Taking a pipe round a corner corridor optimisation question

You have a corridor which has an L-shape in it. The corridor looks like this:

cor

where a and b are the widths of the sections of the corridor. The question is to find the longest pipe that can be carried down this corridor. The word pipe here just means something long and with essentially 0 thickness. There is a huge simplification which is being assumed here, which is that the corridor is only 2 dimensional. Of course in a 3 dimensional corridor we have a lot more room to manoeuvre.

Let’s think about a pipe going round the corner. The longest pipe that can go through is the length of the shortest gap that it has to go through. So let’s think of a pipe at a particular angle with respect to the corner:

cor2

The pipe here is the blue line and \theta is the angle that it makes with respect to the corner.…

By | October 24th, 2018|Courses, First year, MAM1000, Undergraduate|1 Comment

Prove that for every positive integer n, 9^n – 8n -1 is divisible by 64.

Prove that for every positive integer n, 9^n-8n-1 is divisible by 64.

This question screams proof by induction, so we start with the base case, which in this case is n=1:

9^1-8-1 which is indeed divisible by 64.

Now, let’s assume that it holds true for some positive integer n=k. ie:

9^k-8k-1=64p for p\in\mathbb{Z}.

Now let’s see how we can use this to prove that the statement holds true for n=k+1. For n=k+1 we have:

9^(k+1)-8(k+1)-1=9(9^k)-8k-8-1=9(9^k-8k-1)+64k

where we have manipulated the expression to contain the left hand side of the inductive hypothesis. Thereby, plugging in the inductive hypothesis, we get:

9^(k+1)-8(k+1)-1=9(9^k)-8k-8-1=9(64p)+64k=64(9p+k)

but clearly 9p+k is an integer, so this is divisible by 64 and thus the statement holds true for n=k+1, thus it holds true for all positive integers k

How clear is this post?
By | October 24th, 2018|Courses, First year, MAM1000, Uncategorized, Undergraduate|0 Comments

A tricky complex numbers problem

The question is as follows:

If \frac{\pi}{6}\in arg(z+a) and \frac{2\pi}{3}\in arg(z-a) and a\in \mathbb{R}, find z.

So, let’s think about the information given and what we are trying to find. We want to find the complex number z which satisfies this slightly strange set of constraints, and the constraints are given in terms of z and a. So, by the looks of things, the answer will depend on a and so the final expression should be a function of a.

Now let’s explore the constraints. In fact, let’s simply take z+a and z-a as two complex numbers, but importantly two numbers which differ only by a real number 2a, so wherever they lie in the complex plane, they have the same imaginary part and differ only by an real part.

Now, the constraints are about the arguments of the two complex numbers. It doesn’t tell us anything about the magnitude of the numbers, so all the information tells us is the direction are in relation to the origin.…

By | October 24th, 2018|Courses, First year, MAM1000, Uncategorized, Undergraduate|1 Comment

Using polynomials to solve differential equations

One of the aims of MAM1000W isn’t just to teach you individual mathematical topics, but over time to allow you to see the links between these subjects. Sometimes we do this explicitly, and sometimes you should notice the connections yourself simply by seeing one topic pop up in the middle of another. As I’ve written before, so much of it is about noticing patterns.

Today in class I gave a differential equation which wouldn’t be solvable by any of the methods we have looked at.

y''(x)+\cos(x)y'(x)+e^xy(x)=x^2

This is second order linear but its coefficients are not constant. We don’t have any way in to solve this. We actually wanted to solve this with the initial conditions y(0)=1 and y'(0)=-1.

Actually, that’s a lie. We didn’t want to solve it, but we wanted to get an approximation for the solution close to x=0. This is like saying: OK, so we have a differential equation for population dynamics, or climate change, or the heating of an object, and we don’t worry too much about the very far future, but we want to know what it’s going to do in the short term.…

By | October 15th, 2018|Uncategorized|1 Comment

Why did we choose that range for theta when doing trig substitutions?

Remember when we are doing a trig substitution, for instance for an integral with:

 

\sqrt{a^2-x^2}

 

We said that we should choose x=a\sin\theta, which seemed reasonable, but we also said that -\frac{\pi}{2}\le\theta\le\frac{\pi}{2}. Where did this last bit come from?

Well, we want a couple of things to hold true. The first is that any substitution that we make, we have to be able to undo. That is, we will substitute x for a function of \theta but in the end we need to convert back to x and so to do that we have to be able to write the inverse function of, in this case x=a\sin\theta. The \sin function is itself not invertible because it’s not one to one, so we have to choose a range over which it is one to one. We could choose -\frac{\pi}{2}\le\theta\le\frac{\pi}{2} or we could choose \frac{\pi}{2}\le\theta\le\frac{3\pi}{2} (amongst an infinite set of possibilities). That would also be invertible. However, remember that we are going to end up with a term of the form:

 

\sqrt{1-\sin^2\theta}=\sqrt{\cos^2\theta}

 

So if we want this to simplify, we had better choose our range of \theta such that \cos\theta is positive, so that we can write \sqrt{\cos^2\theta}=\cos\theta.…

By | August 2nd, 2018|Courses, English, First year, MAM1000, Undergraduate|2 Comments

Integrals with sec and tan when the power of tan is odd

We went through an example in class today which was

 

\int tan^6\theta \sec^4\theta d\theta

 

In this case we took out two powers of sec and then converted all the other \sec into $latex\ tan$, which left a function of tan times sec^2\theta d\theta. We wanted to do this because the derivative of \tan is \sec^2 and so we can do a simple substitution. If we have an odd power of \tan, we can employ a different trick. Let’s look at:

 

I=\int \tan^5\theta\sec^7\theta d\theta.

