Some more volume visualisations

Here is an animation which may help you imaging a shape which has a circular base, with parallel slices perpendicular to the base being equilateral triangles:movie3

 

The same thing, where the slices are squares.movie4

 

And here is the region in the (x,y) plane between y=\sqrt{x}, the x-axis and the line x=1. rotated about the y-axis. Here a thin shell is drawn in the volume, then pulled out. Then it is replaced, then the volume is filled with shells, and each of them is pulled out of the volume vertically. This is to give you an idea about how to visualise the method of cylindrical shells.

 

movie5

How clear is this post?
By | September 14th, 2017|Courses, First year, MAM1000, Undergraduate|1 Comment

Guidelines for visualising and calculating volumes of revolution

I have seen some people try to blindly use the formulae for volumes of revolution by cylindrical cross-sections and by cylindrical shells, and I thought that I would write a guide as to how I would recommend tackling such problems, as generally just using the formulae will lead you down blind alleys.

I’ve created an example, with an animation, which I hope will help to master this technique.

So, here is a relatively fool-proof strategy:

  1. Draw the region which you are going to have to rotate around some axis. This will generally be a matter of:
    • Drawing the curves that you have been given
    • Finding where they intersect
  2. Draw the line about which you are supposed to rotate the region
  3. Draw the reflection of the region about the line of rotation: This gives you a slice through the volume that will be formed
  4. Now you have to decide which method to use:
    • Take a slice through the volume perpendicular to the axis of rotation.
By | September 13th, 2017|Courses, First year, MAM1000, Uncategorized, Undergraduate|2 Comments

Using integration to calculate the volume of a solid with a known cross-sectional area.

Hi there again, I have not written a post in while, here goes my second post.

I would like us to discuss one of the important applications of integration. We have seen how integration can be used to solve the area problem, in this post we are going to see how we can use a similar idea to solve the volume problem. I suggest that we start by looking at the solids whose volume we know very well. You should be able to calculate the volumes of the cylinders below (yes,  they are all cylinders.)

 

circular cylinder                                 rectangular cylinder                triangular cylinder

Cylinders are nice, we only need to multiply the cross-sectional area by the height/length to find the volume. This is because they have two identical flat ends and the same cross-section from one end to the other. Unfortunately, not all the solid figures that we come across everyday are cylinders. The figures below are not cylinders.…

Introduction to trigonometric substitution

I have decided to start writing some posts here, and this is my first post. I would like to introduce trig substitution by presenting an example that you have seen before. Trig substitution is one of the techniques of integration, it’s like u substitution, except that you use a trig function only.

Let’s get into the example already!

\int_{-1}^{1} \sqrt{1-x^2} dx

If you equate the integrand to y (and get x^2+y^2=1, y\geq 0), you should be able to see that this is the area of the upper half of a unit circle. The answer to this definite integral is therefore the area of the upper half of the unit circle (yes, the definite integral of f(x) from a to b gives you the net area between f(x) and the x-axis from x=a to x=b), is \frac{\pi}{2}.

We relied on the geometrical interpretation of the integral to solve the definite integral, but can we also show this algebraically?…

By | August 27th, 2017|English, First year, MAM1000|3 Comments

Riemann sums to definite integral conversion

In the most recent tutorial there is a question about converting a Riemann sum to a definite integral, and it seems to be tripping up quite a few students. I wanted to run through one of the calculations in detail so you can see how to answer such a question.

 

Let’s look at the example:

 

\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(9\left(4+(i-1)\frac{6}{n}\right)^2-8\left(4+(i-1)\frac{6}{n}\right)+7\right)\frac{13}{n}

 

There are many ways to tackle such a question but let’s take one particular path. Let’s start by the fact that when the limit is defined, the limit of a sum is the sum of the limits. We can split up our expression into 3, which looks like:

 

\lim_{n\rightarrow\infty}\sum_{i=1}^n9\left(4+(i-1)\frac{6}{n}\right)^2\frac{13}{n}-\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(8\left(4+(i-1)\frac{6}{n}\right)\right)\frac{13}{n}+\lim_{n\rightarrow\infty}\sum_{i=1}^n7\frac{13}{n}

 

Let’s tackle each of these separately. Let’s look at the first term:

 

\lim_{n\rightarrow\infty}\sum_{i=1}^n9\left(4+(i-1)\frac{6}{n}\right)^2\frac{13}{n}

 

Well, we can take the factor of 13 outside the front of the whole thing to start with, along with the factor of 9, and this will give

 

13\times 9\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(4+(i-1)\frac{6}{n}\right)^2\frac{1}{n}

 

We see here that we have a sum of terms, and a factor which looks like \frac{1}{n} in each term.…

By | August 23rd, 2017|Courses, First year, MAM1000, Uncategorized, Undergraduate|7 Comments

Some sum identities

During tutorials last week, a number of students asked how to understand identities that are used in the calculation of various Riemann sums and their limits.

