817 times) Tweet Share

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#### Ansaki

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9495 on: June 16, 2019, 03:47:10 pm »
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I edited your LaTeX code a bit to make it look nicer. It takes time to learn, but don't get hung up on it. It's more important that you understand maths first

Here's a starting hint: $\int_{\pi/2}^\pi \sin^3(x)\cos^2(x)\,dx=\int_{\pi/2}^\pi \sin(x)\big(1-\cos^2(x)\big)\cos^2(x)\,dx=\int_{\pi/2}^\pi \sin(x)\big(\!\cos^2(x)-\cos^4(x)\big)dx$ What substitution could you make?

Edit: undefined beat me to it lol

Split Sin^3(x) into Sin(x) Sin^2(x) then use the pythagorean identity to turn sin^2(x) into 1-cos^2(x). Let u=cos(x) which cancels out the remaining sin(x) and you should get an easy to evaluate integral.

hey guys! sorry for the late reply but thanks so much helped me out massively!

#### Tatlidil

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9496 on: June 16, 2019, 07:11:05 pm »
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Whats up AN!
So I have a few questions...(I have a SAC tomorrow, part 1 is vectors/vector calc)
How do I find linear dependency/independency
How do I find perpendicular vectors
Let me know if you need a question

#### DBA-144

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9497 on: June 16, 2019, 08:17:55 pm »
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Whats up AN!
So I have a few questions...(I have a SAC tomorrow, part 1 is vectors/vector calc)
How do I find linear dependency/independency
How do I find perpendicular vectors
Let me know if you need a question

Linear dependency just use the definition A (m) + B (n) = C where A, B and C are vectors as I don't know latex. the definition should be in your textbook. Essentially, you are just showing that you can multiply the two vectors (not necessarily by the same factor) and add them to form the third vector.

In 2 dimensions, 3 vectors will always be linearly dependent and for the 3 dimensions, 4 vectors will always be linearly dependent. We need to use the above definitions when we can't be sure and need to check.

The dot product is a.b= |a| |b| cos theta. Perpendicular --> 90 degree angle at intersection --> a.b = |a| |b| x 0 = 0. that is, perpendicular vectors are such that a.b=0. THe idea here is that cos 90=0, by the way.

Are there any questions specifically that you had?

#### Tatlidil

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9498 on: June 16, 2019, 10:44:14 pm »
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Linear dependency just use the definition A (m) + B (n) = C where A, B and C are vectors as I don't know latex. the definition should be in your textbook. Essentially, you are just showing that you can multiply the two vectors (not necessarily by the same factor) and add them to form the third vector.

In 2 dimensions, 3 vectors will always be linearly dependent and for the 3 dimensions, 4 vectors will always be linearly dependent. We need to use the above definitions when we can't be sure and need to check.

The dot product is a.b= |a| |b| cos theta. Perpendicular --> 90 degree angle at intersection --> a.b = |a| |b| x 0 = 0. that is, perpendicular vectors are such that a.b=0. THe idea here is that cos 90=0, by the way.

Are there any questions specifically that you had?
How do you show independency?

#### S_R_K

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9499 on: June 17, 2019, 05:46:45 pm »
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How do you show independency?

To show that vectors a, b, c are linearly independent, show that it is not possible to find real numbers m, n such that c = ma + nb. This will generally involve finding a set of simultaneous equations in m and n, and showing that there is no simultaneous solution.

#### AlphaZero

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9500 on: June 18, 2019, 08:38:12 pm »
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To show that vectors a, b, c are linearly independent, show that it is not possible to find real numbers m, n such that c = ma + nb. This will generally involve finding a set of simultaneous equations in m and n, and showing that there is no simultaneous solution.

I'm in the middle of exams right now, so I'll keep this brief, but this actually is not true and is a common misconception among Specialist Maths students. I'll be writing a concise guide on linear dependence/independence in the near future.

