First Year Mathematics A (Differentiation & Linear Algebra) (1 Viewer)

InteGrand · Jun 9, 2016

Re: MATH1131 help thread

Flop21 said:
If I use u = 1 + root(x)...

x = (u-1)^2
dx = 2u-2

Then putting into the original thing, integrate (2u-2)/u...

can someone take it from here and finish it off? I get very close to the answer but don't get it, I think because I'm doing incorrect algebra after that last step ^.

$\noindent So the integral becomes $I = \int \frac{1}{u}\times 2\left(u-1\right)\text{ d}u$. Write $\frac{u-1}{u}$ as $1-\frac{1}{u}$, so then $I = 2\int \left(1-\frac{1}{u}\right)\text{ d}u=2u - 2\ln |u| +C$. Since $u=1+\sqrt{x}$, this becomes $2+2\sqrt{x}-2\ln \left(1+\sqrt{x}\right)+C$. Since $2+C$ is just an arbitrary constant again, we can write the final answer as $I=2\sqrt{x}-2\ln \left(1+\sqrt{x}\right)+C$.$

Flop21 · Jun 9, 2016

Re: MATH1131 help thread

Okay soz I'm still very confused,

For this one...

integrate (1-x)/(1+x)^3 dx

So I make u = 1+x, du = dx, x = u-1

so it becomes

(2-u) / u^3 du

and now what do I do? If I'm on the right track that is.

InteGrand · Jun 9, 2016

Re: MATH1131 help thread

Flop21 said:
Okay soz I'm still very confused,

For this one...

integrate (1-x)/(1+x)^3 dx

So I make u = 1+x, du = dx, x = u-1

so it becomes

(2-u) / u^3 du

and now what do I do? If I'm on the right track that is.

$\noindent Split up the fraction, so it becomes $\frac{2}{u^3}-\frac{1}{u^2}$, which is easy to integrate now.$

Red_of_Head · Jun 9, 2016

Re: MATH1131 help thread

http://www.integral-calculator.com/ is pretty useful if you don't know how to solve a particular integral. You can check answers and get a walkthrough of the working out.

Flop21 · Jun 9, 2016

Re: MATH1131 help thread

Does anyone know the mark format of the final exam, 1131?

There's 3 questions it seems from past papers, are they all of equal value? (why don't they show what each question is worth?)

Flop21 · Jun 9, 2016

Re: MATH1131 help thread

1024*e^(-5pi/3) + 1024*e^(5pi/3)

How do you turn that into cos, I get you can go, = 1024 (e^(5pi/3) + e^(-5pi/3)). Now what.

I recall there's a definition for it somewhere but cannot find it, and if it exists do we have to memorise it (well I guess obviously if it's in this q).

iforgotmyname · Jun 9, 2016

Re: MATH1131 help thread

Flop21 said:
1024*e^(-5pi/3) + 1024*e^(5pi/3)

How do you turn that into cos, I get you can go, = 1024 (e^(5pi/3) + e^(-5pi/3)). Now what.

I recall there's a definition for it somewhere but cannot find it, and if it exists do we have to memorise it (well I guess obviously if it's in this q).

2012 paper, e^itheta=cos(theta)+isin(theta)

InteGrand · Jun 9, 2016

Re: MATH1131 help thread

Flop21 said:
1024*e^(-5pi/3) + 1024*e^(5pi/3)

How do you turn that into cos, I get you can go, = 1024 (e^(5pi/3) + e^(-5pi/3)). Now what.

I recall there's a definition for it somewhere but cannot find it, and if it exists do we have to memorise it (well I guess obviously if it's in this q).

$\noindent Recall that $e^{i\theta}+e^{-i\theta}=2\cos \theta$ (which follows from Euler's formula as posted above).$

(And I think you forgot to type the i's in the exponents.)

