Author Topic: VCE Specialist 3/4 Question Thread! (Read 2553884 times) Tweet Share

SocialRhubarb · « **Reply #1995 on:** July 21, 2013, 06:42:20 pm »

Well, you've basically already done all the hard work already.

$M(\frac{3}{4}\cos(\theta),\ -\frac{3}{2}\sin(\theta))$

$x=\frac{3}{4}\cos(\theta),\ y=-\frac{3}{2}\sin(\theta)$

$\cos^2(\theta)=(\frac{4}{3}x)^2,\ \sin^2(\theta)=(-\frac{2}{3}y)^2$

$\frac{16x^2}{9}+\frac{4y^2}{9}=1$ , which is an ellipse.

Looking at the domain of of theta, the sine and cosine functions are both positive. Therefore, the domain of M is positive and its range is neagtive, so it's the bottom right quarter of an ellipse.

saba.ay · « **Reply #1996 on:** July 21, 2013, 06:50:24 pm »

Quote from: SocialRhubarb on July 21, 2013, 06:42:20 pm

Well, you've basically already done all the hard work already.

$M(\frac{3}{4}\cos(\theta),\ -\frac{3}{2}\sin(\theta))$

$x=\frac{3}{4}\cos(\theta),\ y=-\frac{3}{2}\sin(\theta)$

$\cos^2(\theta)=(\frac{4}{3}x)^2,\ \sin^2(\theta)=(-\frac{2}{3}y)^2$

$\frac{16x^2}{9}+\frac{4y^2}{9}=1$ , which is an ellipse.

Looking at the domain of of theta, the sine and cosine functions are both positive. Therefore, the domain of M is positive and its range is neagtive, so it's the bottom right quarter of an ellipse.

Thank you so so much

Aelru · « **Reply #1997 on:** July 21, 2013, 07:16:21 pm »

I'm quite sure on the other processes of working out b/c,
but I'm just struggling with a.

Mainly, with finding the width of that confounded surface area.
got the 480 value, just not sure where the x disappeared.

Answer is : $\frac{dx}{dt} = \frac {-1}{480\sqrt{4-x}}$

Thanks~

b^3 · « **Reply #1998 on:** July 21, 2013, 07:56:13 pm »

Hint: Form an equation for the circle at the end of the tube, and from that find an expression for the width of that rectangle.

Spoiler

Starting off I'm going to let $a$ be in the horizontal direction and $x$ be in the vertical direction, since we are given $x$ as upwards.
Now we need to find an expression for the circle, then we can rearrange for $a$ in terms of $x$ . The width of the rectangle will be double the $a$ coordinate, and as the height decreases, the value of this coordinate will change.
$\begin{alignedat}{1}a^{2}+\left(x-2\right)^{2} & =4 \\ a^{2} & =4-x^{2}+4x-4 \\ a^{2} & =-x^{2}+4x \\ a & =\pm\sqrt{x\left(4-x\right)} \\ l & =2a \\ & =2\sqrt{x\left(4-x\right)} \\ A & =l\times w \\ & =2\sqrt{x\left(4-x\right)}\times6 \\ & =12\sqrt{x\left(4-x\right)} \\ \frac{dx}{dt} & =\frac{-0.025\sqrt{x}}{A} \\ & =\frac{-\sqrt{x}}{40\times12\sqrt{x}\sqrt{\left(4-x\right)}} \text{ (the }\sqrt{x}\text{ cancels out here)} \\ & =\frac{-1}{480\sqrt{4-x}} \end{alignedat}$

Aelru · « **Reply #1999 on:** July 21, 2013, 08:10:51 pm »