In a nutshell though, the correct statement relating linear combinations of 3 vectors $(\mathbf{a},\mathbf{b},\mathbf{c})$ and linear independence/dependence is: $\exists \,m,n\in\mathbb{R}\setminus\{0\}\ \ \text{s.t.}\ \ \mathbf{c}=m\mathbf{a}+n\mathbf{b}\ \implies\ \mathbf{a},\mathbf{b},\mathbf{c}\ \text{are linearly dependent vectors}$ Simply negating both sides of the statement doesn't make it equivalent. The following is an incorrect statement: $\forall\,m,n\in\mathbb{R}\setminus\{0\},\ \ \mathbf{c}\neq m\mathbf{a}+n\mathbf{b}\ \implies\ \mathbf{a},\mathbf{b},\mathbf{c}\ \text{are linearly independent vectors}$A simple counterexample is shown by the vectors $\mathbf{a}=2\mathbf{i}-\mathbf{j}+\mathbf{k},\ \ \mathbf{b}=-4\mathbf{i}+2\mathbf{j}-2\mathbf{k},\ \ \mathbf{c}=3\mathbf{i}+2\mathbf{j}+\mathbf{k}$ It can very easily be shown that  $\mathbf{c}\neq m\mathbf{a}+n\mathbf{b}\ \ \forall\,m,n\in\mathbb{R}$  yet its obvious that  $\mathbf{a},\mathbf{b},\mathbf{c}$  are linearly dependent since $2\mathbf{a}+\mathbf{b}+0\mathbf{c}=\mathbf{0}.$
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#### S_R_K

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9501 on: June 19, 2019, 06:07:20 pm »
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Thanks AlphaZero, fair enough. I should have said that a set of vectors are linearly independent "if, for each vector in the set, it is not possible to write it as a linear combination (where not all scalars are zero) of the others" without choosing a specific linear combination. In the example you've given, while it's not possible to write c as a linear combination of a and b, it is possible to write it a as a linear combination of b and c -> hence, linearly dependent.

#### AlphaZero

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9502 on: June 19, 2019, 08:47:20 pm »
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Thanks AlphaZero, fair enough. I should have said that a set of vectors are linearly independent "if, for each vector in the set, it is not possible to write it as a linear combination (where not all scalars are zero) of the others" without choosing a specific linear combination. In the example you've given, while it's not possible to write c as a linear combination of a and b, it is possible to write it a as a linear combination of b and c -> hence, linearly dependent.

Indeed, but I think it's very tedious showing that none of the vectors can be written as a linear combination of the rest. For a set of 3 vectors, that would involve solving simultaneous equations 3 times.

To prove linear independence, it's probably best to go back to a definition. Guide coming soon
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#### Sine

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9503 on: July 01, 2019, 09:10:11 pm »
+1
For applications of differential equations, do we need to know the logistic equation and will we be required to memorise formulas such as Newton's Law of Cooling or will they be given to us?

Edit: I just realised that Newton's Law of Cooling is not particularly difficult to memorise but what about other equations?
I think Newton's law of cooling was on the very old study designs formula sheets - as for your question, no, you won't need to memorise any specific differential equation - they will give you what is required.
If you want to make sure of this - ask your teacher or have a look at the study design.

#### AlphaZero

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9504 on: July 01, 2019, 09:22:55 pm »
+1
For applications of differential equations, do we need to know the logistic equation and will we be required to memorise formulas such as Newton's Law of Cooling or will they be given to us?

Edit: I just realised that Newton's Law of Cooling is not particularly difficult to memorise but what about other equations?

You do not need to memorise any differential equations for specific types of models (eg: Newton's Law of Cooling, the Logistic equation, etc.).

You do however need to be able to read a description of a model and form a differential equation from the information given (eg: in mixing problems, etc.

Otherwise, you'll be given the differential equation (eg: in beam theory problems)
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#### Antae

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9505 on: July 02, 2019, 09:54:53 pm »
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You do not need to memorise any differential equations for specific types of models (eg: Newton's Law of Cooling, the Logistic equation, etc.).

You do however need to be able to read a description of a model and form a differential equation from the information given (eg: in mixing problems, etc.