Flop21 · Jun 9, 2016

Re: MATH1131 help thread

Implicit differentiation...

x^2 - 5xsiny + y^2 = 4

iforgotmyname · Jun 9, 2016

MATH1131 help thread

turntaker said:
How should I go about doing 1 a)

Just go like point+t(vector of line)

So remember <2,5> + t<1,2> so what this means is that the line L contains the point <2,5> and is parallel to <1,2>

iforgotmyname · Jun 9, 2016

Re: MATH1131 help thread

Flop21 said:
Implicit differentiation...

x^2 - 5xsiny + y^2 = 4

2x-5siny+5xcosyy'+2yy'=0

Flop21 · Jun 9, 2016

Re: MATH1131 help thread

iforgotmyname said:
2x-5siny+5xcosyy'+2yy'=0

Yeah I got the answer, but could someone show working out, thanx

leehuan · Jun 9, 2016

Re: MATH1131 help thread

Flop21 said:
Does anyone know the mark format of the final exam, 1131?

There's 3 questions it seems from past papers, are they all of equal value? (why don't they show what each question is worth?)

Are you sure it's three and not four

Flop21 said:
Implicit differentiation...

x^2 - 5xsiny + y^2 = 4

$x^2-5x\sin{y}+y^2=4\\ \Rightarrow 2x- \underbrace{ \left(5\sin{y}+5x\cos{y}\frac{dy}{dx}\right) }_{\text{product rule}} +2y \frac{dy}{dx} =0\\ \Rightarrow \frac{dy}{dx}=\frac{2x-5\sin{y}}{5x\cos{y}-2y}$

Flop21 · Jun 9, 2016

Re: MATH1131 help thread

turntaker said:
How should I go about doing 1 a)

Knowing the definition of a line in that format, wouldn't it be <2,5> + <1,2>*t ?

turntaker · Jun 9, 2016

Re: MATH1131 help thread

Flop21 said:
Knowing the definition of a line in that format, wouldn't it be <2,5> + <1,2>*t ?

idk

i hate this

Flop21 · Jun 9, 2016

Re: MATH1131 help thread

leehuan said:
Are you sure it's three and not four

$x^2-5x\sin{y}+y^2=4\\ \Rightarrow 2x- \underbrace{ \left(5\sin{y}+5x\cos{y}\frac{dy}{dx}\right) }_{\text{product rule}} +2y \frac{dy}{dx} =0\\ \Rightarrow \frac{dy}{dx}=\frac{2x-5\sin{y}}{5x\cos{y}-2y}$

Haha damn, can you tell I was only up to 2011 past papers when I asked that (they had only 3 questions).

But yeah I do mean that, q still stands.

And thanks for the working out.

Flop21 · Jun 9, 2016

Re: MATH1131 help thread

turntaker said:
idk

i hate this

Vector form of a line:

r = <any point on the line> + t<direction>

leehuan · Jun 9, 2016

Re: MATH1131 help thread

iforgotmyname said:
Just go like point+t(vector of line)

So remember <2,5> + t<1,2> so what this means is that the line L contains the point <2,5> and is parallel to <1,2>

I have no idea why the question got deleted but to expand on this.

$If a line is given in parametric vector form $\textbf{x}=\textbf{a} + \lambda \textbf{u} \quad \lambda \in \mathbb R \\ $then $\textbf{a}$ is called a position vector, and $\textbf{u}$ is called a direction vector.$

$A position vector is just a point which a line passes through. The DIRECTION vector is more significant here, as it shows what way the line points.$

$$So in general, any line parallel to the above line will have equation $\textbf{x} = \textbf{c} + \lambda \textbf{u}$ for a new position vector $\textbf{c}$

Here, we have a=[1;1], t=[1;2], c=[2;5]

leehuan · Jun 9, 2016

Re: MATH1131 help thread

Flop21 said:
Haha damn, can you tell I was only up to 2011 past papers when I asked that (they had only 3 questions).

But yeah I do mean that, q still stands.

And thanks for the working out.

Lol idk we don't do the same course. But I thought you guys also had 4 questions. (2 alg 2 calc)

turntaker · Jun 9, 2016

Re: MATH1131 help thread

I deleted it cause it was sideways

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