Author Topic: VCE Specialist 3/4 Question Thread! (Read 2678687 times) Tweet Share

SocialRhubarb · « **Reply #2535 on:** October 31, 2013, 11:49:58 pm »

There are generally two methods to find minimum distances:
1. differentiating the distance expression
2. using the dot product

The dot product works because it will tell you when the velocity is perpendicular to the position vector, which in most cases in specialist maths will be the minimum distance. Why is this the case?

Think about where the minimum distance will occur. The object will be getting closer, and closer, and closer and then ... further away, and for an instant in between the two, velocity will be perpendicular to the position vector, an instant where it is neither getting further away nor closer to the origin, and the magnitude of the position vector is getting neither bigger nor smaller. Hence, we can equate the dot product of the position vector and the velocity vector to find the points where the velocity is perpendicular to the position vector, which will often correspond to minimums.

The other method is to use Pythagoras' theorem to establish an expression for the distance, and equating its derivative to 0.

$|\textbf{r}(t)|=\sqrt{(4t-20)^2+(2t)^2+(80-2t)^2}$

e^1 · « **Reply #2536 on:** November 01, 2013, 12:51:31 am »

If anyone has done question 1d)ii. for VCAA 2010 exam, is it ok to dot product like below?

$\begin{alignedat}{1} \vec{OQ}\cdot\vec{AB} &=(\frac{1}{2}\mathbf{a} +\frac{1}{2}\mathbf{b})\cdot (-\mathbf{a} +\mathbf{b}) \\ &= -\frac{1}{2} + \frac{1}{2} =0 \\ \text{Hence, } \vec{OQ} \text{ is perpendicular to } \vec{AB}\end{alignedat}$

Also, how would you do multiple choice question 22 in the same exam?

Thank you!

b^3 · « **Reply #2537 on:** November 01, 2013, 01:09:11 am »

You'd need to show $\underset{\sim}{a}.\underset{\sim}{a}=|\underset{\sim}{a}|^{2}$ and then use the fact that $|\underset{\sim}{a}|=|\underset{\sim}{b}|$ to get zero.

Spoiler

$\begin{alignedat}{1}\overrightarrow{OQ}.\overrightarrow{AB} & =\left(\frac{1}{2}\underset{\sim}{a}+\frac{1}{2}\underset{\sim}{b}\right).\left(-\underset{\sim}{a}+\underset{\sim}{b}\right) \\ & =-\frac{1}{2}\underset{\sim}{a}.\underset{\sim}{a}+\frac{1}{2}\underset{\sim}{b}.\underset{\sim}{b} \\ & =\frac{1}{2}\left(|\underset{\sim}{b}|^{2}-|\underset{\sim}{a}|^{2}\right) \\ |\underset{\sim}{a}|=|\underset{\sim}{b}|\text{ as we have an isosceles triangle} \\ \overrightarrow{OQ}.\overrightarrow{AB} & =\frac{1}{2}\left(|\underset{\sim}{b}|^{2}-|\underset{\sim}{b}|^{2}\right) \\ & =0 \\ \text{Hence \ensuremath{\overrightarrow{OQ}} is prependicular to }\overrightarrow{AB} \end{alignedat}$

Two slightly different ways (can't remember if separable ODE's are on the course or not, which is the second method).
1.
$\begin{alignedat}{1}F\left(x\right) & =ma \\ a & =\frac{F\left(x\right)}{m} \\ \frac{d\left(\frac{1}{2}v^{2}\right)}{dx} & =\frac{F\left(x\right)}{m} \\ \frac{1}{2}v^{2} & =\int_{x_{0}}^{x_{1}}\frac{F\left(x\right)}{m}dx+C_{0} \\ x=x_{0},\: C=\frac{1}{2}v_{0}^{2} \\ \frac{1}{2}v_{1}^{2} & =\int_{x_{0}}^{x_{1}}\frac{F\left(x\right)}{m}dx+\frac{1}{2}v_{0}^{2} \\ v_{1} & =\sqrt{\frac{2}{m}\int_{x_{0}}^{x_{1}}F\left(x\right)dx+v_{0}^{2}}\:\left(\text{assuming \ensuremath{v_{1}>0}}\right) \end{alignedat}$

2.

