1. ## 10:0228th Oct 2013

Notes: 2350

Reblogged from whyyesiamafangirl

Here at Qwertee.com we’re not content with selling awesome limited edition tees at the super price of £8/€10/$12 every day, we love to give away great prizes too! So this week we’re giving away no less than a Month of FREE TEES to one Lucky Winner as well as lots of other great prizes!!! All you have to do to enter is: (1) Follow our Tumblr blog: qwertee.tumblr.com (2) Like/Reblog this post (3) Go to the entry form here: http://bit.ly/1hanbQI (or click the picture) and just your Tumblr blog name (so in our case qwertee.tumblr.com) and email in the boxes on the right! That’s it! PS: Every time you you share the competition with your friends using the links on the entry page equals another chance to win and you can share each Tumblr post we make about the competition to increase your chances still further! Good Luck!!! 2. ## 10:3113th Apr 2013 Notes: 2 Reblogged from make-merthur-not-horcruxes We have a curious relationship with passion. It appears to be a highly sought-after commodity; it’s a cliché so overused in CVs, dating sites and university applications that it’s all but undergone complete semantic satiation. The word is thrown around like dollar bills at a strip club*; I… As someone whose family doesn’t always understand my passion for my subject/computers, this is so true. 3. ## 17:1425th Jan 2013 Notes: 43 Reblogged from essaycrisis ### When people ask me questions about Oxford Dedicated to an Durhamite now Exonian (Exeter people, what do you call yourselves?!) Some of us even had our sanity before we started 4. ## 13:4314th Jan 2013 Tags: oxfordsnowst-annes image: Download It’s been snowing in Harry Potter land! 5. ## 09:5622nd Dec 2012 Notes: 4728 Reblogged from wilwheaton image: Download NRA, GTFO A demonstrator held up a banner as Wayne LaPierre, executive vice president of the National Rifle Association, delivered a statement in Washington on Friday. (Photo: Chip Somodevilla / Getty Images via The New York Times) I’m sincerely asking my followers to reblog the shit out of this photo. 6. ## 19:0315th Dec 2012 Notes: 10861 Reblogged from wilwheaton Let me tell you a story. The day after Columbine, I was interviewed for the Tom Brokaw news program. The reporter had been assigned a theory and was seeking sound bites to support it. “Wouldn’t you say,” she asked, “that killings like this are influenced by violent movies?” No, I said, I wouldn’t say that. “But what about ‘Basketball Diaries’?” she asked. “Doesn’t that have a scene of a boy walking into a school with a machine gun?” The obscure 1995 Leonardo Di Caprio movie did indeed have a brief fantasy scene of that nature, I said, but the movie failed at the box office (it grossed only$2.5 million), and it’s unlikely the Columbine killers saw it.
The reporter looked disappointed, so I offered her my theory. “Events like this,” I said, “if they are influenced by anything, are influenced by news programs like your own. When an unbalanced kid walks into a school and starts shooting, it becomes a major media event. Cable news drops ordinary programming and goes around the clock with it. The story is assigned a logo and a theme song; these two kids were packaged as the Trench Coat Mafia. The message is clear to other disturbed kids around the country: If I shoot up my school, I can be famous. The TV will talk about nothing else but me. Experts will try to figure out what I was thinking. The kids and teachers at school will see they shouldn’t have messed with me. I’ll go out in a blaze of glory.”

In short, I said, events like Columbine are influenced far less by violent movies than by CNN, the NBC Nightly News and all the other news media, who glorify the killers in the guise of “explaining” them. I commended the policy at the Sun-Times, where our editor said the paper would no longer feature school killings on Page 1. The reporter thanked me and turned off the camera. Of course the interview was never used. They found plenty of talking heads to condemn violent movies, and everybody was happy.

— Roger Ebert (via squid-skywalker)

7. ## 16:2113th Dec 2012

Notes: 5

Reblogged from thejoysofcode

### When Jeff Atwood likes us

/* by Paul & Olivier */

My goal in life, nothing else

8. ## 18:097th Dec 2012

Notes: 26

Reblogged from make-merthur-not-horcruxes

### Art students in the morning

During term

Out of term

And computer science students, at least out of term

9. ## 11:176th Nov 2012

Notes: 11

Reblogged from make-merthur-not-horcruxes

### When someone makes fun of my college

if they’re from my college

when it’s someone from another college

Everyone’s secretly jealous of my college’s architecture, so I get this far too much

10. ## 09:5124th Aug 2012

I got an e-mail from my college a couple of weeks ago, asking me to work through some problem sheets, so this will be my effort to document my progress through these sheets. I’ve also got no idea how to do the “see more” thing, so I’m afraid you’re going to have to scroll through it all.

# Problem Sheet 1 – Standard Functions and Techniques

1.1  Find the radius and centre of the circle described by the equation$x^{2}+y^{2}-2x-4y+1=0$by writing it in the form $\left(x-a\right)^{2} + \left(y-b\right)^{2} = c^{2}$ for suitable $a$, $b$ and $c$.

\begin{align*} x^{2}-2x+y^{2}-4y&=-1\\ \left(x-1\right)^{2}-1+\left(y-2\right)^{2}-4&=-1\\ \left(x-1\right)^{2}+\left(y-2\right)^{2}&=4=2^{2}\end{align*}The circle therefore has a radius of 2, and centre $\left(1,2\right)$

1.2  Find the equation of the line perpendicular to $y = 3x$ passing through the point $\left(3, 9\right)$.

\begin{align*} \perp\Rightarrow m&=-\dfrac{1}{3}\\ y-y_1&=m\left(x-x_1\right)\\ y-9&=-\dfrac{1}{3}\left(x-3\right)\\ y&=-\dfrac{1}{3}x+8\end{align*}

1.3  Given$\sin\left(A\pm B\right)=\sin A\cos B\pm\cos A\sin B$and $\cos\left(A\pm B\right)=\cos A \cos B \mp \sin A \sin B$show that $\cos A\sin B=\dfrac{1}{2}\left[\sin\left(A+B\right)-\sin\left(A-B\right)\right]$ and$\sin^{2}A=\dfrac{1}{2}\left[1-\cos2A\right]$

\begin{align*} \cos A\sin B&=\dfrac{1}{2}\left[\sin\left(A+B\right)-\sin\left(A-B\right)\right]\\ \cos A\sin B&=\dfrac{1}{2}\left[\sin A\cos B+\cos A\sin B-\left(\sin A\cos B-\cos A\sin B\right)\right]\\ \cos A\sin B&=\dfrac{1}{2}\left[0\sin A\cos B+2\cos A\sin B\right]\\\end{align*}

and

\begin{align*} \cos\left(A+B\right)&=\cos A \cos B-\sin A \sin B\\ \text{Let }A&=B\\ \cos\left(A+A\right)=\cos2A&=\cos A\cos A-\sin A\sin A\\ \cos2A&=\cos^{2}A-\sin^{2}A\\ \sin^{2}\theta+\cos^{2}\theta&\equiv1\\ \cos2A&=1-2\sin^{2}A\\ \sin^{2}A&=\dfrac{1}{2}\left[1-\cos2A\right]\end{align*}

1.4  Show that$4 \cos\left(\alpha t\right) + 3\sin\left(\alpha t\right)= 5\cos\left(\alpha t + \phi\right)$where $\phi=\tan^{-1}\left(-\dfrac{3}{4}\right)$

\begin{align*} 4\cos\left(\alpha t\right)+3\sin\left(\alpha t\right)&=R\sin\left(\alpha t+\beta\right)\\ 4\cos\left(\alpha t\right)+3\sin\left(\alpha t\right)&=R\sin\left(\alpha t\right)\cos\beta+R\sin\beta\cos\left(\alpha t\right)\\ \text{Compare}&\text{ co-efficients}\\ 4&=R\cos\beta\\3&=R\sin\beta\\ \Rightarrow \dfrac{3}{4}&=\dfrac{R\sin\beta}{R\cos\beta}\\ \dfrac{3}{4}&=\tan\beta\\ \tan^{-1}\dfrac{3}{4}&=\beta=\phi\\ R^{2}&=3^{2}+4^{2}\\ R&=5\\ 4 \cos\left(\alpha t\right) + 3\sin\left(\alpha t\right)&= 5\cos\left(\alpha t + \phi\right)\end{align*}

