Author Topic: VCE Methods Question Thread! (Read 6089063 times) Tweet Share

cosine · « **Reply #11655 on:** August 06, 2015, 05:56:59 pm »

Assistance pls

odeaa · « **Reply #11656 on:** August 06, 2015, 06:46:09 pm »

In a certain town, the rate of deaths due to a particular disease has been found to closely fit the mathematical model $\frac{dN}{dt}=\frac{10,000}{(t+3)^{3}}$ , where $t$ is the time in months after the disease was first detected and $N$ is the number of deaths.

Calculate the total number of deaths predicted by the model, giving your answer to the nearest whole number of deaths

How do you go about this? Can someone please explain the theory behind it, I though you would need to antiderive it twice, once to find N(x) and then again to find the area. I'm a bit lost here

cosine · « **Reply #11657 on:** August 06, 2015, 07:16:53 pm »

Quote from: odeaa on August 06, 2015, 06:46:09 pm

In a certain town, the rate of deaths due to a particular disease has been found to closely fit the mathematical model $\frac{dN}{dt}=\frac{10,000}{(t+3)^{3}}$ , where $t$ is the time in months after the disease was first detected and $N$ is the number of deaths.

Calculate the total number of deaths predicted by the model, giving your answer to the nearest whole number of deaths

How do you go about this? Can someone please explain the theory behind it, I though you would need to antiderive it twice, once to find N(x) and then again to find the area. I'm a bit lost here

$\frac{dN}{dt}=\frac{10,000}{(t+3)^{3}}$

$\frac{dN}{dt} = 10,000(t+3)^{-3}$

$N = 10,000 \int (t+3)^{-3}dt$

$N = 10,000 * \frac{(t+3)^{-2}}{-2} + c$

$N = \frac{-5000}{(t+3)^2} + c$

Is there supposed to be a number to plug in? What do you mean just 'number of deaths', im pretty sure this function is defined for unlimited positive numbers...

odeaa · « **Reply #11658 on:** August 06, 2015, 07:34:34 pm »

Quote from: cosine on August 06, 2015, 07:16:53 pm

heres the worked solution, i guess as t aproaches infinity, that part of the equation will be 0

$\frac{dN}{dt}=\frac{10,000}{(t+3)^{3}}$

$N=\int_{0}^{\infty}\frac{10,000}{(t+3)^3}dt$

$[\frac{-5000}{(t+3)^3}]_{0}^{\infty}$

$N=0+\frac{5000}{9}=556$

I just dont get why you have to integrate the derivative, would that just give you the original function (without the c), and then you would have to integrate that?

Is there supposed to be a number to plug in? What do you mean just 'number of deaths', im pretty sure this function is defined for unlimited positive numbers...

This is the worked solutions, which I dont really understand but yah

IntelxD · « **Reply #11659 on:** August 06, 2015, 07:52:31 pm »

Quote from: odeaa on August 06, 2015, 07:34:34 pm

This is the worked solutions, which I dont really understand but yah

The worked solution sort of completes the question in one step which can be confusing, I'll try to break it down for you with a more generic approach.

You're given dN/dT which is the rate of change of deaths with respect to time. The question has asked you to determine the total number of deaths predicted by the model. In order to do so, you must first find the number of deaths as a function of time, N(t). To do this, we integrate the function.Our result is N=-5000(t+3)^-3 + c. The constant here is easy to determine as we know that there should be 0 deaths before the disease has been detected. Sub 0 into both N and t to find c. Now, since the question has asked for the total number of deaths, we need to find the value of N for the long term. We can do this by finding the limit of the function as t approaches infinity. As t approaches infinity, the fraction will approach 0 (as t is in the denominator) leaving us with a value that represents the total number of deaths predicted by the function.

odeaa · « **Reply #11660 on:** August 06, 2015, 08:02:26 pm »

Quote from: IntelxD on August 06, 2015, 07:52:31 pm

The worked solution sort of completes the question in one step which can be confusing, I'll try to break it down for you with a more generic approach.

You're given dN/dT which is the rate of change of deaths with respect to time. The question has asked you to determine the total number of deaths predicted by the model. In order to do so, you must first find the number of deaths as a function of time, N(t). To do this, we integrate the function.Our result is N=-5000(t+3)^-3 + c. The constant here is easy to determine as we know that there should be 0 deaths before the disease has been detected. Sub 0 into both N and t to find c. Now, since the question has asked for the total number of deaths, we need to find the value of N for the long term. We can do this by finding the limit of the function as t approaches infinity. As t approaches infinity, the fraction will approach 0 (as t is in the denominator) leaving us with a value that represents the total number of deaths predicted by the function.

That made it so easy to understand, thanks man! Legend!

cosine · « **Reply #11661 on:** August 06, 2015, 08:21:30 pm »

To antidifferentiate this, is it recognition? Because I have troubles recognising the fractions...

StupidProdigy · « **Reply #11662 on:** August 06, 2015, 08:27:51 pm »

Quote from: cosine on August 06, 2015, 08:21:30 pm

To antidifferentiate this, is it recognition? Because I have troubles recognising the fractions...

Nah just long divide or use whatever method to simplify this fraction, this will break it up into two manageable expressions to integrate. Is that what you wanted to know?

thanks for the help on some of my other questions this week too cosine

Splash-Tackle-Flail · « **Reply #11663 on:** August 06, 2015, 08:29:43 pm »

Quote from: cosine on August 06, 2015, 08:21:30 pm

To antidifferentiate this, is it recognition? Because I have troubles recognising the fractions...
[/quote

When the numerator and denominator are of the same degree, you can divide them. or do:
$\frac{2x+1}{x+1}=\frac{2(x+1)-1}{x+1}=\frac{2(x+1)}{x+1} -\frac{1}{x+1}=2-\frac{1}{x+1}$ Then antidiffrentiate as usual

cosine · « **Reply #11664 on:** August 06, 2015, 08:38:08 pm »

Cheers guys.

With the attached one, I get 5/(1-2x) inside the log, but not sure how they get it... ?

Floatzel98 · « **Reply #11665 on:** August 06, 2015, 08:50:57 pm »

Quote from: cosine on August 06, 2015, 08:38:08 pm

Cheers guys.

With the attached one, I get 5/(1-2x) inside the log, but not sure how they get it... ?

It is inside an absolute value so the 2 expressions are equivalent.

cosine · « **Reply #11666 on:** August 06, 2015, 08:59:59 pm »

Quote from: Floatzel98 on August 06, 2015, 08:50:57 pm

It is inside an absolute value so the 2 expressions are equivalent.

Yes, but I am asking for an algebraic explanation, please.

Floatzel98 · « **Reply #11667 on:** August 06, 2015, 09:15:19 pm »

Quote from: cosine on August 06, 2015, 08:59:59 pm

Yes, but I am asking for an algebraic explanation, please.

I might be a bit off track on what you are actually asking again, but if you take a negative 2 out of the denominator instead of a 2, you get the answer the book gets. So:

$y = 5\int \frac{1}{2 - 4x}dx$

$y = 5\int \frac{1}{-2(-1 + 2x)}dx$

$y = -\frac{5}{2}\int \frac{1}{2x - 1}dx$

Follow the normal procedure from there and you should end up with that answer. Hopefully that helps. Sorry otherwise

cosine · « **Reply #11668 on:** August 06, 2015, 09:16:52 pm »

Quote from: Floatzel98 on August 06, 2015, 09:15:19 pm

I might be a bit off track on what you are actually asking again, but if you take a negative 2 out of the denominator instead of a 2, you get the answer the book gets. So:

$y = 5\int \frac{1}{2 - 4x}dx$

$y = 5\int \frac{1}{-2(-1 + 2x)}dx$

$y = -\frac{5}{2}\int \frac{1}{2x - 1}dx$

Follow the normal procedure from there and you should end up with that answer. Hopefully that helps. Sorry otherwise

My question is why my log function is different from the answer.. not sure what you're doing man?

e^1 · « **Reply #11669 on:** August 06, 2015, 09:46:17 pm »

Quote from: cosine on August 06, 2015, 09:16:52 pm

My question is why my log function is different from the answer.. not sure what you're doing man?