Spoiler

$\begin{alignedat}{1}F\left(x\right) & =ma \\ a & =\frac{F\left(x\right)}{m} \\ v\frac{dv}{dx} & =\frac{F\left(x\right)}{m}\text{ (Separable ODE)} \\ \int_{v_{0}}^{v_{1}}vdv & =\int_{x_{0}}^{x_{1}}\frac{F\left(x\right)}{m}dx \\ \frac{1}{2}\left(v_{1}^{2}-v_{0}^{2}\right) & =\int_{x_{0}}^{x_{1}}\frac{F\left(x\right)}{m}dx \\ v_{1}^{2} & =\frac{2}{m}\int_{x_{0}}^{x_{1}}F\left(x\right)dx+v_{0}^{2} \\ v_{1} & =\sqrt{\frac{2}{m}\int_{x_{0}}^{x_{1}}F\left(x\right)dx+v_{0}^{2}}\:\left(\text{assuming \ensuremath{v_{1}>0}}\right) \end{alignedat}$

lzxnl · « **Reply #2538 on:** November 01, 2013, 10:47:38 am »

Quote from: b^3 on November 01, 2013, 01:09:11 am

You'd need to show $\underset{\sim}{a}.\underset{\sim}{a}=|\underset{\sim}{a}|^{2}$ and then use the fact that $|\underset{\sim}{a}|=|\underset{\sim}{b}|$ to get zero.

Spoiler
$\begin{alignedat}{1}\overrightarrow{OQ}.\overrightarrow{AB} & =\left(\frac{1}{2}\underset{\sim}{a}+\frac{1}{2}\underset{\sim}{b}\right).\left(-\underset{\sim}{a}+\underset{\sim}{b}\right) \\ & =-\frac{1}{2}\underset{\sim}{a}.\underset{\sim}{a}+\frac{1}{2}\underset{\sim}{b}.\underset{\sim}{b} \\ & =\frac{1}{2}\left(|\underset{\sim}{b}|^{2}-|\underset{\sim}{a}|^{2}\right) \\ |\underset{\sim}{a}|=|\underset{\sim}{b}|\text{ as we have an isosceles triangle} \\ \overrightarrow{OQ}.\overrightarrow{AB} & =\frac{1}{2}\left(|\underset{\sim}{b}|^{2}-|\underset{\sim}{b}|^{2}\right) \\ & =0 \\ \text{Hence \ensuremath{\overrightarrow{OQ}} is prependicular to }\overrightarrow{AB} \end{alignedat}$
Two slightly different ways (can't remember if separable ODE's are on the course or not, which is the second method).
1.
$\begin{alignedat}{1}F\left(x\right) & =ma \\ a & =\frac{F\left(x\right)}{m} \\ \frac{d\left(\frac{1}{2}v^{2}\right)}{dx} & =\frac{F\left(x\right)}{m} \\ \frac{1}{2}v^{2} & =\int_{x_{0}}^{x_{1}}\frac{F\left(x\right)}{m}dx+C_{0} \\ x=x_{0},\: C=\frac{1}{2}v_{0}^{2} \\ \frac{1}{2}v_{1}^{2} & =\int_{x_{0}}^{x_{1}}\frac{F\left(x\right)}{m}dx+\frac{1}{2}v_{0}^{2} \\ v_{1} & =\sqrt{\frac{2}{m}\int_{x_{0}}^{x_{1}}F\left(x\right)dx+v_{0}^{2}}\:\left(\text{assuming \ensuremath{v_{1}>0}}\right) \end{alignedat}$

2.
Spoiler
$\begin{alignedat}{1}F\left(x\right) & =ma \\ a & =\frac{F\left(x\right)}{m} \\ v\frac{dv}{dx} & =\frac{F\left(x\right)}{m}\text{ (Separable ODE)} \\ \int_{v_{0}}^{v_{1}}vdv & =\int_{x_{0}}^{x_{1}}\frac{F\left(x\right)}{m}dx \\ \frac{1}{2}\left(v_{1}^{2}-v_{0}^{2}\right) & =\int_{x_{0}}^{x_{1}}\frac{F\left(x\right)}{m}dx \\ v_{1}^{2} & =\frac{2}{m}\int_{x_{0}}^{x_{1}}F\left(x\right)dx+v_{0}^{2} \\ v_{1} & =\sqrt{\frac{2}{m}\int_{x_{0}}^{x_{1}}F\left(x\right)dx+v_{0}^{2}}\:\left(\text{assuming \ensuremath{v_{1}>0}}\right) \end{alignedat}$

b^3, you give VCAA too much credit. Differential equations with two variables are not on the course, even though it would be easy to put on the course.