1.5  Show that, for $-1\le x\le1$,$\cos\left(\sin^{-1}x\right)=\pm\sqrt{1-x^{2}}$ For the triangle with hypotenuse$=1$, opposite$=x$, the other side length (adjacent) must be $\sqrt{1-x^{2}}$. Using $\sin\theta=\dfrac OH$ and $\cos\theta=\dfrac AH$, $\sin\theta=x$ and $\cos\theta=\sqrt{1-x^{2}}$$\sin^{-1} x=\theta=\arccos \sqrt{1-x^{2}}$$\cos\sin^{-1} x=\sqrt{1-x^{2}}$

1.6  Given$\sinh\left(A \pm B\right) = \sinh A \cosh B \pm \cosh A \sinh B$and$\cosh\left(A \pm B\right) = \cosh A \cosh B \pm \sinh A \sinh B$show that $\cosh A \cosh B =\dfrac{1}{2} \left[\cosh\left(A + B\right) + \cosh\left(A - B\right)\right]$ and $\sinh^{2} A = \dfrac{1}{2}\left[\cosh{2A} - 1\right]$

(a)\begin{align*} \cosh A\cosh B&=\dfrac{1}{2} \left[\cosh A \cosh B + \sinh A \sinh B+\cosh A \cosh B-\sinh A \sinh B\right]\\ \cosh A\cosh B&=\dfrac{1}{2} \left[2\cosh A \cosh B + 0\sinh A \sinh B\right]\\\end{align*}

(b)\begin{align*} \cosh{2A}&=\sinh^{2}A+\cosh^{2}A\\ \cosh{2A}&=2\sinh^{2}A+1\\ \cosh{2A}-1&=2\sinh^{2}A\\ \sinh^{2} A &= \dfrac{1}{2}\left[\cosh{2A} - 1\right]\end{align*}

1.7  Given that$\sinh x=\dfrac{1}{2}\left[e^{x}-e^{-x}\right]$find $\sinh^{-1}x$

\begin{align*} \text{Let }y&=\sinh x\\ 2ye^{x}&=e^{2x}-1\\ e^{2x}-2ye^{x}-1&=0\\ \left(e^{x}-y\right)^{2}-y^{2}-1&=0\\ e^{x}-y&=\sqrt{y^{2}+1}\\ x&=\ln\left[y+\sqrt{y^{2}+1}\right]\\ \Rightarrow \sinh^{-1}x&=\ln\left[x+\sqrt{x^{2}+1}\right]\end{align*}

1.8  Express $\dfrac{x}{\left(x-1\right)\left(x-2\right)}$in partial fractions.

\begin{align*} \dfrac{x}{\left(x-1\right)\left(x-2\right)}&=\dfrac{A}{x-1}+\dfrac{B}{x-2}\\ x&=A\left(x-2\right)+B\left(x-1\right)\\ \text{Let }x&=1&\text{Let }x&=2\\ 1=1A+0B&\Rightarrow A=1&2=0A+1B&\Rightarrow B=2\\ \dfrac{x}{\left(x-1\right)\left(x-2\right)}&=\dfrac{1}{x-1}+\dfrac{2}{x-2}\end{align*}

1.9  If $a_n=\dfrac{1}{n}$, find $\sum\limits_{i=1}^{5}a_n$ as a fraction. $\sum\limits_{i=1}^{5}a_n=\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{120+60+40+30+24}{120}=\dfrac{274}{120}$

1.10  If $S=\sum\limits_{i=0}^{N} x^{i}$, show that $xS=\sum\limits_{i=1}^{N+1} x^{i}$. Hence show that $S-xS=1-x^{N+1}$ and therefore that $S=\dfrac{1-x^{N+1}}{1-x}$

\begin{align*} S&=x^{0}+x^{1}+x^{2}+\ldots+x^{N}\\ xS&=x^{1}+x^{2}+x^{3}+\ldots+x^{N+1}\\ xS&=\sum\limits_{i=0}^{N} x^{i+1}\\ &=\sum\limits_{i=1}^{N+1} x^{i}\\ S-xS&=x^{0}+x^{1}+x^{2}+\ldots+x^{N}-\left(x^{1}+x^{2}+x^{3}+\ldots+x^{N+1}\right)\\ &=1+x^{1}-x^{1}+x^{2}-x^{2}+\ldots+x^{N}-x^{N}-x^{N+1}\\ &=1-x^{N+1}\\ S&=\dfrac{1-x^{N+1}}{1-x}\\ \overset{S-xS=1-x^{N+1}}{\Rightarrow}S&=\dfrac{S-xS}{1-x}\\ &=\dfrac{S\left(1-x\right)}{1-x}=S\end{align*}

# Problem Sheet 2 – Differentiation

2.1  Given that $\sinh x=\dfrac{1}{2}\left[x+\sqrt{1+x^{2}}\right]$show that$\dfrac{\mathrm{d}}{\mathrm{d}x} \sinh x=\cosh x$ $\dfrac{\mathrm{d}}{\mathrm{d}x} \sinh x=\dfrac{1}{2}\left[e^{x}—e^{-x}\right]=\dfrac{1}{2}\left[e^{x}+e^{-x}\right]=\cosh x$

2.2  Given that $\cosh x=\dfrac{1}{2}\left[e^{x}+e^{-x}\right]$show that$\dfrac{\mathrm{d}}{\mathrm{d}x} \cosh x=\sinh x$ $\dfrac{\mathrm{d}}{\mathrm{d}x} \cosh x=\dfrac{1}{2}\left[e^{x}+-e^{-x}\right]=\dfrac{1}{2}\left[e^{x}-e^{-x}\right]=\sinh x$

2.3  Let $n$ be a positive integer. Show that $\dfrac{\mathrm{d}^{n}}{\mathrm{d}x^{n}}x^{n}=n!$

\begin{align*} \dfrac{\mathrm{d}}{\mathrm{d}x} x^{n}&=nx^{n-1}\\ \text{for $r < n$}\dfrac{\mathrm{d}^{r}}{\mathrm{d}x^{r}}x^{n}&=\dfrac{n!}{\left(n-r\right)!}x^{n-r}\\ \text{for $r=n$}\dfrac{\mathrm{d}^{r}}{\mathrm{d}x^{r}}x^{n}&=\dfrac{n!}{\left(n-r\right)!}x^{n-r}=\dfrac{n!}1\times x^{0}=n!\end{align*}

2.4  If $y=\ln x$, show that$\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{x}$

\begin{align*} \dfrac{\mathrm{d}y}{\mathrm{d}x}&=\lim_{h\to0}\left[\dfrac{\ln\left(x+h\right)-\ln x}{x+h-x}\right]\\ &=\lim_{h\to0}\left[\dfrac{\ln\left(\dfrac{x+h}{x}\right)}{h}\right]\\ \text{Let $n=\dfrac{1}{h}$} \dfrac{\mathrm{d}y}{\mathrm{d}x}&=\lim_{n\to\infty}\left[n\ln\left(1+\dfrac{1}{nx}\right)\right]\\ &=\lim_{n\to\infty}\left[\dfrac xx\ln\left(1+\dfrac{1}{nx}\right)^{n}\right]\\ &=\lim_{n\to\infty}\left[\dfrac 1x\ln\left(1+\dfrac{1}{nx}\right)^{nx}\right]\\ e&=\lim_{x\to\infty}\left[1+\dfrac{1}{x}\right]^{x}\\ &=\dfrac{1}{x}\lim_{n\to\infty}\left[\ln\left(1+\dfrac{1}{nx}\right)^{nx}\right]\\ &=\dfrac{1}{x}\ln e\\ \dfrac{\mathrm{d}y}{\mathrm{d}x}&=\dfrac{1}{x}\end{align*}

2.5  Find the equation of the tangent to the curve $y=x^{2}$ at $\left(1,1\right)$ $\dfrac{\mathrm{d}y}{\mathrm{d}x}=2x\Rightarrow\dfrac{\mathrm{d}y\left(1\right)}{\mathrm{d}x}=2\Rightarrow y-1=2\left(x-1\right)\Rightarrow y=2x-3$

2.6  Find the slope of the curve $y=4x+e^{x}$ at $\left(0,1\right)$ $\dfrac{\mathrm{d}y}{\mathrm{d}x}=4+e^{x}\Rightarrow\dfrac{\mathrm{d}y\left(0\right)}{\mathrm{d}x}=4+e^{0}=5$

