First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (1 Viewer)

InteGrand · Aug 29, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
Is this different for Pn ?

Yeah, remember P_n has dimension n+1, not n. So P_4 has dimension 5. So five linearly independent vectors in P_4 would form a basis for it.

Flop21 · Aug 29, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
Yeah, remember P_n has dimension n+1, not n. So P_4 has dimension 5. So five linearly independent vectors in P_4 would form a basis for it.

Ohh interesting, thank you!

Flop21 · Aug 29, 2016

Re: MATH1231/1241/1251 SOS Thread

When showing the a function is a linear transformation or not...

what if all they give is this:

T: R->R defined by T(x) = xcosx

all the videos I've seen involve vectors, so I'm confused on how to solve this with just 1 x and a function. Do I choose a random value for x or?

InteGrand · Aug 29, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
When showing the a function is a linear transformation or not...

what if all they give is this:

T: R->R defined by T(x) = xcosx

all the videos I've seen involve vectors, so I'm confused on how to solve this with just 1 x and a function. Do I choose a random value for x or?

$\noindent To show that $T$ is not linear, we just need to show that it fails to satisfy a condition of linearity. For example, it suffices to come up with an example where $T$ doesn't satisfy $T\left(\alpha x\right) = \alpha T\left(x\right)$. We show an example below.$

$\noindent Note $T\left(\pi\right) = \pi \cos \pi = -\pi$, whilst $T\left(2\pi\right) = 2\pi \cos 2\pi = 2\pi$. So $T \left(2\pi\right) \neq 2T\left(\pi\right)$, so $T$ can't be linear.$

leehuan · Aug 29, 2016

Re: MATH1231/1241/1251 SOS Thread

Hmm... I can't do simple harmonic motion.

$\text{Try to ignore the fact that the period is }\frac{2\pi}{\omega}\text{ in }\frac{d^2x}{dt^2}=-\omega^2x$

$$A cylindrical buoy of 80cm in diameter floats in water with its axis vertical. When depressed slightly and released, it bobs up and down according to the differential equation$\\ m\frac{d^2x}{dt^2}=-40^2\pi gx$

$where $m$ is the mass of the buoy in grams, $x$ cm the displacement from the equilibrium position. Take the acceleration due to gravity $g$ to be 980cm/sec$^2$. The period of oscillation is observed to be 2.5 seconds. What is the mass of the buoy?$

Answer: 50^2 g / pi (approx 780) grams

leehuan · Aug 29, 2016

Re: MATH1231/1241/1251 SOS Thread

So much applied maths... @_@

P.S. with the problem above, if there's no other way to do it but to assume T=2pi/omega, please include a brief outline of the derivation of the formula.
______________

$$Find the steady state oscillations of the vibrating system modelled by the equation$\\ \frac{d^2y}{dt^2}+3\frac{dy}{dt}+2y=20\sin t$

What does steady state mean in this context and why does it justify the fact that the correct answer is just a particular solution y=-6cos(t)+2sin(t)

RenegadeMx · Aug 29, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
So much applied maths... @_@

P.S. with the problem above, if there's no other way to do it but to assume T=2pi/omega, please include a brief outline of the derivation of the formula.
______________

$$Find the steady state oscillations of the vibrating system modelled by the equation$\\ \frac{d^2y}{dt^2}+3\frac{dy}{dt}+2y=20\sin t$

What does steady state mean in this context and why does it justify the fact that the correct answer is just a particular solution y=-6cos(t)+2sin(t)

long term solution i think second condition after u finish is take t->inf

hence ur homogenous solutions all go to 0 and ur left with ur particular sol.

InteGrand · Aug 30, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Hmm... I can't do simple harmonic motion.

$\text{Try to ignore the fact that the period is }\frac{2\pi}{\omega}\text{ in }\frac{d^2x}{dt^2}=-\omega^2x$

$$A cylindrical buoy of 80cm in diameter floats in water with its axis vertical. When depressed slightly and released, it bobs up and down according to the differential equation$\\ m\frac{d^2x}{dt^2}=-40^2\pi gx$

$where $m$ is the mass of the buoy in grams, $x$ cm the displacement from the equilibrium position. Take the acceleration due to gravity $g$ to be 980cm/sec$^2$. The period of oscillation is observed to be 2.5 seconds. What is the mass of the buoy?$

Answer: 50^2 g / pi (approx 780) grams

$\noindent The fact that the period is $\frac{2\pi}{\omega}$ follows from the fact that the general solution of the ODE $\ddot{x} = -\omega^{2} x$ is $x(t) = R \cos \left(\omega t +\varphi\right)$, for arbitrary constants $\varphi, R$. This function has period $\frac{2\pi}{\omega}$ (except in trivial case where $R$ happens to be $0$, which usually isn't the case for practical purposes like here).$

leehuan · Aug 30, 2016

Re: MATH1231/1241/1251 SOS Thread

I asked a former student for help... Turns out that the trick was when t=0 x=0 AND when t=2.5 x=0

Taking arcsin on the first case gives ... = 0 whereas taking arcsin on the second case ... = 2pi

Which makes sense. I just forgot how to use the definition of period.