ahat · « **Reply #2539 on:** November 01, 2013, 10:56:15 am »

Hey b^3, I'm having a bit of trouble understanding your method above ^. When I first did it, I did it this way (would you/someone mind commenting?)

$a = \frac{F(x)}{m}$

$\frac{d}{dx}(\frac{1}{2}v^2) = \frac{F(x)}{m}$

Using the Numerical approximation formula

$(x_0,v_0) = (a, b)$

$\therefore \frac{1}{2}v^2 =\int_{x_{0}}^{x_{1}}\frac{F\left(t\right)}{m}dt+v_{0}$

$\therefore v^2 = \frac{2}{m} \int_{x_{0}}^{x_{1}}F(t)dx+v_{0}$

$\therefore v = \sqrt{\frac{2}{m} \int_{x_{0}}^{x_{1}}F(t)dx+v_{0}^2}$

For $v_{0}$ to be $v_{0}^2$ I sort of just assumed it must be outside of the squareroot.

Thanks (and I looked at your latex coding, it is so complicated o_O)
That second method is seriously neat btw. Wish we learnt it.

lzxnl · « **Reply #2540 on:** November 01, 2013, 11:32:24 am »

Quote from: ahat on November 01, 2013, 10:56:15 am

Hey b^3, I'm having a bit of trouble understanding your method above ^. When I first did it, I did it this way (would you/someone mind commenting?)

$a = \frac{F(x)}{m}$

$\frac{d}{dx}(\frac{1}{2}v^2) = \frac{F(x)}{m}$

Using the Numerical approximation formula

$(x_0,v_0) = (a, b)$

$\therefore \frac{1}{2}v^2 =\int_{x_{0}}^{x_{1}}\frac{F\left(t\right)}{m}dt+v_{0}$

$\therefore v^2 = \frac{2}{m} \int_{x_{0}}^{x_{1}}F(t)dx+v_{0}$

$\therefore v = \sqrt{\frac{2}{m} \int_{x_{0}}^{x_{1}}F(t)dx+v_{0}^2}$

For $v_{0}$ to be $v_{0}^2$ I sort of just assumed it must be outside of the squareroot.

Thanks (and I looked at your latex coding, it is so complicated o_O)
That second method is seriously neat btw. Wish we learnt it.

You didn't actually use a numerical approximation formula; you did the same thing as b^3 did above with finding the integration constant.

With one flaw.

$\frac{d}{dx}(\frac{v^2}{2})=\frac{F(x)}{m}$

The integration formula for the solution would have $\frac{v_0^2}{2}$ as the integration constant.

ahat · « **Reply #2541 on:** November 01, 2013, 11:52:08 am »

Quote from: nliu1995 on November 01, 2013, 11:32:24 am

You didn't actually use a numerical approximation formula; you did the same thing as b^3 did above with finding the integration constant.

With one flaw.

$\frac{d}{dx}(\frac{v^2}{2})=\frac{F(x)}{m}$

The integration formula for the solution would have $\frac{v_0^2}{2}$ as the integration constant.

Thanks, why is the constant as such?

lzxnl · « **Reply #2542 on:** November 01, 2013, 11:57:12 am »

Replace v^2/2 with y. It should be clearer then.

When you integrate d/dx (v^2/2) = F(x)/m, you need to integrate both sides from when x=x0 to the final x value; call it s.
Then, you have to integrate d/dx (v^2/2) from x=x0 to s, which is v^2/2 when x=s MINUS v^2/2 when x=x0. Note that v is a function of x. If we let v(x0) be v0, you'll get b^3's result. ETC.