2.7  Find the angle of inclination of the tangent to the curve $y=x^{2}+x+1$ at $\left(0,1\right)$ $\dfrac{\mathrm{d}y}{\mathrm{d}x}=2x+1\Rightarrow\dfrac{\mathrm{d}y\left(0\right)}{\mathrm{d}x}=1$ Let $\theta$ be the angle of inclination. $\tan \theta=1\Rightarrow\theta=\dfrac{\pi}{4}$

2.8  The displacement $y(t) \mathrm{m}$ of a body at time $t \mathrm{s}$ ($t\ge0$) is given by $y(t)=t-\sin t$. At what times is the body at rest? $\dfrac{\mathrm{d}y}{\mathrm{d}x}=1-\cos t\Rightarrow1=\cos t \text{ \dfrac{\mathrm{d}y}{\mathrm{d}x}=0 when at rest}\Rightarrow t=2\pi n; n\in\mathbb{Z}$

2.9  The displacement $y(t) \mathrm{m}$ of a body at time $t\mathrm{s}$ is given by $y(t)=3t^{3}+4t+1$. Find its acceleration when $t=4$ $\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=18t\Rightarrow\dfrac{\mathrm{d}^{2}y\left(4\right)}{\mathrm{d}x^{2}}=18\times4=72$

2.10  If$y=\sum\limits_{n=0}^{N}a_nx^{n}$show that$\dfrac{\mathrm{d}y}{\mathrm{d}x}=\sum\limits_{n=1}^{N}na_nx^{n-1}$

\begin{align*} \dfrac{\mathrm{d}}{\mathrm{d}x} ax^{n}&=a\lim_{h\to0}\left[\dfrac{\left(x+h\right)^{n}-x^{n}}{x+h-x}\right]\\ &=a\lim_{h\to0}\left[\dfrac{x^{n}-x^{n}+{n\choose1}x^{n-1}h+{n\choose2}x^{n-2}h^{2}+\ldots+h^{n}}{h}\right]\\ &=a\lim_{h\to0}\left[\dfrac{nhx^{n-1}+{n\choose2}x^{n-2}h^{2}+\ldots+h^{n}}{h}\right]\\ &=a\lim_{h\to0}\left[nx^{n-1}+{n\choose2}x^{n-2}h+\ldots+h^{n-1}\right]\\ &=anx^{n-1}\end{align*}

This change can then be applied generally to $\sum\limits_{n=0}^{N}a_nx^{n}$ However, the limits must be changed, as $\dfrac{\mathrm{d}}{\mathrm{d}x} kx^{0}=0$ $y=\sum\limits_{n=0}^{N}a_nx^{n}\Rightarrow\dfrac{\mathrm{d}y}{\mathrm{d}x}=\sum\limits_{n=1}^{N}na_nx^{n-1}$

# Problem Sheet 3 – Further Differentiation

3.1  If $y=ln\left(1+x^{2}\right)$, find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$$\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{2x}{1+x^{2}}$

3.2  If $\displaystyle y=\dfrac{x}{1+x^{2}}$, find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1+x^{2}-x\left(2x\right)}{\left(1+x^{2}\right)^{2}}=\dfrac{1-x^{2}}{\left(1+x^{2}\right)^{2}}$

3.3  If $y=\cosh x^{4}$, find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ $\dfrac{\mathrm{d}y}{\mathrm{d}x}=4x^{3}\sinh x^{4}$

3.4  if $y=x^{2}\ln x$, find $\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}$ $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{x^{2}}{x}+2x\ln x=x+2x\ln x$ $\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=1+\dfrac{2x}{x}+2\ln x=3+2\ln x$

3.5  If $y=\left(1+x^{2}\right)^{-1/2}$ $\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{1}{2}\times2x\left(1+x^{2}\right)^{-\dfrac{3}{2}}=-x\left(1+x^{2}\right)^{-\dfrac{3}{2}}$

3.6  If $y=\sinh^{-1}x$, find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$

\begin{align*} \sinh y&=x\\ \cosh y&=\dfrac{\mathrm{d}x}{\mathrm{d}y}\\ \dfrac{\mathrm{d}y}{\mathrm{d}x}&=\dfrac{1}{\cosh y}\\ \cosh^{2}x-\sinh^{2}x&=1\\ \dfrac{\mathrm{d}y}{\mathrm{d}x}&=\dfrac{1}{\sqrt{1+\sinh^{2}y}}\\ \dfrac{\mathrm{d}y}{\mathrm{d}x}&=\dfrac{1}{\sqrt{1+x^{2}}}\end{align*}

3.7  If $y=\ln\left[x+\sqrt{x^{2}+1}\right]$, find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$

\begin{align*} \dfrac{\mathrm{d}y}{\mathrm{d}x}&=\dfrac{1-\dfrac{1}{2}2x\left(x^{2}+1\right)^{-\dfrac{1}{2}}}{x+\sqrt{x^{2}+1}}\\ &=\dfrac{1-x\left(x^{2}+1\right)^{-\dfrac{1}{2}}}{x+\sqrt{x^{2}+1}}\\ &=\dfrac{1}{\sqrt{1+x^{2}}}\left(\sqrt{1+x^{2}}+x\right)\dfrac{1}{x+\sqrt{1+x^{2}}}\\ &=\dfrac{1}{\sqrt{1+x^{2}}}\end{align*}

3.8  If $y=\cos^{-1}\sin x$, find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ $\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{1}{\sqrt{1-\sin^{2}x}}\times{\cos x}=-\cos x\sqrt{\sec^{2}x}$

3.9  A curve is given in polar co-ordinates by $r=1+\sin^{2}\theta$. Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ at $\theta=\dfrac{\pi}{4}$

\begin{align*} x&=r\left(\theta\right)\sin\theta\\ y&=r\left(\theta\right)\cos\theta\\ x&=\sin\theta+\sin^{3}\theta\\ y&=\left(1+\sin\theta\right)\cos\theta\\ \dfrac{\mathrm{d}x}{\mathrm{d}\theta}&=\cos\theta+3\sin^{2}\theta\cos\theta\\ \dfrac{\mathrm{d}y}{\mathrm{d}\theta}&=\cos^{2}\theta-\left(1+\sin\theta\right)\sin\theta\\ \dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{\frac{\mathrm{d}y}{\mathrm{d}\theta}}{\frac{\mathrm{d}x}{\mathrm{d}\theta}}&=\dfrac{\cos\theta+3\sin^{2}\theta\cos\theta}{\cos^{2}\theta-\left(1+\sin\theta\right)\sin\theta}\\ &=\dfrac{\cos\theta\left(1+3\sin^{2}\theta\right)}{\cos^{2}\theta-\sin\theta-\sin^{2}\theta}\\ &=\dfrac{\cos\theta\left(1+3\sin^{2}\theta\right)}{2\cos^{2}\theta-\sin\theta-1}\\ \text{for }\theta=\dfrac{\pi}{4};\dfrac{\mathrm{d}y}{\mathrm{d}x}&=\dfrac{\cos\dfrac{\pi}{4}\left(1+3\sin^{2}\dfrac{\pi}{4}\right)}{2\cos^{2}\dfrac{\pi}{4}-\sin\dfrac{\pi}{4}-1}\\ &=-\dfrac{5}{2}\end{align*}

3.10  Show that if $y=\dfrac{1}{2a}\ln\left|\dfrac{x-a}{x+a}\right|,\text{ then }\dfrac{\mathrm{d}y}{\mathrm{d}x}-\dfrac{1}{x^{2}-a^{2}}$

\begin{align*} y&=\dfrac{1}{2a}\left(\ln|x-a|-\ln|x+a|\right)\\ \dfrac{\mathrm{d}y}{\mathrm{d}x}&=\dfrac{1}{2a}\left(\dfrac{1}{x-a}-\dfrac{1}{x+a}\right)\\ &=\dfrac{1}{2a}\left(\dfrac{x+a-x+a}{\left(x-a\right)\left(x+a\right)}\right)\\ &=\dfrac{1}{2a}\left(\dfrac{2a}{x^{2}-a^{2}}\right)\\ &=\dfrac{1}{x^{2}-a^{2}}\end{align*}

# Problem Sheet 4 – Applications of Differentiation

4.1  Given that $f\left(x-ct\right)$, where $x$ and $c$ are constant, show that$\dfrac{\mathrm{d}^{2}}{\mathrm{d}t^{2}}f\left(x-ct\right)=c^{2}f”\left(x-ct\right)$and calculate this expresssion when $f(u)=\sin u$

\begin{align*} \dfrac{\mathrm{d}}{\mathrm{d}t}f\left(x-ct\right)&=-cf’\left(x-ct\right)\\ \dfrac{\mathrm{d}^{2}}{\mathrm{d}t^{2}}\left(-cf’\left(x-ct\right)\right)&=c^{2}f”\left(x-ct\right)\\ \text{for }f(u)&=\sin u\\ \dfrac{\mathrm{d}^{2}}{\mathrm{d}t^{2}}f\left(x-ct\right)&=-c^{2}\sin\left(x-ct\right)\end{align*}

4.2  Classify the stationary point of $y=x^{-2}\ln x$, where $x>0$

\begin{align*} \dfrac{\mathrm{d}y}{\mathrm{d}x}&=\dfrac{1}{x^{3}}\left(1-2\ln x\right)\\ \dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}&=\dfrac{1}{x^{4}}\left(6\ln x-5\right)\\ \dfrac{\mathrm{d}y}{\mathrm{d}x}&=0=\dfrac{1}{x^{3}}\left(1-2\ln x\right)\\ \dfrac{1}{2}&=\ln x\\ x&=e^{\dfrac{1}{2}}\\ \text{for }x=e^{\dfrac{1}{2}},\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}&=\dfrac{1}{e^{2}}\left(-5+6\ln e^{\dfrac{1}{2}}\right)\\ \dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} < 0\Rightarrow\text{max}\end{align*}

4.3  Classify the stationary points of $y(x)=x^{2}-3x+2$ $\dfrac{\mathrm{d}y}{\mathrm{d}x}=2x-3, \dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=2\Rightarrow\text{min}$

4.4  The numbers $x$ and $y$ are subject to the constraints $x+y=\pi$. Find the values of $x$ and $y$ for which $\cos x\sin y$ takes its minimum value.

\begin{align*} \sin x\cos y&=\sin x\cos\left(\pi-x\right)\\ &=-\sin x\cos x\\ &=\sin x\cos x\\ &=\dfrac{1}{2}\sin2x\\ \text{min of}\sin2x&=-1\\ x&=\dfrac{1}{2}\left(2\pi n-\dfrac\pi2\right);n\in\mathbb{Z}\end{align*}

# Problem Sheet 5 – Taylor Series

5.1  Verify the following Taylor expansions (taking the ranges of validity for granted)

(a)$e^{x}=1+\dfrac{1}{1!}x+\dfrac{1}{2!}x^{2}+\dfrac{1}{3!}x^{3}+\ldots+\dfrac{1}{n!}x^{n}+\ldots\text{ valid for any x}$

\begin{align*} \text{for $x=0$, }e^{x}&=1\\ &=1+\dfrac{1}{2!}0^{2}+\dfrac{1}{3!}0^{3}+\ldots+\dfrac{1}{n!}0^{n}\\&=1\\ \dfrac{\mathrm{d}}{\mathrm{d}x} e^{x}&=e^{x}\\ \dfrac{\mathrm{d}}{\mathrm{d}x} e^{x}&=\dfrac{\mathrm{d}}{\mathrm{d}x}1+\dfrac{\mathrm{d}}{\mathrm{d}x} \dfrac{1}{1!}x+\dfrac{\mathrm{d}}{\mathrm{d}x} \dfrac{1}{2!}x^{2}+\dfrac{\mathrm{d}}{\mathrm{d}x} \dfrac{1}{3!}x^{3}+\ldots\\ &\quad+\dfrac{\mathrm{d}}{\mathrm{d}x} \dfrac{1}{n!}x^{n}+\dfrac{\mathrm{d}}{\mathrm{d}x} \dfrac{1}{\left(n+1\right)!}x^{n+1}\ldots\\ &=1+\dfrac{1}{1!}x+\dfrac{1}{2!}x^{2}+\dfrac{1}{3!}x^{3}+\ldots+\dfrac{1}{\left(n-1\right)!}x^{n-1}+\dfrac{1}{n!}x^{n}\ldots\\ &=e^{x}\end{align*}

(b)$\sin x=x-\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}-\ldots+\dfrac{\left(-1\right)^{n}x^{2n+1}}{\left(2n+1\right)!}+\ldots \text{ valid for any x}$

\begin{align*} \text{for $x=0$, }\sin x&=0\\ &=0-\dfrac{0^{3}}{3!}+\dfrac{0^{5}}{5!}-\ldots+\dfrac{\left(-1\right)^{n}0^{2n+1}}{\left(2n+1\right)!}+\ldots\\ &=0\\ \dfrac{\mathrm{d}^{2}}{\mathrm{d}x}\sin x&=-\sin x\\ \sin x&=x-\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}-\ldots+\dfrac{\left(-1\right)^{n}x^{2n+1}}{\left(2n+1\right)!}\\ \dfrac{\mathrm{d}}{\mathrm{d}x} \sin x&=-1+\dfrac{x^{2}}{2!}-\dfrac{x^{4}}{4!}+\ldots-\dfrac{\left(-1\right)^{n}x^{2n}}{\left(2n\right)!}\\ \dfrac{\mathrm{d}^{2}}{\mathrm{d}x}\sin x&=-x+\dfrac{x^{3}}{3!}-\dfrac{x^{5}}{5!}+\ldots+\dfrac{\left(-1\right)^{n}x^{2n-1}}{\left(2n-1\right)!}\\ &=-\sin x\end{align*}

(c)$\cos x=1-\dfrac{x^{2}}{2!}+\dfrac{x^{4}}{4!}-\ldots+\dfrac{\left(-1\right)^{n}x^{2n}}{\left(2n\right)!}\text{ valid for any x}$

\begin{align*} \text{for $x=0$, }\cos x&=1\\ &=1-\dfrac{0^{2}}{2!}+\dfrac{0^{4}}{4!}-\ldots+\dfrac{\left(-1\right)^{n}0^{2n}}{\left(2n\right)!}\\ &=1\\ \dfrac{\mathrm{d}^{2}}{\mathrm{d}x}\cos x&=-\cos x\\ \cos x&=1-\dfrac{x^{2}}{2!}+\dfrac{x^{4}}{4!}-\ldots+\dfrac{\left(-1\right)^{n}x^{2n}}{\left(2n\right)!}\\ \dfrac{\mathrm{d}}{\mathrm{d}x} \cos x&=-x+\dfrac{x^{3}}{3!}-\dfrac{x^{5}}{5!}+\ldots+\dfrac{\left(-1\right)^{n}x^{2n-1}}{\left(2n-1\right)!}\\ \dfrac{\mathrm{d}^{2}}{\mathrm{d}x}\cos x&=-1+\dfrac{x^{2}}{2!}-\dfrac{x^{4}}{4!}+\ldots-\dfrac{\left(-1\right)^{n}x^{2n-2}}{\left(2n-2\right)!}\\ &=-\cos x\end{align*}

(d)Let $\alpha$ be a constant.$\left(1+x\right)^{\alpha}=1+\alpha x+\dfrac{\alpha\left(\alpha-1\right)}{2!}x^{2}+\dfrac{\alpha\left(\alpha-1\right)\left(\alpha-2\right)}{3!}x^{3}+\ldots\text{ valid for -1<x<1}$

\begin{align*} \text{for $x=0$, }\left(1+x\right)^{\alpha}&=1\\ &=1+\alpha 0+\dfrac{\alpha\left(\alpha-1\right)}{2!}0^{2}+\dfrac{\alpha\left(\alpha-2\right)\left(\alpha-3\right)}{3!}0^{3}+\ldots\\ &=1\\ \dfrac{\mathrm{d}}{\mathrm{d}x}\left(1+x\right)^{\alpha}&=\dfrac{\mathrm{d}}{\mathrm{d}x} \left(1+\alpha x+\dfrac{\alpha\left(\alpha-1\right)}{2!}x^{2}\right.\\ &\left.\qquad\qquad{}+\dfrac{\alpha\left(\alpha-2\right)\left(\alpha-3\right)}{3!}x^{3}+\ldots\right)\\ &=\alpha+\alpha\left(\alpha-1\right)x+\dfrac{\alpha\left(\alpha-1\right)\left(\alpha-2\right)}{2!}x^{2}\\ &\qquad {}+\dfrac{\alpha\left(\alpha-2\right)\left(\alpha-3\right)\left(\alpha-4\right)}{3!}x^{3}+\ldots\\ &=\alpha\left(1+\left(\alpha-1\right)x+\dfrac{\left(\alpha-1\right)\left(\alpha-2\right)}{2!}x^{2}\right.\\ &\left.\quad\qquad {}+\dfrac{\left(\alpha-2\right)\left(\alpha-3\right)\left(\alpha-4\right)}{3!}x^{3}+\ldots\right)\\ &=\alpha\left(1+x\right)^{\alpha-1}\end{align*}

(e)$\ln\left(1+x\right)=x-\dfrac {x^{2}}2+\dfrac{x^{3}}{3}-\ldots+\dfrac{\left(-1\right)^{n-1}x^{n}}{n}+\ldots\text{ valid for -1< x \le 1}$

\begin{align*} \text{for $x=0$, }\ln\left(1+x\right)&=0\\ &=0-\dfrac {0^{2}}2+\dfrac{0^{3}}{3}-\ldots+\dfrac{\left(-1\right)^{n-1}0^{n}}{n}+\ldots\\ &=0\\ \dfrac{\mathrm{d}}{\mathrm{d}x} \ln\left(1+x\right)&=\dfrac{\mathrm{d}}{\mathrm{d}x} \left(x-\dfrac {x^{2}}2+\dfrac{x^{3}}{3}-\ldots+\dfrac{\left(-1\right)^{n-1}x^{n}}{n}+\ldots\right)\\ &=1-x+x^{2}-x^{3}+\ldots+n\dfrac{\left(-1\right)^{n}}{n}x^{n-1}\\ &=1-x+x^{2}-x^{3}+\ldots+{\left(-1\right)^{n}}x^{n-1}\\ &=\left(1+x\right)^{-1}\\&=\dfrac{1}{1+x}\\\end{align*}

5.2 Obtain a four-term Taylor expansion for the following, valid near $x=0$.

(a)$\left(1+x\right)^{1/2}$

\begin{align*} \left(1+x\right)^{1/2}&=1+\dfrac{1}{2} x+\dfrac{\dfrac{1}{2}\left(\dfrac{1}{2}-1\right)}{2!}x^{2}+\dfrac{\dfrac{1}{2}\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}-2\right)}{3!}x^{3}+\ldots\\ &=1+\dfrac{1}{2}x-\dfrac{1}{8}x^{2}+\dfrac{1}{16}x^{3}+\ldots\\\end{align*}

(b)$\sin\left(2x\right)$

\begin{align*} \sin\left(2x\right)&=\left(2x\right)-\dfrac{\left(2x\right)^{3}}{3!}+\dfrac{\left(2x\right)^{5}}{5!}-\dfrac{\left(2x\right)^{7}}{7!}\ldots+\dfrac{\left(-1\right)^{n}\left(2x\right)^{2n+1}}{\left(2n+1\right)!}+\ldots\\ &=2x-\dfrac{4x^{3}}{3}+\dfrac{4x^{5}}{15}-\dfrac{8x^{7}}{315}+\ldots+\dfrac{\left(-1\right)^{n}\left(2x\right)^{2n+1}}{\left(2n+1\right)!}+\ldots\end{align*}

(c)$\ln\left(1+3x\right)$

\begin{align*} \ln\left(1+\left(3x\right)\right)&=\left(3x\right)-\dfrac {\left(3x\right)^{2}}2+\dfrac{\left(3x\right)^{3}}{3}-\dfrac{\left(3x\right)^{4}}{4}+\ldots+\dfrac{\left(-1\right)^{n-1}\left(3x\right)^{n}}{n}+\ldots\\ &=3x-\dfrac{9}{2}x^{2}+9x^{3}-\dfrac{27}8x^{4}-\ldots+\end{align*}

# Problem Sheet 6 – Complex Numbers

6.1  Reduce to standard form

(a)$\dfrac{3+i}{4-i}=\dfrac{3+i}{4-i}\times\dfrac{4+i}{4+i}=\dfrac{\left(3+i\right)\left(4+i\right)}{15}=\dfrac{11+7i}{15}$and

(b)$\left(1+i\right)^{5}=1+5i-10-10i+5-i=-4-6i$

6.2  Prove

(a)$\left|z_1z_2\right|=\left|z_1\right|\left|z_2\right|$

(b)$\left|\!\dfrac{z_1}{z_2}\!\right|=\dfrac{\left|z_1\right|}{\left|z_2\right|}\text{ when }z_2\neq0$

6.3  Given that $e^{i\theta}=\cos{\theta}+i\sin{\theta}$, prove that $\cos{\left(A+B\right)}=\cos{A}\cos{B}-\sin{A}\sin{B}$

\begin{align*} e^{i\left(A+B\right)}&=\cos{\left(A+B\right)}+i\sin{\left(A+B\right)}\\ e^{iA}e^{iB}&=\left(\cos{A}+i\sin{A}\right)\left(\cos{B}+i\sin{B}\right)\\ &=\cos{A}\cos{B}-\sin{A}\sin{B}+i\sin{A}\cos{B}+i\sin{B}\cos{A}\\ \text{By comparing real and imaginary parts}&\\ \cos{\left(A+B\right)}&=\cos{A}\cos{B}-\sin{A}\sin{B}\\ \sin{\left(A+B\right)}&=\sin{A}\cos{B}+\sin{B}\cos{A}\end{align*}

6.4  Let $z=1+i$. Find the following complex numbers in standard form and plot their corresponding points in the Argand diagram:-

a$\bar{z}^{2}$ $\bar{z}^{2}=\left(1-i\right)^{2}=-2i$ and

b

$\dfrac{z}{\bar{z}}$ $\dfrac{1+i}{1-i}=\dfrac{2i}{2}=i$

6.5  Find the modulus and principal arguments of

a

$-2+2i$ $r=\sqrt{2^{2}+2^{2}}=2\sqrt2;\theta=\tan^{-1}\dfrac{-2}2=\dfrac{3\pi}4$

b

$3+4i$$r=5;\theta=\tan^{-1}\dfrac{4}{3}\approx0.9273$

6.6  Find all the complex roots of

a

$\cosh z=1$

\begin{align*} \dfrac{1}{2}\left[e^{z}+e^{-z}\right]&=1\\ e^{2z}+1&=2e^{z}\\ \text{Let }x=e^{z}\\ x^{2}-2x+1&=0\\ x=1\\ e^{z}&=1\\ z&=2\pi n i, n\in\mathbb{Z}\end{align*}

b

$\sinh z=1$

\begin{align*} \dfrac{1}{2}\left[e^{z}-e^{-z}\right]&=1\\ e^{2z}-1&=2e^{z}\\ \text{Let }x=e^{z}\\ x^{2}-2x-1&=0\\ \left(x-1\right)^{2}-2&=0\\ x&=1\pm\sqrt2\\ e^{z}&=1\pm\sqrt2\\ z&=\ln{\left(1\pm\sqrt2\right)}\\ &=i\left(2\pi n-i\ln\left(1+\sqrt2\right)\right)\\ &=i\left(2\pi n +\pi-i\ln\left(\sqrt2-1\right)\right),n\in\mathbb{Z}\end{align*}

(c)$e^{z}=-1$ $z=\pi+2\pi n, n\in\mathbb{Z}$

(d)$\cos z=\sqrt2$

\begin{align*} e^{iz}-e^{-iz}&=2\sqrt2\\ e^{2iz}-1&=2\sqrt2,e^{iz}\\ e^{2iz}-2\sqrt2,e^{iz}-1&=0\\ \left(e^{iz}-\sqrt2\right)^{2}-3&=0\\ e^{iz}&=\sqrt2\pm\sqrt3\\ z&=\dfrac{1}{i}\ln\left(\sqrt2\pm\sqrt3\right)\\\end{align*}

6.7  Show that the mapping $w=z+\dfrac cz$where $z=x+iy$, $w=u+iv$ and $c$ is a real number, maps the circle $\mod{z}=1$ in the $z$ plane into an ellipse in the $w$ plane and find its equation.

