1st Year University Mathematics Thread (1 Viewer)

Carrotsticks · Mar 24, 2014

Hey all!

We have a 3U and 4U Maths Marathon, but I noticed that we don't have such a thing for University students!

I'll start with the basic calculations (just to ease everybody in) before throwing in some harder problems.

$\\ $Consider the triangle in $ \mathbb{R}^3 $ with vertices at $ A(-1,2,3), B(2,-2,0) $ and $ C(3,1,-4). \\\\ $Determine if it has any obtuse angles.$$

Carrotsticks · Mar 24, 2014

Carrotsticks · Mar 24, 2014

Carrotsticks · Mar 24, 2014

Galactic Jam · May 11, 2014

Carrotsticks said:
$\\ $Consider the triangle in $ \mathbb{R}^3 $ with vertices at $ A(-1,2,3), B(2,-2,0) $ and $ C(3,1,-4). \\\\ $Determine if it has any obtuse angles.$$

dead thread

$\\ \overrightarrow{BA} = -3i + 4j + 3k,\ \overrightarrow{BC} = i + 3j - 4k$

$\overrightarrow{BA} \cdot \overrightarrow{BC} = -3$

$As the dot product is negative the vectors make an obtuse angle.$

anomalousdecay · Nov 18, 2014

FrankXie said:
solve or find the ananytic solution of $y^\prime+y^2=y\cos x$

I think I'll have a go at some of these after my last exam.

Here is another one courtesy of FrankXie

Amleops · Nov 18, 2014

Shame this thread hasn't taken off yet. I'm doing Calculus and Linear Algebra at Summer School this year so I could do with some extra practice if we can start this thread up again

.

As for now, haven't really learnt a lot of the content yet, but from cursory readings I think I can do one of the questions:

Carrotsticks said:
$\\ $Use the dot product to prove that $ ||\bold{u}+\bold{v}||^2+||\bold{u}-\bold{v}||^2=2 \left ( ||\bold{u}||^2 + ||\bold{v}||^2 \right )$

$||\bold{u}+\bold{v}||^2+||\bold{u}-\bold{v}||^2=\left (\bold{u}+\bold{v}) \cdot (\bold{u}+\bold{v}) + (\bold{u}-\bold{v}) \cdot (\bold{u}-\bold{v}) \\ &= \bold{u}\cdot(\bold{u}+\bold{v})+\bold{v} \cdot (\bold{u}+\bold{v})+\bold{u}\cdot(\bold{u}-\bold{v})-\bold{v}\cdot(\bold{u}-\bold{v}) \\ &= \bold{u}\cdot\bold{u}+\bold{u}\cdot\bold{v}+\bold{v}\cdot\bold{u}+\bold{v}\cdot\bold{v}+\bold{u} \cdot \bold{u}-\bold{u}\cdot\bold{v}-\bold{v}\cdot\bold{u}+\bold{v}\cdot\bold{v} \\ &= 2 (\bold{u}\cdot\bold{u}+\bold{v}\cdot\bold{v}) \\ &= 2 \left ( ||\bold{u}||^2 + ||\bold{v}||^2 \right )$

And yes, one of my aims this Summer is to learn how to properly use LaTeX haha.

Carrotsticks · Nov 18, 2014

Amleops said:
Shame this thread hasn't taken off yet. I'm doing Calculus and Linear Algebra at Summer School this year so I could do with some extra practice if we can start this thread up again .

