Author Topic: VCE Specialist 3/4 Question Thread! (Read 2547234 times) Tweet Share

tiff_tiff · « **Reply #3105 on:** April 30, 2014, 05:13:39 pm »

oh lord that looks difficult... thank you but what is k1 and k2?

Mieow · « **Reply #3106 on:** April 30, 2014, 05:29:47 pm »

Quote

Find x, such that cosec(2x - pi/3) = (-2√3)/3, for 0≤x≤2pi.

In the worked solution, it says:

Quote

Let θ = 2x - (pi/3), where -pi/3 ≤ θ ≤ (11pi)/3

Can anyone explain to me how they got that restriction? Thank you!

Thorium · « **Reply #3107 on:** April 30, 2014, 05:35:50 pm »

Quote from: tiff_tiff on April 30, 2014, 05:13:39 pm

oh lord that looks difficult... thank you but what is k1 and k2?

Oops, I did it the other way. I found p and q in terms of k1 and k2. Sorry about that.

Phy124 · « **Reply #3108 on:** April 30, 2014, 06:00:29 pm »

Quote from: Mieow on April 30, 2014, 05:29:47 pm

In the worked solution, it says:
Quote
Let θ = 2x - (pi/3), where -pi/3 ≤ θ ≤ (11pi)/3
Can anyone explain to me how they got that restriction? Thank you!

$0\leq x\leq 2\pi$

$2(0)-\frac{\pi}{3} \leq 2(x) - \frac{\pi}{3} \leq 2(2\pi)-\frac{\pi}{3}$

$-\frac{\pi}{3} \leq \theta \leq \frac{11\pi}{3}$

T-Infinite · « **Reply #3109 on:** April 30, 2014, 07:14:35 pm »

Could someone please help me with this question,

a) Given that S= { ${z:\left | z-i \right |\leq 1}$ } and T= { ${z:Im(z)\leq 0) and Re(z)\geq 2}$ }, sketch the graphs of S and T on the one set of axes

b) If z1 $\epsilon$ S and Z2 $\epsilon$ T, find the least value of |z1 - z2|

For part a) I know S is a circle, but for T, when it says "and" do we draw both and only keep the intersection between the two? I don't how to do part b...

Mieow · « **Reply #3110 on:** April 30, 2014, 10:30:22 pm »

Quote from: Phy124 on April 30, 2014, 06:00:29 pm

Can anyone explain to me how they got that restriction? Thank you!

$0\leq x\leq 2\pi$

$2(0)-\frac{\pi}{3} \leq 2(x) - \frac{\pi}{3} \leq 2(2\pi)-\frac{\pi}{3}$

$-\frac{\pi}{3} \leq \theta \leq \frac{11\pi}{3}$

oh thanks

but how come you didn't add pi/3 and divide by 2 like when finding the domain of a graph?

b^3 · « **Reply #3111 on:** May 01, 2014, 05:02:44 pm »

Quote from: Mieow on April 30, 2014, 10:30:22 pm

oh thanks
but how come you didn't add pi/3 and divide by 2 like when finding the domain of a graph?

For the domain of the graph you're finding the values of $x$ which fit the restriction, where as here we want to find the values of $2x-\frac{\pi}{3}$ that fit our restriction (as this restriction will give us the values to ignore after we take the inverse trigonometric function). Remember that the values that $2x-\frac{\pi}{3}$ can take won't give the domain of the function. So we can start with the restriction like Phy has and then work our way back to find what

$2x-\frac{\pi}{3}}$ can actually be between.

Eugenet17 · « **Reply #3112 on:** May 07, 2014, 09:34:28 pm »

13 An ellipse has equation x^2/a^2+y^2/b^2=1 .The tangent at a point P(a costheta, b sintheta) intersects the axes at points M and N and O is the origin.
a Find the area of triangle OMN in terms of a, b and theta.
b Find the values of theta for which the area of triangle OMN is a minimum and state this
minimum area in terms of a and b.

Help please!