Quote from: b^3 on July 21, 2013, 07:56:13 pm

Hint: Form an equation for the circle at the end of the tube, and from that find an expression for the width of that rectangle.
Spoiler
Starting off I'm going to let $a$ be in the horizontal direction and $x$ be in the vertical direction, since we are given $x$ as upwards.
Now we need to find an expression for the circle, then we can rearrange for $a$ in terms of $x$ . The width of the rectangle will be double the $a$ coordinate, and as the height decreases, the value of this coordinate will change.
$\begin{alignedat}{1}a^{2}+\left(x-2\right)^{2} & =4 \\ a^{2} & =4-x^{2}+4x-4 \\ a^{2} & =-x^{2}+4x \\ a & =\pm\sqrt{x\left(4-x\right)} \\ l & =2a \\ & =2\sqrt{x\left(4-x\right)} \\ A & =l\times w \\ & =2\sqrt{x\left(4-x\right)}\times6 \\ & =12\sqrt{x\left(4-x\right)} \\ \frac{dx}{dt} & =\frac{-0.025\sqrt{x}}{A} \\ & =\frac{-\sqrt{x}}{40\times12\sqrt{x}\sqrt{\left(4-x\right)}} \text{ (the }\sqrt{x}\text{ cancels out here)} \\ & =\frac{-1}{480\sqrt{4-x}} \end{alignedat}$

;; Brilliant insight. So in these diagram questions, I should always find a equation to relate things together?
i initially tried a^2 + x^2 = 4, but I guess that wouldn't give me the variables to solve what's needed.
What was your thought process for this b^3? ><

Alwin · « **Reply #2000 on:** July 21, 2013, 08:18:10 pm »

Quote from: Aelru on July 21, 2013, 08:10:51 pm

What was your thought process for this b^3? ><

Answer: Amazing!

b^3 will probably elaborate more later

b^3 · « **Reply #2001 on:** July 21, 2013, 08:18:44 pm »

The key piece of information that people always seem to miss from a question is that which is right in front of them, but not explicitly stated, that is people seem to miss geometric information that can help them with the question, in other words, is the diagram telling me anything that the question hasn't?

On the thought processes, I knew I needed another piece of information to link the surface area of the water with the height. I knew we had to replace the A, so firstly we needed an expression for A, which as it's just an area of a rectangle, would be length times width, for which we already had the length. Now the width will vary, but it will varying depending on the height of the water. As it drains, it will drain down to zero when you get to the bottom of the circle, so that's why I placed the origin at the bottom of this circle, then it's just a matter of forming the equation and manipulating it to get it in terms of $x$ , and substituting the whole thing back in.

So really I think what would help you is, to ask yourself "what do I need to find" or "what do I need to link to each other", and then to look for what you can form there, and in some cases (you won't always need to find another equation), to see where you can place an origin on the diagram to make the equation/new piece of information easiest to manipulate.

Hope that helps

Alwin · « **Reply #2002 on:** July 21, 2013, 08:38:21 pm »

i know this q has been answered (very thoroughly) by b^3, but since Aelru asked about thought process, I thought I'd post my alternate method too:

Its not as imaginative has b^3's approach, and I just pythagoras - nothing too fancy:

$\begin{alignedat}{1} \left(\frac{l}{2}\right)^{2} &= 2^2 - (2-x)^2 \\ \frac{l}{2} &= \sqrt{4 - 4 + 4x - x^2} \\ l &= 2\sqrt{4x - x^2} \\ l &=2\sqrt{x(4-x)} \end{alignedat}$

The rest follows as b^3's method

Aelru · « **Reply #2003 on:** July 21, 2013, 09:12:20 pm »

Gets the results thought

Yep, that explanation helped a lot b^3. My main weakness is analysis, and that's a crucial part of specialist ==

Alwin, that's a rather clever method as well. What program did you used to graphically represent it?

Alwin · « **Reply #2004 on:** July 21, 2013, 09:18:02 pm »

Quote from: Aelru on July 21, 2013, 09:12:20 pm

Gets the results thought

Yep, that explanation helped a lot b^3. My main weakness is analysis, and that's a crucial part of specialist ==

Alwin, that's a rather clever method as well. What program did you used to graphically represent it?

lol Aelru . . . I used microsoft word HAHA

Aelru · « **Reply #2005 on:** July 22, 2013, 06:18:52 pm »

Microsoft word... quite creative use of it.