As for now, haven't really learnt a lot of the content yet, but from cursory readings I think I can do one of the questions:

$||\bold{u}+\bold{v}||^2+||\bold{u}-\bold{v}||^2=\left (\bold{u}+\bold{v}) \cdot (\bold{u}+\bold{v}) + (\bold{u}-\bold{v}) \cdot (\bold{u}-\bold{v}) \\ &= \bold{u}\cdot(\bold{u}+\bold{v})+\bold{v} \cdot (\bold{u}+\bold{v})+\bold{u}\cdot(\bold{u}-\bold{v})-\bold{v}\cdot(\bold{u}-\bold{v}) \\ &= \bold{u}\cdot\bold{u}+\bold{u}\cdot\bold{v}+\bold{v}\cdot\bold{u}+\bold{v}\cdot\bold{v}+\bold{u} \cdot \bold{u}-\bold{u}\cdot\bold{v}-\bold{v}\cdot\bold{u}+\bold{v}\cdot\bold{v} \\ &= 2 (\bold{u}\cdot\bold{u}+\bold{v}\cdot\bold{v}) \\ &= 2 \left ( ||\bold{u}||^2 + ||\bold{v}||^2 \right )$

And yes, one of my aims this Summer is to learn how to properly use LaTeX haha.

Good luck with that! If you have any questions please feel free to ask in our Mathematics section, or perhaps in this thread.

Yep, your proof looks good =)

Also, learning TeX is great, can be quite slow to begin with but it'll become second nature in no time.

FrankXie · Nov 18, 2014

Carrotsticks said:
$\\ $Show that the function $ f(x,y) = \ln (x^2+y^2) + 2 \tan ^{-1} \left ( \frac{y}{x} \right ) $ satisfies Laplace's Equation $ \frac{\partial^2 f}{\partial x ^2} + \frac{\partial ^2 f}{\partial y^2} = 0$

This one is just simple straight forward partial derivatives.

Try this:

If $f$ is differentiable, find the limit

$\lim_{h\rightarrow0}\frac{f^2(x+3h)-f^2(x-h)}{h}$

Carrotsticks · Nov 18, 2014

FrankXie said:
This one is just simple straight forward partial derivatives.

Try this:

If $f$ is differentiable, find the limit

$\lim_{h\rightarrow0}\frac{f^2(x+3h)-f^2(x-h)}{h}$

Frank, this thread is to encourage people to answer ie: start off with the 'straight forward' questions, and then move on to the more difficult ones.

We're not here to flex our Maths muscles.

seanieg89 · Nov 18, 2014

FrankXie said:
This one is just simple straight forward partial derivatives.

Try this:

If $f$ is differentiable, find the limit

$\lim_{h\rightarrow0}\frac{f^2(x+3h)-f^2(x-h)}{h}$

$=2f(x)\cdot\lim_{h\rightarrow 0}\frac{f(x+3h)-f(x-h)}{h}$

$&=& 2f(x)\cdot (\lim_{h\rightarrow 0}\frac{f(x+3h)-f(x)}{h}+\lim_{h\rightarrow 0}\frac{f(x)-f(x-h)}{h})$

$&=&2f(x)\cdot(3f'(x)+f'(x))=8f(x)f'(x).$

Edit: muscles flexed.

Carrotsticks · Nov 18, 2014

seanieg89 said:
Edit: muscles flexed.

Haha oh you.

But I really hope that more ugrads can participate.

FrankXie · Nov 18, 2014

Carrotsticks said:
$\\ $Show that the function $ f(x,y) = \ln (x^2+y^2) + 2 \tan ^{-1} \left ( \frac{y}{x} \right ) $ satisfies Laplace's Equation $ \frac{\partial^2 f}{\partial x ^2} + \frac{\partial ^2 f}{\partial y^2} = 0$

lol so I start with solving it

$\frac{\partial f}{\partial x}=\frac{2x}{x^2+y^2}+2\frac{-yx^{-2}}{1+\frac{y^2}{x^2}}=\frac{2x}{x^2+y^2}-\frac{2y}{x^2+y^2}$

$\frac{\partial^2 f}{\partial x^2}=\frac{2(x^2+y^2)-(2x-2y)(2x)}{(x^2+y^2)^2}=\frac{-2x^2+2y^2+4xy}{(x^2+y^2)^2}$

$\frac{\partial f}{\partial y}=\frac{2y}{x^2+y^2}+2\frac{\frac1x}{1+\frac{y^2}{x^2}}=\frac{2y}{x^2+y^2}+\frac{2x}{x^2+y^2}$

$\frac{\partial^2 f}{\partial y^2}=\frac{2(x^2+y^2)-(2y+2x)(2y)}{(x^2+y^2)^2}=\frac{2x^2-2y^2-4xy}{(x^2+y^2)^2}$

Done!