 

Here, sec is an odd power and so we can’t employ the same trick as before. Now we want to convert everything to a function of \sec and have only a factor which is the derivative of \sec left over. The derivative of \sec is \sec\tan, so let’s try and take this out:

 

I=\int \tan^5\theta\sec^7\theta d\theta=\int \tan^4\theta\sec^6\theta (\sec\theta\tan\theta)d\theta.

 

Now convert the \tan into \sec by \tan^2\theta=\sec^2\theta-1:

 

I=\int (\sec^2\theta-1)^2\sec^6\theta (\sec\theta\tan\theta)d\theta=\int (\sec^{10}\theta-2\sec^8\theta+\sec^6\theta) (\sec\theta\tan\theta)d\theta

 

where here we have just expanded out the bracket and multiplied everything out.…

Fundamental theorem of calculus example

We did an example today in class which I wanted to go through again here. The question was to calculate

 

\frac{d}{dx}\int_a^{x^4}\sec t dt

 

We spot the pattern immediately that it’s an FTC part 1 type question, but it’s not quite there yet. In the FTC part 1, the upper limit of the integral is just x, and not x^4. A question that we would be able to answer is:

 

\frac{d}{dx}\int_a^{x}\sec t dt

 

This would just be \sec x. Or, of course, we can show that in exactly the same way:

 

\frac{d}{du}\int_a^{u}\sec t dt=\sec u

 

That’s just changing the names of the variables, which is fine, right? But that’s not quite the question. So, how can we convert from x^4 to u? Well, how about a substitution? How about letting x^4=u and seeing what happens. This is actually just a chain rule. It’s like if I asked you to calculate:

 

\frac{d}{dx} g(x^4).

 

You would just say: Let x^4=u and then we have:

 

\frac{d}{dx} g(x^4)=\frac{du}{dx}\frac{d}{du}g(u)=4x^3 g'(u).…

Advice for MAM1000W students from former MAM1000W students – part 5

While I resisted Mam1000W every single day, I even complained about how it isn’t useful to myself. Little did I know when it all finally clicked towards the end that even though I wasn’t going to be using math in my life directly, the methodology of thinking and applying helps me to this day.

Surviving Mam1000W isn’t really a miraculous thing. While everyone tends to make it seem like it’s impossible, it is challenging (Not hard) and I said that because I have seen first-hand that practice makes it better each time. Getting to know the principles by actually doing the tuts which is the most important element of the course in my opinion will make sure that even though you feel like you aren’t learning anything when the time comes (usually 2nd semester) it will all click on how you actually are linking the information together.

Another important aspect is playing the numbers game.

By | May 14th, 2018|Courses, First year, MAM1000, Undergraduate|1 Comment

Advice for MAM1000W students from former MAM1000W students – parts 2 and 3

Part 2:
——-

So one thing that really helped me was having a partner in tuts. We would do the tuts as far as we could and we would then try to help one another in the tuts and ask the tutors for help if there was a difference in opinion.

Another thing that helped studying, going through past papers and tuts were so important.

If I was ever stuck and couldn’t really understand the textbook I would go on YouTube and watch a guy named Professor Leonard.  He’s videos are super long but extremely helpful and worth your time.

And last but not least, it’s important that you try your best to work everyday with maths because once you fall behind its difficult to catch up. Even if you do just one problem a day I promise it will help In the future.

and part 3:

——

I would suggest to MAM1 students that they should not fall behind the maths syllabus if they have tests in other subjects because it is very difficult to catch it up and requires much more effort than one thinks.…

By | May 8th, 2018|Uncategorized|1 Comment

Advice for MAM1000W students from former MAM1000W students – part 1

This is the first in a series of posts where I will be putting up the sage words of advice of former MAM1000W. Often, these students struggled their way through the course, before making a breakthrough in their study methods. I hope that maybe it will be easier to listen to students who have been through the struggle, than the advice of lecturers who seem to know it all (though I promise you, we do not!).

Here is the first:

——-

As an Actuarial Science student I was aiming for 70% last year. I clearly remember that at orientation I asked some of the older ActSci students at orientation what they had done when they scored below what they needed to. I was so shocked, and a little scared when the group I asked said they never had. I wasn’t worries at this stage though because I thought I’d done well at maths at school, and I’d do well at maths here.…

By | May 8th, 2018|Uncategorized|3 Comments