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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Chavi on October 16, 2010, 10:55:55 pm

Title: Error is TSFX 2010?
Post by: Chavi on October 16, 2010, 10:55:55 pm: Hey, in question 4, TSFX Exam 1 2010, did anyone find an error with the solutions.

When solving for velocity in terms of displacement,
$a = \frac{d}{dx}(\frac{v^2}{2})$ -is it just me, or do they forget to divide $v^2$ by $2?$
Title: Re: Error is TSFX 2010?
Post by: brightsky on October 16, 2010, 11:08:10 pm: Do you have the paper on you? Don't think many people would have this particular exam.
Title: Re: Error is TSFX 2010?
Post by: Chavi on October 16, 2010, 11:09:46 pm: It's available for free from the TSFX website www.tsfx.com.au
Title: Re: Error is TSFX 2010?
Post by: jasoN- on October 16, 2010, 11:15:51 pm: wait where is it? I can't seem to find it lol
Title: Re: Error is TSFX 2010?
Post by: tcg93 on October 16, 2010, 11:18:33 pm: Quote from: Chavi on October 16, 2010, 11:09:46 pm
It's available for free from the TSFX website www.tsfx.com.au

Whereabouts on the website is it? I've been throughout the whole thing and can't find it
Title: Re: Error is TSFX 2010?
Post by: jasoN- on October 16, 2010, 11:30:14 pm: hmm, I think we need to register 'vcedge online' tab
Quote
Thank You For Registering With VCEdge Online.

A passcode that will enable access to the FREE resources exclusive to VCEdge Online members will be sent by mail on the next
business day.
sigh. why must they do this!

Edit: I only see 2005-2007 spec. exams, are there more when I get my password?
Title: Re: Error is TSFX 2010?
Post by: Chavi on October 16, 2010, 11:43:18 pm: I think you have to register to get the latest ones. I got mine from my teacher.
Anyway, if I type up the question maybe I can get some help:

The acceleration of an object from the origin is given by: $a=\frac{2x}{x^2+1}$

Given that v=2 when x=1 find the velocity of the object when x=5. And place it in the format: $\sqrt{ln(a) + b}$

I got a=169, and b=4
TSFX got a=13 and b=2
The source of confusion relates the integration with (v^2)/2
Title: Re: Error is TSFX 2010?
Post by: jasoN- on October 16, 2010, 11:44:31 pm: Hey I got your answers as well!
man this paper had some tricky questions (completely stuck on the y=f(x) finding curve thing)

I'm yet to be in possession of the solutions
Title: Re: Error is TSFX 2010?
Post by: tcg93 on October 16, 2010, 11:44:46 pm: Quote from: jasoN- on October 16, 2010, 11:30:14 pm
hmm, I think we need to register 'vcedge online' tab
Quote
Thank You For Registering With VCEdge Online.

A passcode that will enable access to the FREE resources exclusive to VCEdge Online members will be sent by mail on the next
business day.
sigh. why must they do this!

Edit: I only see 2005-2007 spec. exams, are there more when I get my password?

no there are not
Title: Re: Error is TSFX 2010?
Post by: brightsky on October 16, 2010, 11:49:42 pm: Wait I'm confused, I'm assuming v is velocity, but then this...

Quote
find the velocity of the object when v=5

...doesn't make sense. :-\
Title: Re: Error is TSFX 2010?
Post by: Chavi on October 16, 2010, 11:53:53 pm: Quote from: brightsky on October 16, 2010, 11:49:42 pm
Wait I'm confused, I'm assuming v is velocity, but then this...

Quote
find the velocity of the object when v=5

...doesn't make sense. :-\
soz, the question said x=5
Title: Re: Error is TSFX 2010?
Post by: jasoN- on October 16, 2010, 11:55:03 pm: yeah, they must have not multiplied the equation by 2, or they may have taken the 2 out (giving 13 and 2) and removed it (<- lol)
Title: Re: Error is TSFX 2010?
Post by: brightsky on October 17, 2010, 12:05:07 am: $v = \int \frac{2x}{x^2 + 1} dx = \log_e|x^2 + 1| + C$

$\because v = 2 \text{ when } x = 1$

$\therefore 2 = \log_e(2) + C$

$C = 2 - \log_e(2)$

$\text{So the equation is } v = \log_e|x^2 + 1| + 2 - \log_e(2)$

$\text{So when } x= 5$ , $v = \log_e(26) + 2 - \log_e(2)$

$\text{So } v = \log_e(13) + 2$

$\text{We let } \sqrt{\log_e(a) + b} = \log_e(13) + 2$

...and it's going to get very complicated. Are you sure there's a square root over that?
Title: Re: Error is TSFX 2010?
Post by: Chavi on October 17, 2010, 12:08:24 am: Quote from: brightsky on October 17, 2010, 12:05:07 am
$v = \int \frac{2x}{x^2 + 1} dx = \log_e|x^2 + 1| + C$