Edit: Answers for a) |ab/sin(2theta)|
b) Min area is ab, theta = (2k+1)pi/4

b^3 · « **Reply #3113 on:** May 07, 2014, 10:15:06 pm »

I put the long working in the spoiler, but the following would be the steps you should try to take before looking at the spoiler.
1. Implicit diff it to find $\frac{dy}{dx}$ .
2. Use this and the point $(a\cos(\theta),b\sin(\theta))$ to find the equation of the tangent.
3. The axes intercepts will give the width and height of the triangle as our ellipse is centred about the origin. Find these
4. Use these and the formula for the area of a triangle to find an expression for the area of the triangle in terms of $a$ , $b$ and $\theta$ .
5. Now you want to minimise this new function you've found, differentiate it with respect to $\theta$ keeping $a$ and $b$ constant.
6. Let this derivative equal to zero and solve for $\theta$
7. Sub this back in to find your minimum area.

https://www.desmos.com/calculator/int977nc4m

Spoiler

Find the tangent of the curve at this point by first using implicit differentiation to find $\frac{dy}{dx}$ .
$\begin{alignedat}{1}\frac{d}{dx}\left(\frac{x^{2}}{a^{2}}\right)+\frac{d}{dx}\left(\frac{y^{2}}{b^{2}}\right) & =\frac{d}{dx}\left(1\right) \\ \frac{2x}{a^{2}}+\frac{2y}{b^{2}}\frac{dy}{dx} & =0 \\ \frac{dy}{dx} & =-\frac{2x}{a^{2}}\div\frac{2y}{b^{2}} \\ \implies\frac{dy}{dx} & =-\frac{b^{2}x}{a^{2}y} \end{alignedat}$
Finding the tangent using $y-y_{1}=m\left(x-x_{1}\right)$
$\begin{alignedat}{1}x & =a\cos\left(\theta\right),\; y=b\sin\left(\theta\right) \\ \frac{dy}{dx} & =-\frac{b^{2}\left(a\cos\left(\theta\right)\right)}{a^{2}\left(b\sin\left(\theta\right)\right)} \\ & =-\frac{b}{a\tan\left(\theta\right)} \\ y-b\sin\left(\theta\right) & =-\frac{b}{a\tan\left(\theta\right)}\left(x-a\cos\left(\theta\right)\right) \\ \implies y & =-\frac{b}{a\tan\left(\theta\right)}\left(x-a\cos\left(\theta\right)\right)+b\sin\left(\theta\right) \end{alignedat}$
Find the $x$ and $y$ intercepts to find the height of the triangle.
$\begin{alignedat}{1}x=0 \\ y & =-\frac{b}{a\tan\left(\theta\right)}\left(-a\cos\left(\theta\right)\right)+b\sin\left(\theta\right) \\ & =b\frac{\cos^{2}\left(\theta\right)}{\sin\left(\theta\right)}+b\sin\left(\theta\right) \\ & =b\left(\frac{\cos^{2}\left(\theta\right)+\sin^{2}\left(\theta\right)}{\sin\left(\theta\right)}\right) \\ & =\frac{b}{\sin\left(\theta\right)} \\ y=0 \\ 0 & =-\frac{b}{a\tan\left(\theta\right)}\left(x-a\cos\left(\theta\right)\right)+b\sin\left(\theta\right) \\ \frac{b}{a\tan\left(\theta\right)}\left(x-a\cos\left(\theta\right)\right) & =b\sin\left(\theta\right) \\ x-a\cos\left(\theta\right) & =a\frac{\sin^{2}\left(\theta\right)}{\cos\left(\theta\right)} \\ x & =a\left(\frac{\sin^{2}\left(\theta\right)+\cos^{2}\left(\theta\right)}{\cos\left(\theta\right)}\right) \\ & =\frac{a}{\cos\left(\theta\right)} \end{alignedat}$
Graph is here: https://www.desmos.com/calculator/int977nc4m
Our $x$ and $y$ intercepts then give the height and width of the triangle.
$\begin{alignedat}{1}A & =\frac{1}{2}bh \\ & =\frac{1}{2}\left(\frac{a}{\cos\left(\theta\right)}\right)\left(\frac{b}{\sin\left(\theta\right)}\right) \\ & =\frac{ab}{2\cos\left(\theta\right)\sin\left(\theta\right)} \\ \text{As }2\cos\left(\theta\right)\sin\left(\theta\right)=\sin\left(2\theta\right) \\ A & =\frac{ab}{\sin\left(2\theta\right)} \end{alignedat}$
This sometimes will give a 'negative area', but it is indeed an area, but just comes out negative due to our coordinate system. So we have
$\begin{alignedat}{1}\implies A & =|\frac{ab}{\sin\left(2\theta\right)}|\end{alignedat}$
To minimise this function we first need to first differentiate it with respect to $\theta$ , leaving $a$ and $b$ constant.
$\begin{alignedat}{1}\frac{dA}{d\theta} & =-\frac{2ab\cos\left(2\theta\right)}{\sin^{2}\left(2\theta\right)} \\ \frac{dA}{d\theta}=0 \\ -\frac{2ab\cos\left(2\theta\right)}{\sin^{2}\left(2\theta\right)} & =0 \\ \cos\left(2\theta\right) & =0 \\ 2\theta & =\frac{\pi}{2},\:\frac{3\pi}{2},\:\frac{5\pi}{2},\dots \\ \implies\theta & =\frac{\pi}{4},\:\frac{3\pi}{4},\:\frac{5\pi}{4},\dots \\ \therefore\theta & =\frac{\left(2k+1\right)\pi}{4},\: k\in\mathbb{Z} \end{alignedat}$
Finally finding the minimum area by substituting one of these values back in.
$\begin{alignedat}{1}\theta & =\frac{\pi}{4} \\ \implies A_{min} & =|\frac{ab}{\sin\left(2\cdot\frac{\pi}{4}\right)}| \\ & =|\frac{ab}{\sin\left(\frac{\pi}{2}\right)}| \\ & =|\frac{ab}{1}| \\ \therefore A_{min} & =ab \end{alignedat}$