As a 'side-question', I find that my calculator (CAS) graphs Slope fields slow. It's so slow, I can make myself some fried fish in 10 minutes before a cubic graph shows. I was wondering if there's a way to speed up the process.

Also, My calculator doesn't seem to show the axis (x,y respectively), when graphing slope fields. This makes it rather...
odd, when trying to interpret graphs with horizontal translations. Is there a way to make the axis show?

Thanks.

sin0001 · « **Reply #2006 on:** July 22, 2013, 06:43:32 pm »

Quote from: Aelru on July 22, 2013, 06:18:52 pm

Microsoft word... quite creative use of it.

As a 'side-question', I find that my calculator (CAS) graphs Slope fields slow. It's so slow, I can make myself some fried fish in 10 minutes before a cubic graph shows. I was wondering if there's a way to speed up the process.

Also, My calculator doesn't seem to show the axis (x,y respectively), when graphing slope fields. This makes it rather...
odd, when trying to interpret graphs with horizontal translations. Is there a way to make the axis show?

Thanks.

How do you even graph slope fields on CAS? And is it possible that we'll need to know this for the exam? because I've heard Slope fields are only ever mentioned in multi-choice

b^3 · « **Reply #2007 on:** July 22, 2013, 06:53:06 pm »

Quote from: sin0001 on July 22, 2013, 06:43:32 pm

How do you even graph slope fields on CAS? And is it possible that we'll need to know this for the exam? because I've heard Slope fields are only ever mentioned in multi-choice

I'm pretty sure that you do need to know it. (well by hand at least), it's been on an exam 1, (2007) Q8 http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths1.pdf

To graph them on the Ti-nspire, go to the graphing window and have the entry bar up, press [ctrl] [menu] [2] [6] ("Graph Type
and "Differential Equations". If you are entering things in terms of y, for the y1 entry you need to use y1 in place of y in the equation.

@Aelru, which calculator?

Aelru · « **Reply #2008 on:** July 22, 2013, 07:03:28 pm »

Casio Classpad 330 (CAS).

sin0001 · « **Reply #2009 on:** July 22, 2013, 10:06:30 pm »

Quote from: b^3 on July 22, 2013, 06:53:06 pm

I'm pretty sure that you do need to know it. (well by hand at least), it's been on an exam 1, (2007) Q8 http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths1.pdf

To graph them on the Ti-nspire, go to the graphing window and have the entry bar up, press [ctrl] [menu] [2] [6] ("Graph Type
and "Differential Equations". If you are entering things in terms of y, for the y1 entry you need to use y1 in place of y in the equation.

@Aelru, which calculator?

Ooh in that case, I should learn how to sketch 'em. Also, I was referring to the Casio Classpad, thanks though!

Search the forums now!

We have moved!

Author Topic: VCE Specialist 3/4 Question Thread! (Read 2553884 times) Tweet Share

SocialRhubarb

Re: Specialist 3/4 Question Thread!

saba.ay

Re: Specialist 3/4 Question Thread!

Aelru

Re: Specialist 3/4 Question Thread!

b^3

Re: Specialist 3/4 Question Thread!

Aelru

Re: Specialist 3/4 Question Thread!

Alwin

Re: Specialist 3/4 Question Thread!

b^3

Re: Specialist 3/4 Question Thread!

Alwin

Re: Specialist 3/4 Question Thread!

Aelru

Re: Specialist 3/4 Question Thread!

Alwin

Re: Specialist 3/4 Question Thread!

Aelru

Re: Specialist 3/4 Question Thread!

sin0001

Re: Specialist 3/4 Question Thread!

b^3

Re: Specialist 3/4 Question Thread!

Aelru

Re: Specialist 3/4 Question Thread!

sin0001

Re: Specialist 3/4 Question Thread!

Recent Posts

User:
Password: