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Uni Stuff => Universities - Victoria => Monash University => Topic started by: alondouek on October 04, 2013, 01:09:13 am

Title: Maths question [MTH1020]
Post by: alondouek on October 04, 2013, 01:09:13 am

So I have a question;

Find $\int_{0}^{3}\sqrt{9-x^2}dx$ using trigonometric substitution.

I got $\frac{9\pi}{4}$ , which I think is right, but I'm not too sure if it is or how I got there lol

Can someone please help?

Title: Re: Maths question [MTH1020]
Post by: b^3 on October 04, 2013, 01:19:59 am

We want to turn the $\sqrt{9-x^{2}}$ into something we can work with, so we will make use of the trigonometric identity $\sin^{2}\left(x\right)+\cos^{2}\left(x\right)=1$ , which becomes $\cos^{2}\left(x\right)=1-\sin^{2}\left(x\right)$ . To make the inside of the square root look like this we will try a substitution. Lets try $x=3\sin(u)$ .

Spoiler

$\begin{alignedat}{1}\text{Let }x=3\sin\left(u\right) \\ 9-x^{2} & =9-\left(3\sin\left(u\right)\right)^{2} \\ & =9\left(1-\sin^{2}\left(u\right)\right) \\ & =9\cos^{2}\left(u\right) \\ \sqrt{9-x^{2}} & =\sqrt{9\cos^{2}\left(u\right)} \\ & =3|\cos\left(u\right)| \\ \cos\left(u\right)\geq0\text{ for }0\leq x\leq3, \left(0\leq u\leq \frac{\pi}{2}\right) \\ \implies\sqrt{9-x^{2}} & =3\cos\left(u\right) \\ dx & =3\cos\left(u\right)du \\ x=0,\: u=\sin^{-1}\left(0\right)=0 \\ x=3,\: u=\sin^{-1}\left(1\right)=\frac{\pi}{2} \\ \int_{0}^{3}\sqrt{9-x^{2}}dx & =\int_{0}^{\frac{\pi}{2}}3\cos\left(u\right)\times3\cos\left(u\right)du \\ & =9\int_{0}^{\frac{\pi}{2}}\cos^{2}\left(u\right)du \\ \cos^{2}\left(u\right) & =\frac{1+2\cos\left(2u\right)}{2} \\ \int_{0}^{3}\sqrt{9-x^{2}}dx & =9\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{2}+\cos\left(2u\right)\right)du \\ & =9\left[\frac{1}{2}u+\frac{1}{2}\sin\left(2u\right)\right]_{0}^{\frac{\pi}{2}} \\ & =9\left(\frac{\pi}{4}+0-0\right) \\ & =\frac{9\pi}{4} \end{alignedat}$

Which gives the same answer you had.

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 04, 2013, 01:28:06 am

Thanks b^3, really appreciate it!

Why exactly are we substituting $x=3sin(u)$ ? Would it be equally valid to use $x=3cos(u)$ , as $sin^2(x)=1-cos^2(x)$ ?

(I'm assuming the 3 is there because of the 9 in the square-root expression? I don't really have a great handle on this theory)

Title: Re: Maths question [MTH1020]
Post by: b^3 on October 04, 2013, 01:31:58 am

$x=3\cos(u)$ should work as well (just be careful of the negative from the derivative). We have the $3$ there because we want to form " $1-\sin^{2}(u)$ ", if we just substituted in $x=\sin(u)$ , then we'd end up with $\sqrt{9-\sin^{2}(u)}$ , which is not what we want as it doesn't satisfy our trigonometric identity. But what do we have to do to satisfy it? We need to get a factor of $9$ out the front of the $\sin^{2}(u)$ , so that we can take a factor of $9$ out of both of the terms, leaving $1-\sin^{2}(u)$ , which is exactly what we want. We can now turn that into our $\cos^{2}(x)$ .

EDIT: Had sines and cosines flipped around, fixed now.

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 04, 2013, 01:36:06 am

Thanks again, great explanation :)

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 05, 2013, 10:05:48 pm

So I was thinking about this over the weekend, and I'm still kind of stumped by it.