Otherwise, you'll be given the differential equation (eg: in beam theory problems)
I think Newton's law of cooling was on the very old study designs formula sheets - as for your question, no, you won't need to memorise any specific differential equation - they will give you what is required.
If you want to make sure of this - ask your teacher or have a look at the study design.

Thanks

#### DBA-144

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9506 on: July 05, 2019, 03:19:26 pm »
+3
Could someone explain why the domain of the solution curve of the differential equation dy/dx=y^2 is restricted to x<2? I understand why the solution curve is y=1/(2-x) but not why x<2.

I have attached an image of the worked solutions. Thanks as always

The domain of the solution curve has to be continuous and also needs to include the initial condition. This is because the solution curves tell us what the values of x and y make the given differential equation 'true'.

Now, consider the case of what you have said here. The function 1/2-x is undefined for x=2. So, our domain can be either x<2 or x>2. The initial condition, y(1)=1, lies in the domain x<2. therefore, the domain is x<2.

Hope this helps.

#### S_R_K

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9507 on: July 05, 2019, 03:36:08 pm »
+2
Thanks DBA-144. Yes, unfortunately the theory of existence / uniqueness of solutions to DEs is brushed under the rug in Spesh.

Just to add to DBA-144's response, one intuitive way to understand why we exclude x > 2 from the domain of the solution is because otherwise we lose uniqueness of the solution. Notice that any function of the form y = 1/(x – 2) + c (for a real constant c) has a derivative of y^2 when x > 2. So if we include x > 2 in the domain, we would have say that the solution to the DE (with that initial condition) is any function of the form: y = 1/(x – 2), when x < 2 and y = 1/(x – 2) + c, when x > 2.

#### S_R_K

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9508 on: July 09, 2019, 12:12:57 pm »
+1
I'm getting some unintelligible behaviour from my CAS (TI-nspire) when solving equations simultaneously.

I was doing Question 4(c), from Section B, from 2018 NHT Exam 2. (Found here).

The method to answer this is straightforward enough: Set the horizontal component of the ball's position equal to 4.5, and the vertical component of the ball's position equal to 1.25, and solve simultaneously for V and t. However, when I attempted to do this by CAS (which surely is required for this question) I got weird results (screenshots attached below).

When I enter the equations exactly as written (ie. use "t" for time and "V" for speed) CAS tells me there is no solution. However, when I swap the labels and use "t" for speed and "V" for time, CAS gives me the desired answer.

Things I have tried, but do not make any difference (ie. the same thing happens, where swapping labels makes a difference for whether CAS finds a solution).

• using "x" and "y", rather than "t" and "v"
• solving for {v,t} rather than {t,v}
• including / omitting a domain restriction
• swapping the positions of the "t" and "v" coefficients of sin and cos
• writing the RHSs as decimals rather than fractions

This occurs on both the handheld and the computer software.

I have no idea what is going on.

#### S_R_K

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9509 on: July 09, 2019, 01:16:29 pm »
+1
I tried what you did on my CAS and it also doesn't work. However, solving it using symbolab.com does give the correct answer of v=7.8. The exact answer is quite complex which suggests it may be a limitation of the CAS, although I'm not sure why it wouldn't give an approximate answer.

Alternatively, you could graph the two relations on your CAS and find the intersection point which I think would be ok for this particular question as it only asks for the answer correct to one decimal point.

Edit: I think I found a better method. Instead of solving system of equations, just use normal solve (Menu,3,1). Type the two equations and separate them with the word 'and', and solve for v and t.

Thanks. With respect to your method of using a normal solve, I got the same result. Using "t" for time and "v" for speed gave no solution; swapping the labels gives the desired solution.

Yes, the approximate solution can be found by using the intersection tool on a graph page (but what if an exact answer is required?). And I'm sure there are other workarounds. My concern is with how predictable this behaviour is, especially given that this is from a VCAA exam. I would not want a student to use the simultaneous equations solver, be told there is "no solution" and then start wasting time / freaking out when they've done everything absolutely correctly.

I guess my question is whether there is something that we should know about how the TI-nSpire treats variables so that this issue can be avoided / predicted.