6.8  Show that$\cos^{6}\theta=\dfrac{1}{32}\left(\cos6\theta+6\cos4\theta+15\cos2\theta+10\right)$

\begin{align*} \cos^{6}\theta&=\left(\dfrac{e^{i\theta}+e^{-i\theta}}{2}\right)^{6}\\ &=\dfrac{\left(e^{i\theta}\right)^{6}+6\left(e^{i\theta}\right)^{5}\left(e^{-i\theta}\right)+15\left(e^{i\theta}\right)^{4}\left(e^{-i\theta}\right)^{2}20\left(e^{i\theta}\right)^{3}\left(e^{-i\theta}\right)^{3}}{64}\\ &\qquad{}+\dfrac{15\left(e^{i\theta}\right)^{2} \left(e^{-i\theta}\right)^{4}+6\left(e^{i\theta}\right)\left(e^{-i\theta}\right)^{5}+\left(e^{-i\theta}\right)^{6}}{64}\\ &=\dfrac{2\cos{6\theta}+12\cos{4\theta}+30\cos{2\theta}+20}{64}\\ &=\dfrac{1}{32}\left(\cos6\theta+6\cos4\theta+15\cos2\theta+10\right)\end{align*}

# Problem Sheet 7 – Matrices

7.1  The matrix $A=\left(a_{ij}\right)$ is given by $A =\begin{pmatrix} 1 &2 &3\\ -1 &0 &1\\ 2 &-2 &4\\ 1 &5 &-3 \end{pmatrix}$Identify the elements $a_{13}$ and $a_{31}$.

$a_{13}=3$ and $a_{31}=2$

7.2  Given that$A=\begin{pmatrix}1&3&0\\2&1&1\end{pmatrix}, B=\begin{pmatrix}1&0\\2&1\\-1&-1\end{pmatrix}, C=\begin{pmatrix}2&1\\-1&1\\-0&1\end{pmatrix}$verify the distributive law $A\left(B+C\right)=AB+AC$ for the three matrices.

\begin{align*} AB&=\begin{pmatrix} 1 & 3 & 0 \\ 2 & 1 & 1 \\\end{pmatrix}.\begin{pmatrix} 1 & 0 \\ 2 & 1 \\ -1 & -1 \\\end{pmatrix}= \begin{pmatrix}7&3\\3&0\end{pmatrix}\\ AC&=\begin{pmatrix} 1 & 3 & 0 \\ 2 & 1 & 1 \\ \end{pmatrix}.\begin{pmatrix} 2 & 1 \\ -1 & 1 \\ 0 & 1 \\\end{pmatrix}=\begin{pmatrix}-1&4\\3&4\end{pmatrix}\\ AB+AC&=\begin{pmatrix}6&7\\6&4\end{pmatrix}\\ B+C&=\begin{pmatrix}3&1\\1&2\\-1&0\end{pmatrix}\\ A\left(B+C\right)&=\begin{pmatrix}1&3&0\\2&1&1\end{pmatrix}.\begin{pmatrix}3&1\\1&2\\-1&0\end{pmatrix}= \begin{pmatrix} 6&7\\ 6&4 \end{pmatrix}\end{align*}

7.3  Let $A=\begin{pmatrix}4&2\\2&1\end{pmatrix},B=\begin{pmatrix}-2&-1\\4&2\end{pmatrix}$Show that $AB=0$, but that $BA\neq0$

\begin{align*} AB&=\begin{pmatrix}4&2\\2&1\end{pmatrix}.\begin{pmatrix}-2&-1\\4&2\end{pmatrix}&BA&=\begin{pmatrix}-2&-1\\4&2\end{pmatrix}.\begin{pmatrix}4&2\\2&1\end{pmatrix}\\ &=\begin{pmatrix}0&0\\0&0\end{pmatrix}&&=\begin{pmatrix} -1&-5\\ 20&10 \end{pmatrix}\end{align*}

7.4  A general $n\times n$ matrix is given by $A = \left(a_{ij}\right)$. Show that $A+A^\top$ is a symmetric matrix, and that $A - A^\top$ is skew-symmetric. Express the matrix $A = \begin{pmatrix} 2 &1 &3\\ -2& 0& 1\\ 3 &1 &2 \end{pmatrix}$ as the sum of a symmetric matrix and a skew-symmetric matrix.

\begin{align*} A^{\top}&=\begin{pmatrix} 2 &-2 &3\\ 1& 0& 1\\ 3 &1 &2 \end{pmatrix}\\ A+A^\top&=\begin{pmatrix} 2 &1 &3\\ -2& 0& 1\\ 3 &1 &2 \end{pmatrix}+\begin{pmatrix} 2 &-2 &3\\ 1& 0& 1\\ 3 &1 &2 \end{pmatrix}\\ &=\begin{pmatrix} 4&-1&6\\ -1&0&2\\ 6&2&4 \end{pmatrix}\\ A-A^\top&=\begin{pmatrix} 2 &1 &3\\ -2& 0& 1\\ 3 &1 &2 \end{pmatrix}-\begin{pmatrix} 2 &-2 &3\\ 1& 0& 1\\ 3 &1 &2 \end{pmatrix}\\ &=\begin{pmatrix} 0&3&0\\ -3&0&0\\ 0&0&0 \end{pmatrix}\\ A&=\begin{pmatrix} 2&-1/2&3\\ -1/2&0&1\\ 3&1&2 \end{pmatrix}+\begin{pmatrix} 0&3/2&0\\ -3/2&0&0\\ 0&0&0 \end{pmatrix}\\\end{align*}

7.5  Let the matrix $A = \begin{pmatrix} 1 &0 &0\\ a &-1 &0\\ b &c &1 \end{pmatrix}$ Find $A^2$. For what relation between $a$, $b$, and $c$ is $A^2 = I$ (the unit matrix)? In this case, what is the inverse matrix of $A$? What is the inverse matrix of $A^{2n-1}$ ($n$ a positive integer)?

\begin{align*} A^{2}&=\begin{pmatrix} 1&0&0\\ 0&1&0\\ 2b+ac&0&1 \end{pmatrix}\\ \text{when }2b&=-ac,A^{2}=I\end{align*}

7.6  Using the rule for inverses of $2\times2$ matrices, write down the inverse of$A=\begin{pmatrix} 1&1\\2&-1 \end{pmatrix}$ $A^{-1}=\dfrac{1}{-1-2}\begin{pmatrix} 1&1\\2&-1 \end{pmatrix}=-\dfrac{1}{3}\begin{pmatrix} 1&1\\2&-1 \end{pmatrix}$

7.7  If $A$ and $B$ are both $n \times n$ matrices with $A$ non-singular, show that $\left(A^{-1}BA\right)^2 = A^{-1}B^2A$

# Problem Sheet 8 – Vectors

8.1  Obtain the components of the vectors below where $L$ is the magnitude and $\theta$ the angle made with the positive direction of the $x$ axis $\left(-180◦ (a)\(L = 3, \theta = 60^\circ$; $\textbf{r}=3\begin{pmatrix}\cos60\\\sin60\end{pmatrix}=\begin{pmatrix}3/2\\3\sqrt{3}/2\end{pmatrix}=\dfrac{3}{2}\textbf{i}+\dfrac{3\sqrt3}{2}\textbf{j}$

(b)$L = 3, \theta = -150^\circ$. $\textbf{r}=3\begin{pmatrix}\cos210\\\sin210\end{pmatrix}=\begin{pmatrix}-3\sqrt3/2\\-3/2\end{pmatrix}=-\dfrac{3\sqrt3}{2}\textbf{i}-\dfrac{3}{2}\textbf{j}$

8.2  Two ships, $S_1$ and $S_2$ set off from the same point $Q$. Each follows a route given by successive displacement vectors. In axes pointing east and north, $S_1$ follows the path to $B$ via $\overrightarrow{QA} = \left(2, 4\right)$, and $\overrightarrow{AB} = \left(4, 1\right)$. $S_2$ goes to $E$ via $\overrightarrow{QC} = \left(3, 3\right)$, $\overrightarrow{CD} = \left(1, 1\right)$ and $\overrightarrow{DE} = \left(2,-3\right)$. Find the displacement vector $\overrightarrow{BE}$ in component form.

\begin{align*} B&=\left(2+4\right)\textbf{i}+\left(4+1\right)\textbf{j}=6\textbf{i}+5\textbf{j}\\ E&=\left(3+1+2\right)\textbf{i}+\left(3+1-3\right)\textbf{j}=6\textbf{i}+1\textbf{j}\\ \overrightarrow{BE}&=-4\textbf{j}\end{align*}

8.3  Sketch a diagram to show that if $A,B,C$ are any three points, then $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = 0$. Formulate a similar result for any number of points.