How do we know which trigonometric expression to substitute? Is there a defined method of selecting one, or is it just something you ascertain from what you're substituting?

tbh I'm finding trig substitution as a whole quite puzzling

Title: Re: Maths question [MTH1020]
Post by: brightsky on October 05, 2013, 10:14:06 pm

just think logically, and try stuff out. say you got an integrand sqrt(a^2-x^2). now won't it be nice if you could get rid of that square root? to do that, we need some square number under the square root. how do we obtain that? use a trig identity! which one? well the expression looks roughly like 1-something^2. ring any bells? use identity cos^2 (x) = 1-sin^2(x) or the like.

trig substitution doesn't always work though. when you get more complicated integrands, you'll just have to think about all the integration techniques in your toolbox and try stuff out.

Title: Re: Maths question [MTH1020]
Post by: b^3 on October 05, 2013, 10:14:48 pm

You want to pick a substitution that changes the square root into something that you can work with. We make use of the relevant trigonometric identity to get the square root of a trig squared, which gives us a mod. Then normally we'd be integrating over a domain where the sign of this mod doesn't change, allowing us to remove the mod and pick the appropriate sign.
It pretty much boils down to this in the end.

(http://calculus.nipissingu.ca/tutorials/integralgifs/trig_sub_table.gif)

EDIT: Beaten, why'd I decided to spend time centering the image :P

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 05, 2013, 10:22:05 pm

Amazing! Huge thanks to both of you :)

Title: Re: Maths question [MTH1020]
Post by: lzxnl on October 05, 2013, 11:10:28 pm

Quote from: alondouek on October 04, 2013, 01:09:13 am

So I have a question;

Find $\int_{0}^{3}\sqrt{9-x^2}dx$ using trigonometric substitution.

I got $\frac{9\pi}{4}$ , which I think is right, but I'm not too sure if it is or how I got there lol

Can someone please help?

I have an easier way of doing that.
We want the area of half a semicircle as your function is that of a semicircle, while the integration bounds span half the domain. So you just have pi*r^2/4 = 9pi/4
Is this considered cheating?

Alternatively you CAN do this integral by parts...although that might be overkill for this question.

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 06, 2013, 03:06:26 am

Don't know if anyone is still awake, but how can I tell when to use either the cylindrical-shell method or the disk method to find a volume by integration?

Title: Re: Maths question [MTH1020]
Post by: BigAl on October 06, 2013, 06:44:08 am

You can use a right angled triangle to figure out the identities. I didn't do mth1020 but from what I've seen on YouTube, you often see fractions in this kind of problems...check prof burgers videos on YouTube. They are the best for introductory calculus

Title: Re: Maths question [MTH1020]
Post by: brightsky on October 06, 2013, 10:27:05 am

Quote from: alondouek on October 06, 2013, 03:06:26 am

Don't know if anyone is still awake, but how can I tell when to use either the cylindrical-shell method or the disk method to find a volume by integration?

shell method is used when the terminals are parallel to the axis about which you are rotating. disk method is used when the terminals are perpendicular to the axis about which you are rotating.

for example, say you want to calculate the volume of the solid produced when you rotate the area bounded by lnx, x = 2, x=4 and the x-axis about the y-axis. do you see how the terminals x=2 and x=4 are parallel to the y-axis (the axis about which you are rotating)? in cases like this, use shell integration. however, if you wanted to calculate the volume of the solid produced when you rotate this area about the x-axis, you would use disk method, as per usual. do you see how the terminals x= 2 and x=4 are perpendicular to the x-axis?

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 06, 2013, 01:48:47 pm

Thanks brightsky! :)

So if I'm rotating around the x-axis the function y=x^3, enclosed by y=1, then I should use the shell method as y=1 is parallel to the axis of rotation?

Title: Re: Maths question [MTH1020]
Post by: lzxnl on October 06, 2013, 02:59:58 pm

Quote from: alondouek on October 06, 2013, 01:48:47 pm

Thanks brisghtsky! :)

So if I'm rotating around the x-axis the function y=x^3, enclosed by y=1, then I should use the shell method as y=1 is parallel to the axis of rotation?

That's not quite enough info...what else are you given? You can't just have the one bound.

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 06, 2013, 03:14:31 pm

Haha yeah, I forgot to mention the y-axis is also a bound

Title: Re: Maths question [MTH1020]
Post by: lzxnl on October 06, 2013, 04:00:14 pm

In that case, you could do it using VCE methods as well; large cylinder formed by rotating y=1 about the x axis minus the solid formed by rotating the curve bound by x=0, x=1 and y=x^3 around the x axis.
Which from my quick calculations seems to yield pi-pi/7=6pi/7

While the other way requires you to find x in terms of y; the integral is 2pi*integral of xy dy = 2pi*integral of y^4/3 dy from 0 to 1 = 6pi/7 as well.

Doesn't really matter which method you use.

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 06, 2013, 04:14:32 pm

hmm, I got something a little different; for the question, I need to take the area enclosed by y=x³, y=1 and the y-axis, then revolve that area around the x-axis.

So, using the shell method, I got:

$V=\int_{0}^{1}2\pi x (x^3) dx$
$=2\pi\int_{0}^{1}x^4dx$
$=2\pi\times \left [\frac{x^5}{5} \right ]_{0}^{1}$
$=2\pi (\frac{1}{5}-0)$
$=\frac{2\pi}{5} units^3$

Title: Re: Maths question [MTH1020]
Post by: lzxnl on October 06, 2013, 04:45:34 pm

Quote from: alondouek on October 06, 2013, 04:14:32 pm

hmm, I got something a little different; for the question, I need to take the area enclosed by y=x³, y=1 and the y-axis, then revolve that area around the x-axis.

So, using the shell method, I got:

$V=\int_{0}^{1}2\pi x (x^3) dx$
$=2\pi\int_{0}^{1}x^4dx$
$=2\pi\times \left [\frac{x^5}{5} \right ]_{0}^{1}$
$=2\pi (\frac{1}{5}-0)$
$=\frac{2\pi}{5} units^3$

http://en.wikipedia.org/wiki/Shell_integration

Says that rotating around the x axis means you integrate with respect to y.
Which makes sense. The formula looks to me like integrating 2pi*r*h.
When revolving around the x axis, if you cut up your volume into tiny slices:
Each slice can be seen as a thin, tiny cylinder for a given y value, with radius y, length x and width dy. The area of one of these slices is 2pi*y*x, so integrating yields 2pi*y*x dy.

Draw up a diagram and it'll make sense.

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 06, 2013, 04:53:10 pm

Okay, that makes sense :) Thanks!

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 06, 2013, 06:06:18 pm

Problem though, even when I try to use the formula from wiki: $V=\int_{a}^{b}2\pi y \times f(y)dy$

I still end up with my prior (presumably wrong) solution;

$f(y)=y^3 as f(x)=y=x^3$

$=2\pi \int_{0}^{1} y \times y^3 dy$

$=2\pi \left [ \frac{y^5}{5} \right ]_{0}^{1}$

$=\frac{2\pi}{5}$

What am I doing wrong?

Title: Re: Maths question [MTH1020]
Post by: lzxnl on October 06, 2013, 06:31:21 pm

Quote from: alondouek on October 06, 2013, 06:06:18 pm

Problem though, even when I try to use the formula from wiki: $V=\int_{a}^{b}2\pi y \times f(y)dy$

I still end up with my prior (presumably wrong) solution;

$f(y)=y^3 as f(x)=y=x^3$

$=2\pi \int_{0}^{1} y \times y^3 dy$

$=2\pi \left [ \frac{y^5}{5} \right ]_{0}^{1}$

$=\frac{2\pi}{5}$

What am I doing wrong?

y=y
But f(y)=x=y^1/3

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 06, 2013, 06:40:58 pm

Quote from: nliu1995 on October 06, 2013, 06:31:21 pm

y=y
But f(y)=x=y^1/3

So to be clear, I'm taking the inverse of f(x), rather than just subbing y in as x?

Title: Re: Maths question [MTH1020]
Post by: lzxnl on October 06, 2013, 10:18:05 pm

According to the article I showed you, if you have something of the form y=f(x) and you revolve that, y=a and y=b, you integrate 2pi*xy dy from y=a to y=b. Remember it in that form; it's probably easier IMO to remember. Now y=f(x), so x=f^-1(y). Sub that in.

In short, just remember the formulas in terms of x and y, or even better, remember where they came from to minimise chances of error.

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 06, 2013, 10:34:18 pm

Brilliant, thanks for the elaboration :)

Title: Re: Maths question [MTH1020]
Post by: BigAl on October 07, 2013, 10:07:38 pm

The way I remembered this formula in spesh was that I formed little differential element, dx or dy, depending on what axis the function was being rotated. Youre essentially adding up little circles separated by dx or dy. by making x or y the subject youre actually changing dependent and independent value

Title: Re: Maths question [MTH1020]
Post by: lzxnl on October 07, 2013, 11:12:49 pm

This formula actually isn't in spesh, which is sort of stupid. Although the reasoning is quite simple, yeah.

Title: Re: Maths question [MTH1020]
Post by: BigAl on October 07, 2013, 11:19:38 pm

You mean it isn't on the formula sheet?

Title: Re: Maths question [MTH1020]
Post by: lzxnl on October 07, 2013, 11:30:43 pm

It's not in the course. Ask a student how to do that and they'll use the wrong formula. Or give you a blank stare.

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 07, 2013, 11:32:02 pm

Quote from: nliu1995 on October 07, 2013, 11:30:43 pm

It's not in the course. Ask a student how to do that and they'll use the wrong formula. Or give you a blank stare.

Hah! I didn't even do spesh and I can do that :P

Title: Re: Maths question [MTH1020]
Post by: b^3 on October 07, 2013, 11:33:33 pm

Quote from: alondouek on October 07, 2013, 11:32:02 pm

Hah! I didn't even do spesh and I can do that :P

That's because from what I've seen the methods and spesh 'equivalent' units at uni actually cover a bit more, and in a bit more detail as well (which is good).

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 07, 2013, 11:37:25 pm

haha I meant the blank stare part, I'm no good with the actual maths

Title: Re: Maths question [MTH1020]
Post by: BigAl on October 07, 2013, 11:38:19 pm

Sorry I didn't read the shell integration bit and assumed this was a volume of revolution question as seen in spesh

Title: Re: Maths question [MTH1020]
Post by: lzxnl on October 07, 2013, 11:44:56 pm

Quote from: b^3 on October 07, 2013, 11:33:33 pm

That's because from what I've seen the methods and spesh 'equivalent' units at uni actually cover a bit more, and in a bit more detail as well (which is good).

That shows how much more content VCE could and probably should cover. I mean, how long are the equivalent units at uni? A quarter to half the length?

Title: Re: Maths question [MTH1020]
Post by: b^3 on October 07, 2013, 11:53:35 pm

Quote from: nliu1995 on October 07, 2013, 11:44:56 pm

That shows how much more content VCE could and probably should cover. I mean, how long are the equivalent units at uni? A quarter to half the length?

Off Topic

Normally a semester, so 12 weeks+exams. They don't look that much harder, but from looking at lectures notes from friends, they go through things more thoroughly.

I remember something I said last year, in relation to a few of the units (this is after spesh units). "It's 2-3 times the amount of content, twice as hard and done in half the time!" Although that may have something to do with engineering units trying to cram a few maths units into one, minu proofs"

Now we probably shouldn't take the thread off topic anymore.

Back to maths questions.

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 07, 2013, 11:58:39 pm

Quote from: b^3 on October 07, 2013, 11:53:35 pm

Back to maths questions.

In that case, what's the best step-by-step procedure to solving differential equations? I have an assignment with a DE question due Thursday, and I don't even know what a DE is let alone how to solve one.

Help please?

Title: Re: Maths question [MTH1020]
Post by: BigAl on October 08, 2013, 12:04:53 am

10 mins before I go to sleep

Title: Re: Maths question [MTH1020]
Post by: lzxnl on October 08, 2013, 12:14:51 am

Quote from: alondouek on October 07, 2013, 11:58:39 pm

In that case, what's the best step-by-step procedure to solving differential equations? I have an assignment with a DE question due Thursday, and I don't even know what a DE is let alone how to solve one.

Help please?

Methods will seriously depend on the DE in question. There isn't a one-size-fits-all procedure for DEs. For example, y'=xy is easy to solve, but y'=x+y requires a bit more work and I'm not even sure how to solve y'=y^2+x. And these are just first-order ordinary differential equations. There are higher order DEs and partial DEs...which are horrible...

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 08, 2013, 12:23:42 am

So I need to 1. solve $\frac{dy}{dx}=\frac{4xy}{1+x^2}$

and 2. "Find the unique function $y(x)$ satisfying the following differential equation
with the given initial condition: $\frac{dy}{dx} - x^2 y = 0$ , $y(1)=1$

like, I have absolutely no idea what I'm looking at

Title: Re: Maths question [MTH1020]
Post by: lzxnl on October 08, 2013, 12:29:03 am

Quote from: alondouek on October 08, 2013, 12:23:42 am

So I need to 1. solve $\frac{dy}{dx}=\frac{4xy}{1+x^2}$

and 2. "Find the unique function $y(x)$ satisfying the following differential equation
with the given initial condition: $\frac{dy}{dx} - x^2 y = 0$ , $y(1)=1$

like, I have absolutely no idea what I'm looking at

OK these ones are part of a particular variety of DE called "separable equations".
Basically, you can rewrite these DEs as dy/dx = f(x)*g(y)

For the first one, we have 1/y*dy/dx = 4x/(1+x^2)
Integrating both sides, we get integral of dy/y = integral of 4x/(1+x^2) dx
I believe you can see pretty quickly that we get ln|ky|=2ln|1+x^2| where I've just put the arbitrary integration constant into the log.
Exponentiating both sides we get |ky|=(1+x^2)^2

For the second one, rewrite as dy/dx = x^2 y
dy/dx * 1/y = x^2
integral of dy/y = integral of x^2 dx
ln|ky|=1/3*x^3
|ky|=e^(1/3*x^3)

y=1 when x=1 so use that to find the particular solution.

The point is that you separate your DE into g(y) dy/dx = f(x)
Integrate both sides so integral of g(y) dy = integral of f(x) dx
and solve

Title: Re: Maths question [MTH1020]
Post by: BigAl on October 08, 2013, 12:29:43 am

Alright, the basic idea is to collect dx and x in one side and y and dy in one side. So for the first question you can write it as (dy/y)=(4xdx/1+x^2) now you can integrate both sides and make y as a function of x. Don't forget +c
Second one is the same but you have to solve for c this time given initial values

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 08, 2013, 12:34:20 am

Thanks! I'll write this up and see if I can make heads or tails of it :)

Quote from: nliu1995 on October 08, 2013, 12:29:03 am

...where I've just put the arbitrary integration constant into the log.

So is this arbitrary integration constant just the "+c" used with indefinite integrals?

Also, what's the point of these differential equations? What are they actually telling me? (I'm trying to understand the concept behind this as well as doing the question)

Title: Re: Maths question [MTH1020]
Post by: BigAl on October 08, 2013, 10:51:05 am

Quote from: alondouek on October 08, 2013, 12:34:20 am

Thanks! I'll write this up and see if I can make heads or tails of it :)

So is this arbitrary integration constant just the "+c" used with indefinite integrals?

Also, what's the point of these differential equations? What are they actually telling me? (I'm trying to understand the concept behind this as well as doing the question)

Yes k is the arbitrary constant. You can't always express something in the form a=cv+d sometimes you have to define relative rate of changes in one equation..I don't know what this equation is but it could be anything. You've probably heard of acceleration from F=ma. Now this is a differential equation and there are several ways you can express a such as vdv/dx

Title: Re: Maths question [MTH1020]
Post by: lzxnl on October 08, 2013, 08:12:42 pm

Quote from: alondouek on October 08, 2013, 12:34:20 am

Thanks! I'll write this up and see if I can make heads or tails of it :)

So is this arbitrary integration constant just the "+c" used with indefinite integrals?

Also, what's the point of these differential equations? What are they actually telling me? (I'm trying to understand the concept behind this as well as doing the question)

The point? Sometimes physical models are given in terms of rates. An example is, of course, F=ma=m*d^2x/dt^2
So for a spring, where F=-kx, you have d^2 x/dt^2 = -kx/m

There are also spesh questions like you have a salt solution of a given volume and concentration, liquid coming in at v1 L/min containing concentration of the same salt c1, and the solution is well mixed and poured off at v2 L/min. That's another DE question.

Title: Re: Maths question [MTH1020]
Post by: b^3 on October 08, 2013, 08:23:25 pm

As nlui has said, mostly for physical models, when you have one thing changing respect to another thing, or even a few variables changing with respect to a few other variables.

There's a few things to do with heat transfer and such, probably above the level of what you're learning now but
$\begin{alignedat}{1}\frac{\partial T}{\partial t} & =\kappa\frac{\partial^{2}T}{\partial x^{2}}\end{alignedat}$ which has a general solution $\begin{alignedat}{1}T\left(x,t\right) & =\underset{l=1}{\overset{\infty}{\sum}}B_{l}e^{-\left(\frac{l^{2}\pi^{2}\kappa t}{L^{2}}\right)}\sin\left(\frac{l\pi x}{L}\right)\end{alignedat}$ .

i.e. Given initial conditions and boundary conditions you can tell how the temperature along the 1-D rod varies in time and space. This can be extended to 2 and 3 dimensions.

Or for example a differential equation governing freefall.
The diff equation you have is $\frac{dv}{dt}=g-\frac{c_{d}v^{2}}{m}$ , which has a partly ugly solution, $v(t)=\sqrt{\frac{gm}{c_{d}}}\tanh \left(\sqrt{\frac{gc_{d}}{m}}\dot t\right)$ .

There's stuff to do with springs and damping, i.e. you have a mass on a spring, and a damper and a force which is applied to the system, i.e. a 'forcing function'. In different situations the system will act differently, i.e. it could be underdamped, meaning it keeps oscillating but the oscillations eventually die down to zero. It could be critically damped at which the mass returns to the equilibrium position in the shortest possible time, or even overdamped, at which the damping force takes longer to return to the original position. Then with the forcing function, if the frequency of oscillation is correct, relative to the natural frequency then you could have the system going off to resonance, at which things like this happen: http://www.youtube.com/watch?v=j-zczJXSxnw

I guess there's also stuff in life sciences, I just don't know as much about it. Like there's stuff governing growth rates of populations and radioactive decay, and there's a lot to do with fluid flow and such as well, (I've heard about some life sciences students doing stuff with fluid flow through blood vessels and such. It annoys me that MBBS kids do stuff and don't denote partial derivatives correctly! arrrg)

There's a lot of applications in engineering for differential equations, even electrical circuits and such.

So yeah, really anytime there is a relationship of something changing with respect to something else changing really.

EDIT: Added a few more examples, ahh procrastination :P

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 08, 2013, 08:34:19 pm

Quote from: b^3 on October 08, 2013, 08:23:25 pm

As nlui has said, mostly for physical models, when you have one thing changing respect to another thing, or even a few variables changing with respect to a few other variables.

There's a few things to do with heat transfer and such, probably above the level of what you're learning now but
$\begin{alignedat}{1}\frac{\partial T}{\partial t} & =\kappa\frac{\partial^{2}T}{\partial x^{2}}\end{alignedat}$ which has a general solution $\begin{alignedat}{1}T\left(x,t\right) & =\underset{l=1}{\overset{\infty}{\sum}}B_{l}e^{-\left(\frac{l^{2}\pi^{2}\kappa t}{L^{2}}\right)}\sin\left(\frac{l\pi x}{L}\right)\end{alignedat}$ .

i.e. Given initial conditions and boundary conditions you can tell how the temperature along the 1-D rod varies in time and space. This can be extended to 2 and 3 dimensions.

Or for example a differential equation governing freefall.
The diff equation you have is $\frac{dv}{dt}=g-\frac{c_{d}v^{2}}{m}$ , which has a partly ugly solution, $v(t)=\sqrt{\frac{gm}{c_{d}}}\tanh \left(\sqrt{\frac{gc_{d}}{m}}\dot t\right)$ .

There's a lot of applications in engineering for differential equations, even electrical circuits and such.

So yeah, really anytime there is a relationship of something changing with respect to something else changing really.

Thanks for this explanation, really puts the concept into context! If DEs measure change between variables, I wonder if they can be used to model population growth and decline in a biological setting?

Also, when I'm writing Riemann sums, I use notation such as $\sum_{i=0}^{n-1}$ . What exactly does this mean (I mean, I know sigma means "the sum of", but what are the above and below parts telling me)?

Title: Re: Maths question [MTH1020]
Post by: b^3 on October 08, 2013, 08:45:02 pm

I couldn't get the edit in quick enough :P

Quote from: b^3 on October 08, 2013, 08:23:25 pm

There's stuff to do with springs and damping, i.e. you have a mass on a spring, and a damper and a force which is applied to the system, i.e. a 'forcing function'. In different situations the system will act differently, i.e. it could be underdamped, meaning it keeps oscillating but the oscillations eventually die down to zero. It could be critically damped at which the mass returns to the equilibrium position in the shortest possible time, or even overdamped, at which the damping force takes longer to return to the original position. Then with the forcing function, if the frequency of oscillation is correct, relative to the natural frequency then you could have the system going off to resonance, at which things like this happen: http://www.youtube.com/watch?v=j-zczJXSxnw

I guess there's also stuff in life sciences, I just don't know as much about it. Like there's stuff governing growth rates of populations and radioactive decay, and there's a lot to do with fluid flow and such as well, (I've heard about some life sciences students doing stuff with fluid flow through blood vessels and such. It annoys me that MBBS kids do stuff and don't denote partial derivatives correctly! arrrg)

There's a lot of applications in engineering for differential equations, even electrical circuits and such. So normally with Riemann sums you're relating it t integration, so you're summing up rectangles under the curve, that you're using to approximate the curve. You've split the curve into $n-1$ rectangles as I'm assuming you had $n$ points, and the $i$ is just an index.

So yeah, really anytime there is a relationship of something changing with respect to something else changing really.

EDIT: Added a few more examples, ahh procrastination :P

Basically you're summing up whatever is 'inside' the sigma from $i=0$ to $i=n-1$ (which for Riemann sums normally means rectangles approximating the curve). I have a feeling it should be $i=1$ , not $i=0$ though, but basically you're summing up those rectangles to get the area under the curve.

EDIT: I probably should add, since I was doing engineering maths at the time that we would have covered Riemann sums, I never really actually covered them at uni, so yeah the above is just based on extra reading and such. Correct me if I've said anything incorrect. Although the concept is kinda hidden away inside the VCE course.

Title: Re: Maths question [MTH1020]
Post by: BigAl on October 08, 2013, 10:17:37 pm

One of the examples the lecturer gave us last semester was relative population growth of foxes and rabbits...while one is decreasing exponentially, other one is increasing...so you may have two rate of change in one equation...to solve such equation, you typically need one more just like solving simulteneous equations

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 09, 2013, 10:50:42 pm

So for the question $Solve \frac{\partial y}{\partial x}=\frac{4xy}{1+x^2}$ , I did:

$\frac{\frac{\partial y}{\partial x}}{y}=\frac{\frac{4xy}{1+x^2}}{y}$

$=\frac{4x}{1+x^2}$

$\therefore \int \frac{\frac{\partial y}{\partial x}}{y} dx = \int \frac{4x}{1+x^2}$

$\therefore ln(y)+c=2ln(x^2+1)+c$

$\therefore ln(y)=2ln(x^2+1)+c$

$\therefore e^{ln(y)}=e^{2ln(x^2+1)+c}$

$\therefore y=e^c(x^2+1)^2$

Am I even close? :'(

Title: Re: Maths question [MTH1020]
Post by: b^3 on October 09, 2013, 11:00:10 pm

Firstly, is it meant to be partial derivatives? It looks like it should just be the single derivative, i.e. $d$ not $\partial$ , since we only have a function of one variable.

