Author Topic: Mathematics Question Thread (Read 1301851 times) Tweet Share

RuiAce · « **Reply #315 on:** August 13, 2016, 05:30:46 pm »

Quote from: olivercutbill on August 13, 2016, 04:39:01 pm

Hey guys-

Can anyone help me understand this question?

$\text{The differential equation is the equation demonstrating the behaviour of logistical growth,}\\ \text{which is a growth that decreases in rate as the population goes to a maximum carrying capacity.}$
$\text{We are asked to show that the ugly thing satisfies the equation. So we just do the same process.}\\ \textit{The only difference is that this question is appreciably harder than that of natural growth due to the messy algebra.}$
$\text{Factorise the solution so that what we're trying to prove is slightly clearer.}\\ \frac{dN}{dt}=N(k-bN)\\ \text{Keep in mind we want to FORCE out the }N$
$\begin{align*}N&=\frac{kN_0}{bN_0+(k-bN_0)e^{-kt}}\\ &= kN_0\left({bN_0+(k-bN_0)e^{-kt}}\right)^{-1}\\ \frac{dN}{dt}&=-kN_0\left({bN_0+(k-bN_0)e^{-kt}}\right)^{-2} \times -k\left(k-bN_0\right)e^{-kt}\\ &= \frac{k^2N_0(k-bN_0)e^{-kt}}{\left({bN_0+(k-bN_0)e^{-kt}}\right)^{-2}}\end{align*}$
$\text{If we're careful enough, we can perfectly factorise out the }N\\ \begin{align*}\implies \frac{dN}{dt}&=\left(\frac{kN_0}{bN_0+(k-bN_0)e^{-kt}}\right)\left(\frac{k(k-bN_0)e^{-kt}}{bN_0+(k-bN_0)e^{-kt}}\right)\\ &= N\left(\frac{k(k-bN_0)e^{-kt}}{bN_0+(k-bN_0)e^{-kt}}\right)\end{align*}$
$\text{So now, we just deal with what's in the brackets.}\\ \text{For the sake of my sanity, I worked backwards on paper. But once you figure out how to work backwards, it's easy to work forwards.}\\ \text{The add-subtract trick comes in handy here in working forwards.}$
$\begin{align*}\frac{dN}{dt}&= N\left(\frac{k(k-bN_0)e^{-kt}}{bN_0+(k-bN_0)e^{-kt}}\right)\\ &=N\left(\frac{kbN_0+k(k-bN_0)e^{-kt}}{bN_0+(k-bN_0)e^{-kt}}-\frac{kbN_0}{bN_0+(k-bN_0)e^{-kt}}\right)\\ &= N\left(k\left(\frac{bN_0+(k-bN_0)e^{-kt}}{bN_0+(k-bN_0)e^{-kt}}\right)-b\left(\frac{kN_0}{bN_0+(k-bN_0)e^{-kt}}\right)\right)\\ &= N(k-bN)\\ &= kN - bN^2\end{align*}$

olivercutbill · « **Reply #316 on:** August 15, 2016, 07:17:27 pm »

(from above)

Thanks for that guys!

conic curve · « **Reply #317 on:** August 17, 2016, 10:52:50 am »

Help with iv and v

RuiAce · « **Reply #318 on:** August 17, 2016, 11:13:02 am »

Quote from: conic curve on August 17, 2016, 10:52:50 am

Help with iv and v

$\text{Using trigonometry in }\triangle PQS\text{ we have that }\\ \sin \theta = \frac{PS}{\ell} \implies PS=\ell \sin \theta$
$\text{Then the area of the triangle can be computed using just }A=\frac{1}{2}bh\\ A=\frac{1}{2} \times (\ell \sin \theta) \times (2\ell \cos \theta) = \ell^2 \cos \theta \sin \theta$
$\text{We keep in mind that }0 < \ell < \frac{\pi}{2}\text{ when we do our calculus, as otherwise we have no triangle.}\\ \begin{align*}A&=\ell^2 \cos \theta \sin \theta\\\frac{dA}{d\ell}&= \ell^2 \left(\cos^2 \theta - \sin^2 \theta\right)\end{align*}$
$\text{To maximise the area, set }\frac{dA}{d\ell}=0\\ \ell^2 (\cos^2\theta - \sin^2\theta) = 0 \implies \tan^2 \theta = 1\\ \text{Keeping in mind }0<\theta < \frac{\pi}{2}\text{, solving this gives }\theta = \frac{\pi}{4}$
$\text{You can test that it is a maximum (and not a minimum) yourself}$

conic curve · « **Reply #319 on:** August 17, 2016, 01:04:05 pm »

Need help for sketching the following:

y= -x^3/6 + x^2/3 - x/6

I did y' and got -3x^2/6 +4x/6 -1/6=0 and when I try to solve for the x intercepts, I can't seem to find them

Also I'm having trouble drawing a y dash table in order to graph it, like what are the steps to doing that and how do I fill it in?

Thanks guys

jamonwindeyer · « **Reply #320 on:** August 17, 2016, 01:42:19 pm »

Quote from: conic curve on August 17, 2016, 01:04:05 pm

Need help for sketching the following:

y= -x^3/6 + x^2/3 - x/6

I did y' and got -3x^2/6 +4x/6 -1/6=0 and when I try to solve for the x intercepts, I can't seem to find them

Also I'm having trouble drawing a y dash table in order to graph it, like what are the steps to doing that and how do I fill it in?

Thanks guys

Hey conic! Derivative looks good, remember that solving for y'=0 isn't solving for x-intercepts, it is solving for turning points:

$\frac{dy}{dx}=\frac{-3x^2}{6}+\frac{4x}{6}-\frac{x}{6} \\ \frac{dy}{dx}=0 \implies -3x^2+4x-1=0 \\ (1-3x)(x-1)=0\\\therefore x=\frac{1}{3}\text{ or }1$

We now test their nature. You can use one of the y' tables as you call it, or more conveniently, we can use the second derivative test:

$\frac{d^2y}{dx^2}=-x+\frac{2}{3} \\\text{At }x=1\text{, }\frac{d^2y}{dx^2}<0\text{, therefore a maximum exists at x=1}\\\text{At }x=\frac{1}{3}\text{, }\frac{d^2y}{dx^2}>0\text{, therefore a minimum exists at }x=\frac{1}{3}$

Is this a test you are familiar with? What you suggested is comparing the sign of the derivative either side of the turning point, but there is an easier way. Basically, it is using a concavity test to determine the nature of a stationary point. If it is concave down, it is a maximum, if it is concave down, it is a minimum

Finding the x-intercepts is done by equating the initial expression to 0. Remember, the x intercept is where y=0:

$\frac{-x^3}{6}+\frac{x^2}{3}-\frac{x}{6}=0 \\\therefore -x^3+2x^2-x=0\\-x(x^2-2x+1)=0 \\\therefore x=0,1$

Putting all of that (x-intercepts and turning points) together, you should be able to sketch something like this

Hope that makes sense

anotherworld2b · « **Reply #321 on:** August 19, 2016, 12:58:22 am »

Hi

Could I please have help with solving these three questions? I apologise if i have put too many questions at once

conic curve · « **Reply #322 on:** August 19, 2016, 06:34:13 am »

RuiAce · « **Reply #323 on:** August 19, 2016, 08:08:23 am »

Quote from: conic curve on August 19, 2016, 06:34:13 am

(Image removed from quote.)

Nice, but I have a few comments.
$\text{All of your }x\text{'s look like }u\text{'s...}\\ \text{For the first one, I'm sure you saw and just didn't bother writing it but }\frac{2}{8}=\frac{1}{4}\\ \text{The most important one is part 3. This is only half correct. The answer is }x=\pm 3\\ \text{Note that for an even power, the plus or minus sign is important.}$
Also for part 3, I'm not sure if you picked up on it but you don't have to FOIL out the whole thing. It's just difference of two squares.

anotherworld2b · « **Reply #324 on:** August 19, 2016, 11:22:38 pm »

thank you for the help conic curve

I was wondering how exactly did the surds disappear to equal 2/8?
I also wanted to ask how do you know when to use a plus or minus sign? I've only learnt that its only for when the power is to 2?

Quote from: conic curve on August 19, 2016, 06:34:13 am

(Image removed from quote.)

Mod edit: Swapped over the quote.

RuiAce · « **Reply #325 on:** August 19, 2016, 11:50:18 pm »

Quote from: anotherworld2b on August 19, 2016, 11:22:38 pm

thank you for the help
I was wondering how exactly did the surds disappear to equal 2/8?
I also wanted to ask how do you know when to use a plus or minus sign? I've only learnt that its only for when the power is to 2?

Mod edit: Took out the quote. I'm not the one who deserves the thanks here.

$\text{This is just because }2\sqrt{2}=\sqrt8\\ \text{So by rewriting:}\\ \begin{align*}2\sqrt{2x}&=\sqrt{2}\\ \sqrt{8x}&=\sqrt{2}\\ 8x&=2\\ x&=\frac{2}{8}=\frac{1}{4}\end{align*}$

anotherworld2b · « **Reply #326 on:** August 20, 2016, 12:31:07 am »

I was also wondering how would you do these 2 questions? I am not sure how to approach them

jamonwindeyer · « **Reply #327 on:** August 20, 2016, 12:59:06 am »

Quote from: anotherworld2b on August 20, 2016, 12:31:07 am

I was also wondering how would you do these 2 questions? I am not sure how to approach them

You might make both a bit easier by making a substation for yourself!

In the first:

$m=5^x\implies (25m-1)(m-1)=0\\\therefore m=\frac{1}{25},1\\\therefore5^x=\frac{1}{25},1\\x=-2,1$

Have a go at a similar thing in the second one! Make a substitution to turn it into a quadratic equation, and see if you can follow the same procedure as me

let me know if you need more of a hand with it

anotherworld2b · « **Reply #328 on:** August 20, 2016, 01:48:05 am »

Thank you for your help

i was able to get the answer for the second one by following your example.

I also wanted to ask for a question like this how should you approach it? Im not sure where i should and how to begin

Quote from: jamonwindeyer on August 20, 2016, 12:59:06 am

You might make both a bit easier by making a substation for yourself!

In the first:

$m=5^x\implies (25m-1)(m-1)=0\\\therefore m=\frac{1}{25},1\\\therefore5^x=\frac{1}{25},1\\x=-2,1$

Have a go at a similar thing in the second one! Make a substitution to turn it into a quadratic equation, and see if you can follow the same procedure as me let me know if you need more of a hand with it

conic curve · « **Reply #329 on:** August 20, 2016, 07:55:48 am »

Quote from: anotherworld2b on August 19, 2016, 11:22:38 pm

thank you for the help conic curve
I was wondering how exactly did the surds disappear to equal 2/8?
I also wanted to ask how do you know when to use a plus or minus sign? I've only learnt that its only for when the power is to 2?Mod edit: Swapped over the quote.

nws

Usually if there is a constant (number) to the power of an even index (i.e. power) it is usually even

e.g. -1 to the power of 4 is equal to 1 right because -1 x -1 x -1 x-1 =1 (remember two negatives make a positive)

if it was -1 to the power of 5 the answer would be -1 (obviously)

2Squareroot of 2x=8

Search the forums now!

We have moved!

Author Topic: Mathematics Question Thread (Read 1301851 times) Tweet Share

RuiAce

Re: Mathematics Question Thread

olivercutbill

Re: Mathematics Question Thread

conic curve

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

conic curve

Re: Mathematics Question Thread

jamonwindeyer

Re: Mathematics Question Thread

anotherworld2b

Re: Mathematics Question Thread

conic curve

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

anotherworld2b

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

anotherworld2b

Re: Mathematics Question Thread

jamonwindeyer

Re: Mathematics Question Thread

anotherworld2b

Re: Mathematics Question Thread

conic curve

Re: Mathematics Question Thread

Recent Posts

User:
Password: