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Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164404 times)  Share 

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AlphaZero

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9480 on: June 02, 2019, 11:01:07 pm »
+1
Ok I understand, sqrt 2 will also be its minimum?
Thanks for your help!

If \(\mathbf{r}(t)\) was defined for \(t\in [0,\,\pi]\) say, then the minimum value of \(|\mathbf{r}(t)|\) would be \(0\), and occurs when \(t=0,\,\pi\). Sketching a graph of \(|\mathbf{r}(t)|\) against \(t\) would be wise here...
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Mitze5

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9481 on: June 03, 2019, 05:02:33 pm »
0
How would you prove/explain that an ellipse of equation x^2/9 + y^2/4 = 1 is inscribed by a rhombus which has sides tangent to the ellipse.

AlphaZero

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9482 on: June 03, 2019, 10:25:24 pm »
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How would you prove/explain that an ellipse of equation x^2/9 + y^2/4 = 1 is inscribed by a rhombus which has sides tangent to the ellipse.

This is a great question. Of course, this would never be asked in an exam since this is more of an investigation question. Had to dig pretty deep for this one. One way to go about your question is to just try to find coordinates of a rhombus for which the ellipse is inscribed in. But this could be tedious, and so it might be wise to prove that for ANY ellipse, you could find a rhombus for which that ellipse is inscribed.

First, let's start with a circle of radius \(1\) centered at the origin. It should be pretty easy to see that we could form a square using 4 points on the coordinate axes whose sides are tangential to the circle. That is, let \[\mathcal{C}:\ x^2+y^2=1\] and let \(\mathcal{S}\) be a square with vertices at \((0,\,\pm p)\) and \((\pm p,\,0)\),  where \(p>0\).

It is clear by symmetry that the line segment joining the points \((0,\,p)\) and \((p,\,0)\) is tangential to \(\mathcal{C}\) at the point \(\left(\dfrac{1}{\sqrt{2}},\,\dfrac{1}{\sqrt{2}}\right)\), and so by equating gradients we have \[\frac{p-1/\sqrt{2}}{0-1/\sqrt{2}}=-1\implies p=\sqrt{2}.\] Now, we leverage the fact that properties like tangency are preserved under dilations from the coordinate axes.

That is, the ellipse given by \[\mathcal{E}:\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad a,b>0\] can be obtained from the circle \(\mathcal{C}\) via \begin{align*}&1.\ \ \text{a dilation by factor }a\text{ from the }x\text{-axis}\\
&2.\ \ \text{a dilation by factor }b\text{ from the }y\text{ -axis}, \end{align*} and so the parallelogram \(\mathcal{R}\) with vertices at \((\pm a\sqrt{2},\,0)\), \((0,\,\pm b\sqrt{2})\) will necessarily inscribe the ellipse \(\mathcal{E}\).

All that remains is to show that \(\mathcal{R}\) has equal side lengths to prove that the parallelogram we formed is indeed a rhombus, which is a pretty trivial task, since the expressions for the length of any of its sides is identical: \[d=\sqrt{(a\sqrt{2})^2+(b\sqrt{2})^2}=\sqrt{2a^2+2b^2}.\]
So, going back to the original problem we can say that the ellipse given by \(\dfrac{x^2}{3^2}+\dfrac{y^2}{2^2}=1\)  is inscribed by a rhombus with vertices at \[(\pm 3\sqrt{2},\,0)\ \ \text{and}\ \ (0,\,\pm 2\sqrt{2}).\]
Going to finish off this post with a quick note: This is by no means the most general result. In fact, it can be shown that the are infinitely many rhombi that could inscribe any ellipse.
« Last Edit: June 03, 2019, 11:23:40 pm by AlphaZero »
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Duelmaster22

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9483 on: June 08, 2019, 05:49:30 pm »
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Hi guys,

Came across a "solve the differential equation" question, where dy/dx = -x/y and y(1) = 1. Managed to get the correct answer (mostly) which was y^2 = 2 -  x^2, but it also states that y>0. What is the reason that this restriction occurs?
,

RuiAce

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9484 on: June 08, 2019, 06:01:47 pm »
+3
Hi guys,

Came across a "solve the differential equation" question, where dy/dx = -x/y and y(1) = 1. Managed to get the correct answer (mostly) which was y^2 = 2 -  x^2, but it also states that y>0. What is the reason that this restriction occurs?
,

Technically speaking in the differential equations framework, \(y\) is still assumed to be an explicit function of \(x\). This is the answer to your question; the rest is just for completeness sake.

Once you have \(y^2 = 2-x^2\) you technically arrive at \( y =\pm \sqrt{2-x^2} \), so to make it a function, the correct sign needs to be chosen. Because we know that \(y(1) = 1\), i.e. \(y > 0\) when \(x=1\), we take the positive case: \(y = \sqrt{2-x^2}\). And hence end up with a semi-circle.

(Without it, you'd obtain \(x^2+y^2=2\), which is the entire circle with radius \(\sqrt{2}\) centred at the origin. But of course, circles are not functions, because they fail the horizontal line test.)

I've no clue how specific VCAA are on this matter though, so someone else may comment further.

(Note that if you didn't have the initial value you wouldn't be able to comment on \(y\geq 0\) or \(y \leq 0\) though.)

Duelmaster22

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9485 on: June 08, 2019, 06:28:28 pm »
0
Technically speaking in the differential equations framework, \(y\) is still assumed to be an explicit function of \(x\). This is the answer to your question; the rest is just for completeness sake.

Once you have \(y^2 = 2-x^2\) you technically arrive at \( y =\pm \sqrt{2-x^2} \), so to make it a function, the correct sign needs to be chosen. Because we know that \(y(1) = 1\), i.e. \(y > 0\) when \(x=1\), we take the positive case: \(y = \sqrt{2-x^2}\). And hence end up with a semi-circle.