Secondly, when you integrate it, you need a $dx$ or $dy$ or whatever on the line with your integrals. So you want
$\begin{alignedat}{1}\frac{1}{y}\frac{dy}{dx} & =\frac{4x}{1+x^{2}} \\ \int\frac{1}{y}dy & =\int\frac{4x}{1+x^{2}}dx \end{alignedat}$

Thirdly, since we're not restricting the domain, you need to keep the modulus when you bring the log in, it can be removed later because of the constant though, as long as you define $A=e^{c}$ , $A\in \mathbb{R}$ , since the $A$ then will account for the $\pm e^{c}$ when you undo the modulus. As you have done, we won't need a mod on the right hand log since the inside of the log is always positive.

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 09, 2013, 11:05:39 pm

Yep, I've got the integrals written down as you have them haha, I'm just trying to learn Latex (for what it's worth). Definitely meant to be single derivatives.

If I'm bringing in the modulus for the left-hand side of the equation, should it be $\left | ln(y) \right |$ ?

Also, what exactly is $A$ ? Is it just a constant? (I'm really, really iffy on the theory for DEs)

Title: Re: Maths question [MTH1020]
Post by: b^3 on October 09, 2013, 11:07:20 pm

$\int \frac{1}{y}dy=\ln |y| \left(+C\right)$ .

Yeah $A$ is just a constant. Hence the $A\in \mathbb{R}$ .

EDIT: Also being a little bit more picky (this will depend on your lecturer though). Don't put therefore, $\therefore$ on every line. Apparently it's more for when you're finishing with a result, i.e. the last line at the end. You can have an implies, $\implies$ though for the earlier lines, if you really want to though.

Although with that being said, I'm sure if we're being really picky that part of what I've just added isn't right as well :P

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 09, 2013, 11:12:17 pm

Thanks! Does that mean I can express my final answer as $y=A(x^2+1)^2$ ?

Title: Re: Maths question [MTH1020]
Post by: BigAl on October 09, 2013, 11:23:20 pm

Quote from: alondouek on October 09, 2013, 11:12:17 pm

Thanks! Does that mean I can express my final answer as $y=A(x^2+1)^2$ ?

Yep :) make sure you specify the domain of A ie A belongs to R...some people are really picky about this
Edit

Title: Re: Maths question [MTH1020]
Post by: BigAl on October 09, 2013, 11:26:30 pm

Wait did you make y subject...i mean did you raise both sides by e?

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 09, 2013, 11:28:35 pm

Quote from: BigAl on October 09, 2013, 11:26:30 pm

Wait did you make y subject...i mean did you raise both sided by e?

Sure did!

-->

Quote from: alondouek on October 09, 2013, 10:50:42 pm

$e^{ln(y)}=e^{2ln(x^2+1)+c}$

Title: Re: Maths question [MTH1020]
Post by: BigAl on October 09, 2013, 11:28:40 pm

Im not feeling really well now..cant follow the steps..but im sure youve done it correctly

Title: Re: Maths question [MTH1020]
Post by: BigAl on October 09, 2013, 11:32:28 pm

Sorry my bad..but i think you need to have e after A

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 10, 2013, 12:19:56 am

Okay, spent half an hour on the phone to a friend who tried to explain the question to me from a theory standpoint, this is what I ended up with:

(http://i.imgur.com/edQUaQV.jpg)

Is this better?

Title: Re: Maths question [MTH1020]
Post by: alondouek on October 10, 2013, 01:35:54 am

Also, in the case where I need to find a unique function for $\frac{\mathrm{dy} }{\mathrm{d} x}-x^2y=0 and y(1)=1$ ,

would $y(x)=e^{\frac{1}{3}(x^3-1)}$ be a valid answer?

Title: Re: Maths question [MTH1020]
Post by: b^3 on October 10, 2013, 01:01:36 pm

Just add in $y\neq0$ when you divide by $y$ , since your assuming this when you integrate through. You get this included solution back at the end when you look at the situation.

And yeah that function would be fine. Something that you might find useful :P http://www.wolframalpha.com/input/?i=y'-x%5E2%20y%3D0%2C%20y(1)%3D1&t=crmtb01

~ In Wolfram We Trust ~

But seriously, it's good for checking if you've worked through things right, and a lot of other things.

EDIT: 3000^th post, wow I've been here for a while...

ATAR Notes: Forum

Uni Stuff => Universities - Victoria => Monash University => Topic started by: alondouek on October 04, 2013, 01:09:13 am