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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: yang_dong on July 19, 2014, 04:09:27 pm

Title: random question
Post by: yang_dong on July 19, 2014, 04:09:27 pm: if the question says that it is decreasing proportionally? what does that mean? I'm not sure proportional to what?
so there is a population infected by a disease and with the aid of the vaccination the disease decrease proportionally? If originally the rate was such that for every one person inflected, he/she infects five people, how many people are required to be vaccinated?
Title: Re: random question
Post by: keltingmeith on July 19, 2014, 04:28:31 pm: You really haven't given us enough information.

A decreasing proportionality could mean two things - let's say a is proportional to b. It could mean that as a decreases, so does b, in which case there is a DIRECT proportionality between the two. Or, it could mean that as a increases, b decreases, which means that there's an INVERSE proportionality between the two. In both cases, there's something decreasing, but they have two different situations:

if there's a direct proportionality, we have $a\propto b \implies a=kb$ . If there's an inverse proportionality, we have $a\propto\frac{1}{b} \implies a=\frac{k}{b}$ . Are you giving us the exact wording of the question, or just picking bits of it out and hoping you're asking us the right things?
Title: Re: random question
Post by: yang_dong on July 19, 2014, 05:28:13 pm: that's why im so confused with this question, but then I guess we have to make the assumption that it is a inverse proportionality because when you vaccinate the disease, the more vaccination, the less the disease???

how would I solve the question then???
Title: Re: random question
Post by: yang_dong on August 03, 2014, 06:01:22 pm: quick question

how did they get that? (please see attachment)

i'm trying to graph the graph y = (1/x^2)/(1+x^2)

thank you
Title: Re: random question
Post by: Nato on August 03, 2014, 06:29:18 pm: Quote from: yang_dong on August 03, 2014, 06:01:22 pm
quick question

how did they get that? (please see attachment)

i'm trying to graph the graph y = (1/x^2)/(1+x^2)

thank you

Hey,
you can use long division.
Perhaps to make it more visible, see it as -x^2+1 divided by x^2+1 and you can use long division from there.
Title: Re: random question
Post by: keltingmeith on August 03, 2014, 08:10:41 pm: Another, quicker method:

$\frac{1-x^2}{1+x^2}=\frac{-(x^2+1)+2}{x^2+1}=\frac{-(x^2+1)}{x^2+1}+\frac{2}{x^2+1}=-1+\frac{2}{x^2+1}$
Title: Re: random question
Post by: yang_dong on August 04, 2014, 11:59:25 am: thank you

i have another quick question

how would i differentiate tan^-1(c/x)?

i know that if it as tan^-1(x/c) i could go 1/(x^2+c^2), but this is the other way around?
Title: Re: random question
Post by: keltingmeith on August 04, 2014, 12:44:21 pm: tan^-1(c/x)=tan^-1(cx^-1)

Then let u= cx^-1, giving us tan^-1(u) and apply the chain rule.
Title: Re: random question
Post by: IndefatigableLover on August 04, 2014, 12:47:58 pm: Quote from: yang_dong on August 04, 2014, 11:59:25 am
thank you

i have another quick question

how would i differentiate tan^-1(c/x)?

i know that if it as tan^-1(x/c) i could go 1/(x^2+c^2), but this is the other way around?
Alright I'll do my best to explain what I'm doing so bear with me.

You know that the derivative of $\tan^{-1}(\frac{x}{c})=\frac{c}{c^2+x^2}$

However if we switched it around then we would get:

$y=\tan^{-1}(\frac{c}{x})$

$\therefore \frac{c}{x}=\tan(y)$

$\therefore x=\frac{c}{\tan(y)}$

From here we'll derive our equation which you should get as:

$\frac{dx}{dy}=\frac{-c}{\sin(y)^2}$

Here we use a trig identity to get rid of the sin(y)^2 part (so change it into cosec^2(y) before changing it into 1+cot^2(y))

$\therefore \frac{dx}{dy}=-c\cdot(1+\frac{1}{\tan(y)^2})$

I think it helps but draw a triangle out and then label the sides accordingly for what tan(y) was originally (I'll upload an image later if you don't get it) but we can see that tan(y)^2 is equal to (c/x)^2

$\therefore \frac{dx}{dy}=-c\cdot(1+\frac{1}{(\frac{c}{x})^2})$

Expand it all out:

$\therefore \frac{dx}{dy}=-c-\frac{cx^2}{c^2}$

$\therefore \frac{dx}{dy}=\frac{-(x^2+c^2)}{c}$

Flip it around and that's your final answer!

$\therefore \frac{dy}{dx}=\frac{-c}{x^2+c^2}$

(Damn takes a long time typing it all up in $\LaTeX$ but beaten by EulerFan101)
Title: Re: random question
Post by: yang_dong on August 04, 2014, 09:55:59 pm: thank you so much

how would i find the antidiff of x/sqrt(25-x^2)
Title: Re: random question
Post by: Jacyan on August 04, 2014, 10:12:56 pm: substitute u = 25 - x^2, hence du = -2x dx. Then integrate -0.5/sqrt(u) and sub back in u. You should get -sqrt(25 - x^2) + c
Title: Re: random question
Post by: Conic on August 04, 2014, 10:17:56 pm: Beaten. :-\

$\textrm{Let }u= 25-x^2.$

$u=25-x^2 \implies \frac{du}{dx}=-2x.$

$\int \frac{x}{\sqrt{25-x^2}}\:dx=\int-\frac{1}{2}\cdot\frac{du}{dx}\cdot \frac{1}{\sqrt{u}}\:du$

$=-\frac{1}{2} \int u^{-\frac{1}{2}}\:du$

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

$=-\sqrt{25-x^2}+C\quad(C\in\mathbb{R}).$