These identities are:

 

\sum_{i=1}^n 1=n

\sum_{i=1}^n i=\frac{n(n+1)}{2}

\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}

\sum_{i=1}^n i^3=\left(\frac{n(n+1)}{2}\right)^2

 

Let’s go through these one by one. We must first remember what the sigma notation means. If we have:

 

\sum_{i=1}^n f(i)

 

It means the sum of terms of the forms f(i) for i starting with 1 and going up to i=n. Sometimes n will actually be an integer, and sometimes it will be left arbitrary. So, the above sum can be written as:

 

\sum_{i=1}^n f(i)=f(1)+f(2)+f(3)+f(4)+....+f(n-2)+f(n-1)+f(n)

 

We haven’t specified what f is, but that’s because this statement is general and applies for any time of function of i. In the first of the identities above, the function is simply f(i)=1, which isn’t a very interesting function, but it still is one. It says, whatever i we put in, output 1. So this sum can be written as:

 

\sum_{i=1}^n 1=1+1+1+1+....+1

 

Where there are n terms.…

By | August 20th, 2017|Courses, First year, MAM1000, Uncategorized, Undergraduate|1 Comment

MAM1000W 2017 semester 2, lecture 1 (part ii)

The distance problem

If I want to know how far I walked during an hour, I can ask how far I walked in the first five minutes, and how far I walked in the second five minutes, and how far I walked in the third five minutes, etc. and add them all together. ie. I could write:

 

d=d_1+d_2+d_3+d_4+...d_{12}

 

Where d_i is the distance walked in the i^{th} five minutes. To calculate a distance, we need to know how fast we are going, and for how long. In fact:

 

distance=velocity \times time

 

where you can think of velocity as the same thing as speed (though there are subtle differences which you will find out about later). This formula works if the velocity is constant, but what if it is changing. Well, if we have a graph of velocity against time, then we can think about splitting the graph into intervals (like the five minute intervals above), and approximating that during a small interval of time, the velocity is roughly constant.…

MAM1000W 2017 semester 2, lecture 1 (part i)

I wanted to put up a little summary of some of the most important things to remember from the end of last semester. There was a sudden input of new concepts, so let’s put some of them down here to get a clear reminder of what we need to know. A few things in this post:

  • The antiderivative
  • Sigma notation
  • Areas under curves

Antiderivatives

An antiderivative of a function f on an open interval I is a function F such that:

 

F'(x)=f(x) for every x\in I

 

Note that we say an antiderivative, not the antiderivative. There can be many functions whose derivatives give the same thing. While we know that:

 

\frac{d}{dx}\sin x=\cos x

 

And therefore we can say that \sin x is an antiderivative of \cos x. However, we can also say that:

 

\frac{d}{dx}(\sin x+3)=\cos x

 

So \sin x+3 is also an antiderivative of \cos x. In fact for any constant c it is true that \sin x+c is an antiderivative of \cos x. We write this statement as:

 

\int\cos x dx=\sin x+c

 

This is called the indefinite integral of \cos x with respect to x.…

Checking direction fields

I was recently asked about how to spot which direction field corresponds to which differential equation. I hope that by working through a few examples here we will get a reasonable intuition as to how to do this.

Remember that a direction field is a method for getting the general behaviour of a first order differential equation. Given an equation of the form:

 

\frac{dy}{dx}=f(x,y)

 

For any function of x and y, the solution to this differential equation must be some function (or indeed family of functions) where the gradient of the function satisfies the above relationship.

The first such equation that we looked at was the equation:

 

\frac{dy(x)}{dx}=x+y(x).

 

We are trying to find some function, or indeed family of functions y(x) which satisfy this equation. We need to find a function whose derivative (y'(x)) at each point x is equal to the value of the function (ie. y(x)), plus that value of x.…

By | October 11th, 2016|Courses, First year, MAM1000, Undergraduate|1 Comment

Radius of convergence of a series, and approximating polynomials

I hinted today that there were sometimes issues when you did a polynomial approximation, that if you tried to find the value of a function a long way from the region about which you’re approximating, that sometimes you wouldn’t be able to do it. This is related to an idea called the radius of convergence of a series. In the following we are just plotting polynomials, but you can see that whereas in the polynomial approximation for sin(x) (on the right), as we get more and more terms, we approximate the function better and better far away from the point x=1 (which is the point about which we are approximating the function). However, for the function \sqrt{1+x}, after x=3, the approximations are nowhere near the function itself. This is because that function has a radius of convergence of 2, when expanded about x=1. This is due to the behaviour of the function at x=-1, which is a distance 2 away.…

By | August 19th, 2016|Courses, First year, MAM1000, Undergraduate|0 Comments