Brytz · « **Reply #2543 on:** November 01, 2013, 02:39:57 pm »

2009 Q7. Can someone explain how to tackle this question, I don't know how to this particular question effectively.

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2009specmath1-w.pdf

Also, explain where each of the 4 marks of the question will likely be? I'm not sure why it's a 4 mark question.

Stevensmay · « **Reply #2544 on:** November 01, 2013, 02:45:37 pm »

Quote from: Brytz on November 01, 2013, 02:39:57 pm

2009 Q7. Can someone explain how to tackle this question, I don't know how to this particular question effectively.

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2009specmath1-w.pdf

Also, explain where each of the 4 marks of the question will likely be? I'm not sure why it's a 4 mark question.

Turns out I can't read, thought a 7 was a 6.
Start by replacing the $y \mbox{ and }\frac{dy}{dx} \mbox{ and } \frac{d^2y}{dx^2}$ with the actual expressions.

$m^2e^{mx}-3me^{mx}-10e^{mx}=0$

Factorise.

$e^{mx}(m^2-3m-10)=0$

$e^{mx} \neq 0$ so no solutions here.

Look for others.

$(m^2-3m-10)=0$

Factorise to find values of m.

$(m+2)(m-5)=0$
$\therefore m = -2\mbox{ or }5$

SocialRhubarb · « **Reply #2545 on:** November 01, 2013, 02:52:04 pm »

Stevensmy, I think you're looking at question 6.

In this question, I think you'd probably get one mark for recognising that you need to express a in terms of v and x,

$a=v\frac{dv}{dx}$

Stevensmay · « **Reply #2546 on:** November 01, 2013, 02:55:50 pm »

Socialrhubarb, yes I was.

From memory, I think there was a mark for correctly determining which solution was correct.

We end up getting $|v^2 - 3| = e^{2(x-1)}$

$v^2 -3 = e^{2(x-1)} \mbox{ or } -e^{2(x-1)}$

You had to show that one of these was incorrect (with substitution) to get a mark I believe.

Make sure you don't get caught out by leaving the negative out in your answer. Seeing as we square root $v^2$
we end up with $v=\pm \sqrt{3+e^{2(x-1)}}$

If we forgot about this and instead just said it was the positive square root, we would be wrong. Substitution will show which sign is needed.

Even though this doesn't look like a 4 mark question, there are some intricacies that will stop us getting full marks if we aren't careful.

One mark for recognizing $a = v \frac{dv}{dx} \mbox{ or } \frac{1}{2} \frac{d(v^2)}{dx}$

One mark for correct integration/integral statement.

One mark for disproving alternate solutions.

One mark for final answer.

ahat · « **Reply #2547 on:** November 01, 2013, 05:00:48 pm »

Quote from: nliu1995 on November 01, 2013, 11:32:24 am

You didn't actually use a numerical approximation formula; you did the same thing as b^3 did above with finding the integration constant.

With one flaw.

$\frac{d}{dx}(\frac{v^2}{2})=\frac{F(x)}{m}$

The integration formula for the solution would have $\frac{v_0^2}{2}$ as the integration constant.

Hey! I think it dawned on me - please tell me if this is correct or not:
If say, it was $\frac{d}{dx}(a^2)$ then the integration constant would be $a^2__0$ ?

zvezda · « **Reply #2548 on:** November 01, 2013, 05:40:46 pm »

Hey,
How would i go about graphing the path of a particle whose position vector relation has an exponential function for the i direction and a quadratic for the j direction??
Help is much appreciated

Alwin · « **Reply #2549 on:** November 01, 2013, 05:46:40 pm »

Quote from: stankovic123 on November 01, 2013, 05:40:46 pm

Hey,
How would i go about graphing the path of a particle whose position vector relation has an exponential function for the i direction and a quadratic for the j direction??
Help is much appreciated

Do you mind giving an example? For the exponential, you can always isolate and solve for t, then sub it into your quadratic equation for y. But the thing is, you then have a quadratic equation for y, in terms of a log function of x.

The only "easy" method I can think of is to have a log scale on your x axis, but that might not go down well in spesh so you'd have to sketch it, "unlogarithm" the x-axis then draw it properly.
These are just my general thoughts on the problem, if you have a question post it and I can be more specific

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