For a series of $n$ points, ($o$ (the origin) to $N$ (tHe $n^\text{th}$ point)) $\overrightarrow{OA}+\overrightarrow{AB}\ldots+\overrightarrow{\left(n-1\right)N}+\overrightarrow{NO}=0$ as it forms a cycle.

8.4  Sketch a diagram to show that if $A,B,C,D$ are any four points, then $\overrightarrow{CD}=\overrightarrow{CB}+\overrightarrow{BA}+\overrightarrow{AD}$. Formulate a similar result for any number of points.

For a series of $n$ points, ($o$ (the origin) to $N$ (the $n^\text{th}$ point)) $\overrightarrow{OA}+\overrightarrow{AB}\ldots+\overrightarrow{\left(n-1\right)N}=\overrightarrow{ON}$ as it forms a cycle.

8.5  Two points $A$ and $B$ have position vectors $\textbf{a}$ and $\textbf{b}$ respectively. In terms of $\textbf{a}$ and $\textbf{b}$ find the position vectors of the following points on the straight line passing through $A$ and $B$.

(a)The mid-point $C$ of $AB$; $\overrightarrow{AB}=\textbf{b}-\textbf{a}\Rightarrow\overrightarrow{OC}=\dfrac{1}{2}\left(\textbf{b}-\textbf{a}\right)$

(b)a point $U$ between $A$ and $B$ for which $\overrightarrow{AU}/\overrightarrow{UB} = 1/3$. $3\overrightarrow{AU}=\overrightarrow{UB}\Rightarrow\overrightarrow{AU}=\dfrac{1}{4}\left(\overrightarrow{AB}\right)=\dfrac{1}{4}\left(\textbf{b}-\textbf{a}\right)$

8.6  Suppose that $C$ has position vector $\textbf{r}$ and $\textbf{r} = \lambda \textbf{a} +\left(1 - \lambda\right)\textbf{b}$ where $\lambda$ is a parameter, and $A$, $B$ are points with $\textbf{a}$, $\textbf{b}$ as position vectors. Show that $C$ describes a straight line. Indicate on a diagram the relative positions of $A$, $B$, $C$, when $\lambda 1$.
$\textbf{r}=\textbf{b}+\lambda\left(\textbf{a}-\textbf{b}\right)$ as $\textbf{a}$ and $\textbf{b}$ are constants, the equation can be seen to be in the form $\textbf{r}=\textbf{c}+\mu\textbf{d}$

8.7  Find the shortest distance from the origin of the line given in vector parametric form by $\textbf{r} = \textbf{a}+t\textbf{b}$,where $\textbf{a} = \left(1, 2, 3\right)$ and $\textbf{b} = \left(1, 1, 1\right)$ ,and $t$ is the parameter (Hint: use a calculus method, with $t$ as the independent variable.)

\begin{align*} |\textbf{r}|&=\sqrt{\left(1+t\right)^2+\left(2+t\right)^2+\left(3+t\right)^2}\\ &=\sqrt{3t^2+12t+14}\\ \dfrac{\mathrm{d}}{\mathrm{d}t}\left(3t^2+12t+14\right)&=6t+12\\ &=0\text{ for }t=-2\\ |\textbf{r}|&=\sqrt{\left(1-2\right)^2+\left(2-2\right)^2+\left(3-2\right)^2}\\ &=\sqrt{2}\end{align*}

8.8  $ABCD$ is any quadrilateral in three dimensions. Prove that if $P$, $Q$, $R$, $S$ are the mid-points of $AB$, $BC$, $CD$, $DA$ respectively, then $PQRS$ is a parallelogram.

8.9  $ABC$ is a triangle, and $P$, $Q$ and $R$ are the mid-points of the respective sides $BC$, $CA$, $AB$. Prove that the medians $AP$, $BQ$, $CR$ meet at a single point $G$ (called the centroid of $ABC$; it is the centre of mass of a uniform triangular plate.)

8.10  Show that the vectors $\overrightarrow{OA} = \left(1, 1, 2\right)$, $\overrightarrow{OB} = \left(1, 1, 1\right)$, and $\overrightarrow{OC} = \left(5, 5, 7\right)$ all lie in one plane.

# Problem Sheet 9 – Vector Dot Product

9.1  The figure $ABCD$ has vertices at $\left(0, 0\right), \left(2, 0\right), \left(3, 1\right)\text{ and }\left(1, 1\right)$. Find the vectors $\overrightarrow{AC}$ and $\overrightarrow{BD}$. Find $\overrightarrow{AC}.\overrightarrow{BD}$. Hence show that the angles between the diagonals of $ABCD$ have cosine $-1/\sqrt5$.

\begin{align*} \overrightarrow{AC}&=\left(3,1\right)\\ \overrightarrow{BD}&=\left(-1,1\right)\\ \cos\theta&=\dfrac{\textbf{a}\cdot\textbf{b}}{|\textbf{a}||\textbf{b}|}\\ &=\dfrac{\left(3,1\right).\left(-1,1\right)}{\sqrt{10}\sqrt{2}}\\ &=\dfrac{-3+1}{2\sqrt{5}}\\ &=-\dfrac{1}{\sqrt{5}}\end{align*}

9.2  Show that the vectors $\textbf{a} = \textbf{i} + 3\textbf{j} + 4\textbf{k}$ and $\textbf{b} = -2\textbf{i} + 6\textbf{j} - 4\textbf{k}$ are perpendicular. Obtain any vector $\textbf{c} = c_1\textbf{i} + c_2\textbf{j} + c_3\textbf{k}$ which is perpendicular to both $a$ and $b$.

\begin{align*} \perp\Rightarrow\textbf{a}\cdot\textbf{b}&=0\\ \textbf{a}\cdot\textbf{b}&=-2+18-16\\ &=0\\ c_1+3c_2+4c_3&=0\\ -2c_1+6c_2-4c_3&=0\\ c_1+3c_2+4c_3&=-2c_1+6c_2-4c_3=0\\ 3c_1-3c_2&=8c_3\\ c_1+3c_2+\dfrac{3}{2}\left(c_1-c_2\right)&=0\\ \dfrac{5}{2}c_1+\dfrac{3}{2}c_2&=0\\ 5c_1&=-3c_2\\ 3c_1+5c_1&=8c_3\\ c_1&=c_3\\ \begin{pmatrix}c\\-\dfrac35c\\c\end{pmatrix}&\perp\textbf{a},\textbf{b}\\ c\begin{pmatrix}1\\-\dfrac35\\1\end{pmatrix}&\perp\textbf{a},\textbf{b};c\neq0\end{align*}

9.3  Find the value of $\lambda$ such that the vectors $\left(\lambda, 2, -1\right)$ and $\left(1, 1, -3\lambda\right)$ are perpendicular.

\begin{align*} \perp\Rightarrow\begin{pmatrix}\lambda\\2\\-1\end{pmatrix}\cdot\begin{pmatrix}1\\1\\-3\lambda\end{pmatrix}&=0\\ \lambda+2+3\lambda&=0\\ \lambda&=-\dfrac12\end{align*}

9.4  Find a constant vector parallel to the line given parametrically by $x = 1- \lambda, y = 2+3\lambda, z = 1+\lambda$ $\overrightarrow{r}=\begin{pmatrix}1\\2\\1\end{pmatrix}+\lambda\begin{pmatrix}-1\\3\\1\end{pmatrix}$

9.5  A circular cone has its vertex at the origin and its axis in the direction of the unit vector $\hat{\textbf{a}}$. The half-angle at the vertex is $\alpha$. Show that the position vector $\textbf{r}$ of a general point on its surface satisfies the equation $\hat{\textbf{a}}\cdot\textbf{r} = |\textbf{r}| \cos\alpha$ Obtain the cartesian equation when $\textbf{a} = \left(2/7, -3/7, -6/7\right)$ and $\alpha = 60$.