P.S. I did not mean to flex muscles. LOL

HeroicPandas · Nov 18, 2014

there is just a small mistake on the fourth line of your working out haha (should be partial y)

FrankXie · Nov 18, 2014

thanks. corrected it. haha

RenegadeMx · Nov 20, 2014

let x be a real number. prove x=floor(x/2)+ceiling(x/2) iff x is an integer

HeroicPandas · Nov 26, 2014

RenegadeMx said:
let x be a real number. prove x=floor(x/2)+ceiling(x/2) iff x is an integer

IF x is an integer, then it can be odd or even

Suppose x is odd, then there exists an integer M such that x = 2M + 1

$x=\left \lfloor \frac{2M+1}{2} \right \rfloor + \left \lceil \frac{2M+1}{2} \right \rceil$

$= \left \lfloor M+\frac{1}{2} \right \rfloor + \left \lceil M + \frac{1}{2}\right \rceil$

$= M + (M+1)$

$= 2M+1$

$= x$

Suppose x is even, then there exists an integer M such that x = 2M

So
$x=\left \lfloor \frac{2M}{2} \right \rfloor + \left \lceil \frac{2M}{2} \right \rceil = \left \lfloor M \right \rfloor + \left \lceil M\right \rceil = M + M= 2M= x$

ONLY IF, not sure atm, thinking about the contrapositive

I have this:

Let x be p/q such that p and q are integers and q is not 0 or 1

LHS = x = p/q

RHS = floor(x/2) + ceil(x/2)
= floor(p/2q) + ceil(p/2q)

sirable1 · Nov 26, 2014

Note: Sorry, I'm not sure how to use Latex at the moment!

Suppose that two vectors A and B are the following A(0,1,-1), B(2,1,M)

Find the value of 'M' if the angle between A and B is Pi/4

I know the formula for computing angles between vectors is (Cos(theta) = a . b/|a||b|) but I am unsure how to compute the inverse (like finding the value of 'M' here).

This appeared in the one of the past papers for MATH1131, S1 2014 and was quite clueless how to approach this question. Any thoughts?

nightweaver066 · Nov 26, 2014

sirable1 said:
Note: Sorry, I'm not sure how to use Latex at the moment!

Suppose that two vectors A and B are the following A(0,1,-1), B(2,1,M)

Find the value of 'M' if the angle between A and B is Pi/4

I know the angle formula for computing angles between vectors is (Cos(theta) = a . b/|a||b|) but I am unsure how to compute the inverse (like finding the value of 'M' here).

This appeared in the one of the past papers for MATH1131, S1 2014 and was quite clueless how to approach this question. Any thoughts?

Lets hope this is right after having not touched first year maths for so long haha

$\text{The dot product is defined as } \mathbf{a}\cdot \mathbf{b} = \|\mathbf{a}\| \|\mathbf{b}\| \cos\theta$

$\text{Applying this,}$

$LHS = (0,1,-1) \cdot (2, 1, M) = 1 - M$

$\begin{align*} RHS &= \|(0,1,-1)\| \| (2, 1, M) \| \cos \frac{\pi}{4} \\ &= \sqrt{2} \times \sqrt{2^2 + 1^2 + M^2} \times \frac{1}{\sqrt{2}}\\ &= \sqrt{5 + M^2}\end{align*}$

$\text{So equating both sides,}$

$\begin{align*} 1 - M &= \sqrt{5 + M^2}\\1 - 2M + M^2 &= 5 + M^2\\2M &= -4\\M &= -2 \end{align*}$

RealiseNothing · Nov 26, 2014

RenegadeMx said:
let x be a real number. prove x=floor(x/2)+ceiling(x/2) iff x is an integer

RHS is defined to be integer lol.

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1st Year University Mathematics Thread (1 Viewer)

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