(Still, that made the diameter bit quite useless)

InteGrand · Aug 30, 2016

Re: MATH1231/1241/1251 SOS Thread

Why not just use the 2pi/(omega) method? Seems faster.

leehuan · Aug 30, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
Why not just use the 2pi/(omega) method? Seems faster.

It's not exactly a part of my course to assume that

InteGrand · Aug 30, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
It's not exactly a part of my course to assume that

$\noindent Oh. It's a pretty standard SHM fact, kind of surprising you can't assume it I guess. But what was the other method you mentioned? Something about an arcsin? Did it require you to find the solution in the form $R \sin \left(\omega t +\varphi\right)$?$

leehuan · Aug 30, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
$\noindent Oh. It's a pretty standard SHM fact, kind of surprising you can't assume it I guess. But what was the other method you mentioned? Something about an arcsin? Did it require you to find the solution in the form $R \sin \left(\omega t +\varphi\right)$?$

Solve the ODE making use of the characteristic equation, then finding the coefficient on cos to be 0. So the coefficient on sin is non zero and thus when t=0, 1.25 we have x=0

It's cause it's not intended to really focus on the properties of SHM. They had to throw in applications of ODEs but because it's not intended to share that many traits with physics they only said "oh btw this is what we call SHM now moving on"

Sent from my iPhone using Tapatalk

InteGrand · Aug 30, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Solve the ODE making use of the characteristic equation, then finding the coefficient on cos to be 0. So the coefficient on sin is non zero and thus when t=0, 1.25 we have x=0

It's cause it's not intended to really focus on the properties of SHM. They had to throw in applications of ODEs but because it's not intended to share that many traits with physics they only said "oh btw this is what we call SHM now moving on"

Sent from my iPhone using Tapatalk

Well you'd be imposing initial conditions yourself wouldn't you (like starting at x = 0)? Because the question didn't seem to give any initial conditions. Anyway, if we did end up solving the ODE, we'd see it's something like sin(omega*t), which we know has period 2pi/(omega). (So I don't think saying T = 2pi/(omega) is very big an assumption.)

(Well I see now it said something about getting depressed slightly and released, which may imply initial velocity of 0 and intial starting position at the extreme negative position. But really we don't need to find the actual solution, once we see it's of the form A*cos(omega*t) + B*sin(omega*t), we can see the period is 2pi/omega. Finding the actual constants A and B isn't important for this question.)

leehuan · Aug 30, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
Well you'd be imposing initial conditions yourself wouldn't you (like starting at x = 0)? Because the question didn't seem to give any initial conditions. Anyway, if we did end up solving the ODE, we'd see it's something like sin(omega*t), which we know has period 2pi/(omega). (So I don't think saying T = 2pi/(omega) is very big an assumption.)

I might have to look at my tutor's working to deal with the initial conditions. (Yeah like I said these came from a former student.)

Flop21 · Aug 31, 2016

Re: MATH1231/1241/1251 SOS Thread

I thought no leading columns on a row meant NO solution.

What's going on here? And has it got something to do with the way you put your polynomals into a matrix?

leehuan · Aug 31, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
I thought no leading columns on a row meant NO solution.

What's going on here? And has it got something to do with the way you put your polynomals into a matrix?

Go back to 1131.

You have three unknowns. But you only have two pivots in your row echelon form. Hence, you'd have to parametrise to get the solution.

$To make it clearer, you would let $\alpha_3=\lambda \qquad (\lambda \in \mathbb{R})\\ $Then, evaluating the rows:$

$R_2: 2\alpha_2-\alpha_3=3 \iff \alpha_2=\frac{1}{2}(3+\alpha)\\ $and similarly for $R_1$

Flop21 · Aug 31, 2016

Re: MATH1231/1241/1251 SOS Thread

oh i think im getting confused

InteGrand · Aug 31, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
I thought no leading columns on a row meant NO solution.

What's going on here? And has it got something to do with the way you put your polynomals into a matrix?

Flop21 said:
oh i think im getting confused

A zero row being present only means no solutions if the right-hand column in that row has a non-zero entry (in other words, the right-hand column is leading).

Here, the right-hand column is non-leading, so solutions exist. Since there is a non-leading column in the left-hand matrix, there are infinitely many solutions. And what do you mean by the way we put the polynomials into a matrix? (The order we put them in as columns won't affect the answer, if that's what you were wondering about.)

Red_of_Head · Aug 31, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
I thought no leading columns on a row meant NO solution.

What's going on here? And has it got something to do with the way you put your polynomals into a matrix?

Because you only have two leading columns and 3 unknowns, you have infinite possible values for a1,a2,a3.

For example, look at row two.

2*a2-a3=3

You could have a2=3,a3=3 or a2=2,a1=1 or a2=50,a3=97 etc.

First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (1 Viewer)

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