(Without it, you'd obtain \(x^2+y^2=2\), which is the entire circle with radius \(\sqrt{2}\) centred at the origin. But of course, circles are not functions, because they fail the horizontal line test.)

I've no clue how specific VCAA are on this matter though, so someone else may comment further.

(Note that if you didn't have the initial value you wouldn't be able to comment on \(y\geq 0\) or \(y \leq 0\) though.)

Ah okay, I think I understand, thanks a lot! So if the question didn't state that y(1) = 1, but instead said something along the lines of "the curve passes through the point (1,1)", y isn't said to be an explicit function of x and the restriction wouldn't occur?

AlphaZero

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9486 on: June 08, 2019, 07:17:01 pm »
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In the exam, VCAA will make it very clear how to express your answer.

If they write  "Express \(y\) as a function of \(x\)",  you must pick the appropriate solution branch so that \(y\) is given by a function of \(x\).

However, note that solutions to ODEs need not be functions, so if the question doesn't specify a form for the answer, you may leave your answer in a somewhat readable form. Eg:  \(x^2+y^2=2\).

Below, I've put two examples from VCAA past exams.  Note that it's very clear what they want you to do.



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Duelmaster22

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9487 on: June 09, 2019, 04:21:05 pm »
0
In the exam, VCAA will make it very clear how to express your answer.

If they write  "Express \(y\) as a function of \(x\)",  you must pick the appropriate solution branch so that \(y\) is given by a function of \(x\).

However, note that solutions to ODEs need not be functions, so if the question doesn't specify a form for the answer, you may leave your answer in a somewhat readable form. Eg:  \(x^2+y^2=2\).

Below, I've put two examples from VCAA past exams.  Note that it's very clear what they want you to do.

(Image removed from quote.)

(Image removed from quote.)

That makes things a bit easer! Thanks!

AlphaZero

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9488 on: June 09, 2019, 06:02:06 pm »
+1
On the 2017 NHT Exam 1, question 3, we are asked to find the gradient of a curve (x in terms of y).

My teacher solved this question by implicit differentiation but I did it by rearranging the equation to find y in terms of x. I did end up with the same answer (in fact, I personally found my method easier) but would I lose marks in the exam for not solving it by implicit diff?

Thanks  :)

One thing that many students fail to notice is that \[\sin(\theta)=a\ \ \text{is }\textbf{not}\ \text{equivalent to}\ \ \theta=\arcsin(a).\] The first statement allows for many possible values of \(\theta\) but the second statement places the restriction  \(-\pi/2\leq \theta \leq \pi/2\).

Your method is fine, but you were quite lucky to obtain the correct answer. Essentially, because you were told that \(a,b\in\mathbb{Z}^+\),  using  \(y=15\arcsin(x)\)  was fine since its gradient is always positive anyway. However, if you sketch the graph of \(x=\sin(y/15)\), you would see that there are actually two possible answers for the gradient where \(x=1/4\).
« Last Edit: June 09, 2019, 06:03:44 pm by AlphaZero »
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JeKnYan

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9489 on: June 10, 2019, 08:16:01 pm »
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If you use partial differentiation on an imp diff question instead of using the product rule, will you still get full marks? I can't seem to get the hang of imp dif product rules...
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AlphaZero

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9490 on: June 10, 2019, 09:24:41 pm »
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If you use partial differentiation on an imp diff question instead of using the product rule, will you still get full marks? I can't seem to get the hang of imp dif product rules...

I'm a little confused about your question. Partial differentiation has nothing to do with the product rule or implicit differentiation. They are not the same thing. Could you perhaps provide an example of where you've used 'partial differentiation' instead of the product rule?

As for your second question about using the product rule in implicit differentiation, I've provided an example below that should hopefully help.

Example

Find \(\dfrac{dy}{dx}\), in terms of \(x\) and \(y\), given the curve \(2x+xy^2-y^3=5\).

Solution
\begin{align*}&\frac{d}{dx}\!\big[x^2\big]+\frac{d}{dx}\!\big[x\big]y^2+x\frac{d}{dy}\!\big[y^2\big]\frac{dy}{dx}-\frac{d}{dy}\!\big[y^3\big]\frac{dy}{dx}=\frac{d}{dx}\big[5\big] \\
\implies &2x+y^2+2xy\frac{dy}{dx}-3y^2\frac{dy}{dx}=0\\
\implies &\frac{dy}{dx}\left(2xy-3y^2\right)=-2x-y^2\\
\implies &\frac{dy}{dx}=\frac{2x+y^2}{3y^2-2xy} \end{align*}
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Ansaki

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9491 on: June 13, 2019, 10:43:57 pm »
0
hey guys i need help with a question, thanks!
question 1:  Evaluate   \[ \int_{π/2}^{π}  sin^{3}xcos^{2}x\,dx \]
btw my first time using the latex system and it was sooo difficult lol


undefined

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9492 on: June 13, 2019, 11:18:50 pm »
0
hey guys i need help with a question, thanks!
question 1:  Evaluate   \[ \int_{π/2}^{π}  sin^{3}xcos^{2}x\,dx \]
btw my first time using the latex system and it was sooo difficult 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.
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AlphaZero

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9493 on: June 13, 2019, 11:23:18 pm »
+1
hey guys i need help with a question, thanks!
question 1:  Evaluate   \[ \int_{\pi/2}^\pi \sin^3(x)\cos^{2}(x)\,dx \]
btw my first time using the latex system and it was sooo difficult lol

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
« Last Edit: June 13, 2019, 11:24:54 pm by AlphaZero »
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Ansaki

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9494 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!