Hope that helps

Zealous · « **Reply #3114 on:** May 07, 2014, 10:20:45 pm »

Quote from: Eugenet17 on May 07, 2014, 09:34:28 pm

13 An ellipse has equation x^2/a^2+y^2/b^2=1 .The tangent at a point P(a costheta, b sintheta) intersects the axes at points M and N and O is the origin.
a Find the area of triangle OMN in terms of a, b and theta.
b Find the values of theta for which the area of triangle OMN is a minimum and state this
minimum area in terms of a and b.

Help please!

Edit: Answers for a) |ab/sin(2theta)|
b) Min area is ab, theta = (2k+1)pi/4

[EDIT] Beaten by Mr Cubed! My goodness you are fast at LaTex haha

Eugenet17 · « **Reply #3115 on:** May 07, 2014, 10:29:40 pm »

oh boy so much detail, thank you both so much it helped incredibly!

Eugenet17 · « **Reply #3116 on:** May 09, 2014, 07:48:21 pm »

Can someone help me with Q2 f) Short Answer Exam 2 2009?

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2009specmaths2-w.pdf

Edit: Also, Q1 e, ii)Short answer Exam 2 2011 as well

alchemy · « **Reply #3117 on:** May 11, 2014, 01:43:43 pm »

Need help with the attached question.

Conic · « **Reply #3118 on:** May 11, 2014, 01:51:41 pm »

Let $u=\mathrm{ln}(3x).$

$\implies \frac{du}{dx}=\frac{1}{x}.$

$\therefore \int \frac{\mathrm{ln}(3x)}{2x}dx=\int \frac{du}{dx}\frac{u}{2}dx$

$=\int \frac{u}{2} du$

$=\frac{1}{4}u^2+C\quad (C\in\mathbb{R})$

$=\frac{1}{4}(\mathrm{ln}(3x))^2+C.$

Also that isn't really a question

alchemy · « **Reply #3119 on:** May 11, 2014, 02:37:55 pm »

Quote from: Conic on May 11, 2014, 01:51:41 pm

Also that isn't really a question

LOL, it was meant to read Find:... But that part got cropped out accidentally
Sorry for the confusion

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