$\because v = 2 \text{ when } x = 1$

$\therefore 2 = \log_e(2) + C$

$C = 2 - \log_e(2)$

$\text{So the equation is } v = \log_e|x^2 + 1| + 2 - \log_e(2)$

$\text{So when } x= 5$ , $v = \log_e(26) + 2 - \log_e(2)$

$\text{So } v = \log_e(13) + 2$

$\text{We let } \sqrt{\log_e(a) + b} = \log_e(13) + 2$

...and it's going to get very complicated. Are you sure there's a square root over that?
Are you sure in the first line it's not (v^2)/2 = f(x)
Title: Re: Error is TSFX 2010?
Post by: jasoN- on October 17, 2010, 12:08:54 am: is it not $a = \frac{d}{dx} (\frac{v^2}{2}) = \frac{2x}{x^2+1}$
hence $\frac{v^2}{2} = \int \frac{2x}{x^2+1}$
Title: Re: Error is TSFX 2010?
Post by: Mao on October 17, 2010, 12:10:20 am: @Brightsky, $\Delta v\neq \int_{\Delta x} a \; dx$ , $\Delta v = \int_{\Delta t} a \; dt$

In this case of $a = f(x)$ , we use $a = v \frac{dv}{dx}$ , which yields $\int_{\Delta v} v\; dv = \int_{\Delta x} a\; dx \implies \frac{v_f^2 - v_i^2}{2} = \int_{x_i}^{x_f} f(x)\; dx$
Title: Re: Error is TSFX 2010?
Post by: brightsky on October 17, 2010, 12:15:05 am: Ahh, so x is not the time? :( ...I should get used to these pronumerals.
Title: Re: Error is TSFX 2010?
Post by: Chavi on October 17, 2010, 12:17:36 am: Quote from: Mao on October 17, 2010, 12:10:20 am
@Brightsky, $\Delta v\neq \int_{\Delta x} a \; dx$ , $\Delta v = \int_{\Delta t} a \; dt$

In this case of $a = f(x)$ , we use $a = v \frac{dv}{dx}$ , which yields $\int_{\Delta v} v\; dv = \int_{\Delta x} a\; dx \implies \frac{v_f^2 - v_i^2}{2} = \int_{x_i}^{x_f} f(x)\; dx$
What is the purpose of this flowery notation?
Title: Re: Error is TSFX 2010?
Post by: jasoN- on October 17, 2010, 12:20:29 am: nope, x is the position

How would you do this question from the same exam? (if my memory serves correctly)

A function y=f(x) has a tangent at point A(x, y) which crosses the x-axis at coordinate (x-1/2, 0)
The function has a y-intercept at coordinate (0, 1/e)
Find the equation of the curve.

Chavi can you confirm this question please :)
Title: Re: Error is TSFX 2010?
Post by: Chavi on October 17, 2010, 12:34:07 am: Quote from: jasoN- on October 17, 2010, 12:20:29 am
nope, x is the position

How would you do this question from the same exam? (if my memory serves correctly)

A function y=f(x) has a tangent at point A(x, y) which crosses the x-axis at coordinate (x-1/2, 0)
The function has a y-intercept at coordinate (0, 1/e)
Find the equation of the curve.

Chavi can you confirm this question please :)
Hey , I did this question - it is tricky, but once you understand what you're asking for it's pretty simple.

OK, so you have points A(x,y) and (x-1/2, 0) - for the tangent to the curve - (don't confuse the y incpt of the function with the tangent to the curve)

then, find the gradient of the tangent by rise/run: dy/dx = y/(x-(x/1/2)) = 2y and thus, dx/dy = 1/2y

From this, x = (ln(y))/2 + c and at the y intercept, x=0 and y = 1/e

so you have 0 = 1/2(-1) + c, and c = 1/2

So, y = e^(2x-1)

Hint: Draw a diagram.
Title: Re: Error is TSFX 2010?
Post by: Mao on October 17, 2010, 12:41:24 am: Quote from: Chavi on October 17, 2010, 12:17:36 am
Quote from: Mao on October 17, 2010, 12:10:20 am
@Brightsky, $\Delta v\neq \int_{\Delta x} a \; dx$ , $\Delta v = \int_{\Delta t} a \; dt$

In this case of $a = f(x)$ , we use $a = v \frac{dv}{dx}$ , which yields $\int_{\Delta v} v\; dv = \int_{\Delta x} a\; dx \implies \frac{v_f^2 - v_i^2}{2} = \int_{x_i}^{x_f} f(x)\; dx$
What is the purpose of this flowery notation?

A subtle point I am trying to push, that these kind of questions should be solved with definite integrals rather than indefinite integrals. :)
Title: Re: Error is TSFX 2010?
Post by: jasoN- on October 17, 2010, 10:42:16 am: Oh that's awesome, thanks Chavi :D wasn't as hard as I thought.