I'm a bit confused on what is being asked, so I will explain the process as well.
Continuing on with what has been said, we know that $\int {\frac{f'(x)}{f(x)}} dx = \ln(|f(x)|)+ C$ .

In this case, we have $I(x) = \frac{5}{2 - 4x} = -\frac{5}{2}\int {\frac{1}{2x - 1}} dx$ and so we see that $f(x)=2x-1$ . However to use the above integral, we need $f'(x) = 2$ . To get this, multiply both numerator and denominator by 2. Since $\frac{2}{2} = 1$ , it is safe to say that $I(x)$ still equals the right hand side multiplied by $\frac{2}{2}$ .

$I(x) = -\frac{5}{2}\int {\frac{2}{2} \cdot\frac{1}{2x - 1}} dx$

$= -\frac{5}{4}\int {\frac{2}{2x - 1}} dx$
$= -\frac{5}{4}\ln(\left|2x - 1\right|)} + C$
$= \frac{5}{4}\ln\left(\frac{1}{\left|2x - 1\right|}}\right) + C$ (log rules)

Substituting x = -2 and y = 10 should give you $C= 10 + \frac{5}{4}\ln(5)$ , resulting in:

$= \frac{5}{4}\ln\left(\frac{1}{\left|2x - 1\right|}}\right) + 10 + \frac{5}{4}\ln(5)$

$= \frac{5}{4}\ln\left(\frac{5}{\left|2x - 1\right|}}\right) + 10$ (with use of log rules).

$= \frac{5}{4}\ln\left(\left|\frac{5}{2x - 1}}\right|\right) + 10$ (since $|ab| = |a||b|$ and $|5| = 5$ ), as desired.

Moreover, for any $a, b \in \mathbb{R}$ . We have $|ab| = |a||b|$ .
Thus $|-a| = |-1||a| = |a|$ . So we can write $|2x - 1| = |-(1 - 2x)| = |1 - 2x|$ and thus:

$\frac{5}{4}\ln\left(\left|\frac{5}{2x - 1}}\right|\right) + 10 = \frac{5}{4}\ln\left(\left|\frac{5}{1 - 2x}}\right|\right) + 10$

If you wanted some explanation/proof on why $|ab| = |a||b|$ works, it is in spoiler.

Proof that |ab| = |a||b|

Proof that $|ab| = |a||b|$ :

Let $x\in \mathbb{R}$ and define $|x| = \begin{cases}x, & \text{if } x\geq 0 \\ -x, & \text{if } x < 0\end{cases}$ .

Given $a,b \in \mathbb{R}$ , we will proof by cases, namely:

1. $a \geq 0, b \geq 0$
2. $a \geq 0, b < 0$
3. $a < 0, b \geq 0$
4. $a < 0, b < 0$

In each case, assume these inequalities. We will show that in each case, $|ab| = |a||b|$ holds in that case.

Case 1: $a \geq 0, b \geq 0$
This implies that also $(ab) \geq 0$ . Therefore $|(ab)| = (ab)$ .
Moreover $|a||b| = ab$ and thus $|ab| = |a||b|$ .

Case 2: $a \geq 0, b < 0$
The two inequalities implies that $(ab) \leq 0$ . Therefore $|(ab)| = -(ab)$ .
Furthermore $|a||b| = a(-b) = -(ab)$ and thus $|ab| = |a||b|$ , in this case.
Case 2 also proves case 3, by symmetry.

Case 4: $a < 0, b < 0$
The two inequalities imply that $(ab) > 0$ . Thus $|(ab)| = (ab)$ .
Moreover $|a||b| = ab$ and therefore $|ab| = |a||b|$ , for case 4.

Since we have proved the four cases to be true, the property that $|ab| = |a||b|$ holds for all $a, b \in \mathbb{R}$ .

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Author Topic: VCE Methods Question Thread! (Read 6089063 times) Tweet Share

cosine

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