Part a) $\hat{\textbf{a}}$ is the unit vector $\Rightarrow|\hat{\textbf{a}}|=1$, $\hat{\textbf{a}}\cdot\textbf{r} = |\hat{\textbf{a}}||\textbf{r}| \cos\alpha\Rightarrow\dfrac{\hat{\textbf{a}}\cdot\textbf{r}}{|\textbf{r}||\hat{\textbf{a}}|}=\cos\alpha$, which is the relating the cosine of the angle between two vectors, their dot product and moduli.

Part b)

\begin{align*} \dfrac17\begin{pmatrix}2\\-3\\-6\end{pmatrix} \textbf{r}&=\dfrac12|\textbf{r}|\\ \dfrac27\left(2x-3y-6z\right)&=\sqrt{x^2+y^2+z^2}\\ \dfrac{4}{49}\left(2x-3y-6z\right)^2&=x^2+y^2+z^2\end{align*}

# Problem Sheet 10 – Vector Cross Product

10.1  For vectors $a$ and $b$, show $|a+b|^2 + |a - b|^2 = 2\left(|a|^2 + |b|^2\right)\text{ and }a \cdot b =\dfrac{1}{4}\left(|a + b|^2 - |a - b|^2\right)$

$\begin{gathered} \text{Let }a=\begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix} \text{ and }b=\begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}\\ \begin{split} |x_1+x_2+y_1+y_2+z_1+z_2|^2+|x_1-x_2+y_1-y_2+z_1-z_2|^2=\\ 2\left(|x_1+y_1+z_1|^2+|x_2+y_2+z_2|^2\right)=\\ 2\left(x_1^2+y_1^2+z_1^2+x_2^2+y_2^2+z_2^2\right)\end{split}\\ \begin{split} \left(x_1+x_2\right)^2+\left(y_1+y_2\right)^2+\left(z_1+z_2\right)^2+\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2+\left(z_1-z_2\right)^2=\\ 2\left(x_1^2+y_1^2+z_1^2+x_2^2+y_2^2+z_2^2\right)\end{split}\\ \begin{split} 2\left(x_1x_2+y_1y_2+z_1z_2\right)+x_1^2+y_1^2+z_1^2+x_2^2+y_2^2+z_2^2-2\left(x_1x_2+y_1y_2+z_1z_2\right)=\\ x_1^2+y_1^2+z_1^2+x_2^2+y_2^2+z_2^2\end{split}\\ \begin{split} x_1^2+y_1^2+z_1^2+x_2^2+y_2^2+z_2^2-2\left(x_1x_2+y_1y_2+z_1z_2\right)=\\ x_1^2+y_1^2+z_1^2+x_2^2+y_2^2+z_2^2-2\left(x_1x_2+y_1y_2+z_1z_2\right)\end{split}\end{gathered}$

10.2  In component form, let $a = \left(1, -2, 2\right), b = \left(3, -1, -1\right),\text{ and }c = \left(-1, 0, -1\right)$. Evaluate $a \times b, a\times \left(b \times c\right) , c\times \left(a \times b\right)$ .

10.3  What is the geometrical significance of $a \times b = 0$?

10.4  Show that the vectors $a = 2i +3j +6k\text{ and }b = 6i +2j - 3k$ are perpendicular. Find a vector which is perpendicular to a and b.

10.5  Let a, b, c be three non-coplanar vectors, and $v$ be any vector. Show that $v$ can be expressed as $v = Xa + Y b + Zc$ where $X, Y, Z,$ are constants given by $X =\dfrac{v. \left(b \times c\right)}{a. \left(b \times c\right)}, Y =\dfrac{v. \left(c \times a\right)}{a. \left(b \times c\right)} , Z =\dfrac{v.\left(a \times b\right)}{a.\left(b \times c\right)}$ (Hint: start by forming, say, $v. \left(b \times c\right)$.)

# Problem Sheet 11 – Integration

11.1  Integrate $\cos\left(3x + 4\right)$ $\int\!\cos\left(3x+4\right)\,\mathrm{d}x=-\dfrac13\sin\left(3x+4\right)+c$

11.2  Integrate $\left(1 - 2x\right)^{10}$ $\int\!\left(1 - 2x\right)^{10}\,\mathrm{d}x=-\dfrac12\left(1-2x\right)^{11}+c$

11.3  Integrate $e^{4x}-1$ $\int\!\left(e^{4x}-1\right)\,\mathrm{d}x=\dfrac14e^{4x}-x+c$

11.4  Integrate $\left(4x + 3\right)^{-1}$ $\int\!\dfrac{1}{4x+3}\,\mathrm{d}x=\dfrac14\int\!\dfrac{4}{4x+3}\,\mathrm{d}x=\dfrac{1}{4}\ln{\left(4x+3\right)}+c$

11.5  Find the equation of the curve passing through the point $\left(1, 2\right)$ satisfying $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x$.

\begin{align*} \dfrac{\mathrm{d}y}{\mathrm{d}x}&=2x\\ \int\!\dfrac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x&=\int\!2x\,\mathrm{d}x\\ y&=x^2+c\\ 2&=1^2+c\\ c&=1\\ y&=x^2+1\end{align*}

11.6  A particle has acceleration $\left(3t^2+4\right)\mathrm{m}\mathrm{s}^{-2}$ at time $t\mathrm{s}$ . If its initial speed is $5\mathrm{m}\mathrm{s}^{-1}$, what is its speed at time $t = 2\mathrm{s}$?

\begin{align*} \dfrac{\mathrm{d} v}{\mathrm{d}t}&=3t^2+4\\ \int\limits_{5}^{v_1}\!\mathrm{d} v&=\int\limits\!_0^{t_1}\left(3t^2+4\right)\,\mathrm{d}t\\ \left[v\right]_5^{v_1}&=\left[t^3+4t\right]_0^{t_1}\\ v_1-5&=t_1^3+4t_1\\ \Rightarrow v&=t^3+4t+5\\ v\left(2\right)&=2^3+4\times2+5=21\mathrm{m}\mathrm{s}^{-1}\end{align*}

11.7  Find the area between the graph of $y =\sin x$ and the $x$-axis from $x = 0$ to $x =\pi/2$. $\int\limits_{0}^{\pi/2}\!\sin x\,\mathrm{d}x=\left[\cos x\right]_0^{\pi/2}=\cos\left(\pi/2\right)-\cos\left(0\right)=-1$

11.8  Find the area between the graph $y=\dfrac{1}{x-1}$and the $x$-axis between $x = 2$ and $x = 3$. $\int\limits_{2}^{3}\!\dfrac{1}{x-1}\,\mathrm{d}x=\left[\ln\left(x-1\right)\right]_2^3=\ln2$

11.9  Find the signed area between the graph $y = 2x+1$ and the x-axis between $x = −1$ and $x = 3$. $\int\limits_{1}^{3}\!\left(2x+1\right)\,\mathrm{d}x=\left[x^2+x\right]^3_1=\left(3^2+3\right)-\left(1^2+1\right)=10$

11.10  Find $y$, given that$\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=\sin x-\dfrac{4}{x^3}$

\begin{align*} \int\!\int\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\mathrm{d}x\,\mathrm{d}x&=\int\!\int\left(\sin x-\dfrac{4}{x^3}\right)\mathrm{d}x\,\mathrm{d}x\\ y&=\int\!\left(-\cos x+\dfrac{2}{x^{2}}+c\right)\,\mathrm{d}x\\ y&=-\sin x-\dfrac{2}{x}+cx+d\end{align*}

11. ## 15:0017th Aug 2012

Notes: 89

Reblogged from purepazaak

Shot of Riften from TES V: Skyrim

12. ## 20:317th Aug 2012

Notes: 318

Reblogged from emptyage

So maybe you saw my Twitter going nuts tonight. Or you saw Gizmodo’s Twitter account blow up. Or you saw this in AllThingsD. Or this in the DailyDot. Although embarrassing, Twitter was the least of it. In short, someone gained entry to my iCloud account, used it to remote wipe all of my…

13. ## 12:0530th Jul 2012

Notes: 6

Reblogged from itsbedlambaby

14. ## 14:0826th Jul 2012

Notes: 2

Reblogged from fail-chicken-deactivated2012081

I’ve given up trying to explain how to properly use the word, ‘I’ to people

It’s just too complex for them

15. ## 08:5021st Jul 2012

Notes: 14658

Reblogged from itsbedlambaby

Tumblr always needs more Star Wars