Arc Length of a Curve

Arc Length of a Curve

with Anna, Chelsea, and Claire

https://docs.google.com/document/d/1cXAeMFuCnEbcgDYZE16NXayhOy5J3gEyBHdcpVDKm3g/edit?usp=sharing

Taylor Series

This is the Taylor Series.

Presented by Caroline Albertson, Lexi Doudera, and Eben Kopp.

On the Thirtieth of June, in the year of 2013, as we know it.

The concept of the Taylor Series was formally introduced by the mathematician Brook Taylor in 1715.

If the Taylor series is centered at zero, then that series is also called a Maclaurin series, named after the Scottish mathematician Colin Maclaurin, who made extensive use of this special case of Taylor series in the 18th century. 

The Taylor series is commonly used to approximate a function, using the series specifically. A Taylor polynomial is any finite number of initial terms of the Taylor series of a function. The Taylor series of a function is the limit of that function's Taylor polynomials, provided that the limit exists.

When you are approximating a function, the Taylor series goes like this:

Where it can be written as a more compact notation:

where n! denotes the factorial of n and ƒ ⁽ⁿ⁾(a) denotes the nth derivative of ƒ evaluated at the point a. 

Example:

We will approximate the graph of  using the Taylor series. 

As the Taylor polynomial increases with each degree, it approaches the correct function.

 

This graph shows  and it's Taylor approximations, polynomials of degree 1, 3, 5, 7,9, 11 and 13.

Here is another example of the Taylor series and the graph of e^x.

This graph shows e^x and it's Taylor approximations, polynomials of degree 1, 2, 3, 4, 5, 6, 7, 8, and 9.

Here are the calculations which go along with the graph:

Now that you know what the Taylor series is, here is a video which shows just how much we love math:

http://www.youtube.com/watch?v=FjGk4QY31Pc&feature=autoshare

Here is a link to our quiz:

https://docs.google.com/forms/d/1N1mYLjqP0XA6LuSDvQCPUIV_jHZxd9FSJHoiJrw56IA/viewform

Integration by Partial Fractions

Integration by partial fractions can be difficult - it's a good thing we're here to help. Let James and Alex guide you through the tricky world of partial fraction decomposition as it pertains to integrals.

We begin with a difficult example of a function to integrate: let's say you wanted to find something like:

Looks time-consuming, no? Well, there is a way to more easily reduce integrals via the magic principle of decomposition! Decomposition of a fraction is the process of breaking down more complex equations into their component parts. The first thing you should know is that this method of integration is based off an old algebra trick, simply applied to calculus. We all know the Fundamental Thorem of Algebra which states that all polynomials can be factored into linear, irreducible quadratic and/or complex (containing imaginary components) factors. This means that polynomials will always have roots which are either linear (such as x+4, x-7, or 12x etc) or are irreducible quadratics which when "reduced" are shown to be made themselves of factors which contain imaginary numbers, always in pairs. Let's say we're not getting into imaginary numbers in our integrating, and we are happy with leaving our decomposed polynomials in terms of their basic building blocks - linear and irreducible quadratic factors.

An important note, however, is that this method about to be described only applies to equations being integrated who take the form of a fraction (technically all equations are fractions) but with the caveat that the denominator (bottom) of the fraction must have a higher power of x than the numerator (top). In cases where the function being integrated has a higher degree numerator, simple long division of your  integral will suffice to find your answer. 

Lets look at the denominator of the above integration.

The reducible quadratic shown in the bottom of the fraction being integrated above is factored into its component parts, two linear factors (x+3) and (x-2). This is nothing new - we've been factoring polynomials for a long time now. Our new integral looks like this: 

This is the first of the four major cases which apply to integration by partial fractions - the decomposition (breaking down into factors for easier integration) of Distinct Linear Factors (when the fraction denominator being decomposed breaks down into simple linear factors). 

1. Take a look at the following equation and see if you can follow the logic involved.

Why have these constants A and B suddenly been introduced to this process? For the equation in question, it is necessary for some value of A and B to exist so as to make the equality above an accurate statement, right? Some A must exist and some B as well which make this a true statement since the denominators of the two parts of the second side of the equality are factors of the first. We'll demonstrate with a very simplified version of this rule:

If we suppose, this makes sense right? We know that.

If we multiply both sides by a funny form of one, we can create a common denominator to match the one we factored apart. When we do this we can see that the above equation is true. 

Lets apply this simple rule to our more complex function being integrated. We'll use the same multiplication by one strategy to create a common denominator, and use that to create a more simple form of our function so as to integrate without as many problems - we'll be decomposing it. 

 ...all dx, of course.

So how does this help us? Well, now both sides of our equation share the same denominator - let's take it right out. Also, lets drop the integration brackets. They're obscuring our goal right now - to find the values for those two constants A and B and create a more simple equation in order to integrate.

Now, we made a fairly large leap of faith in beginning this problem - mostly, that this equation we're using is differentiable. If we continue to hold this to be true, we assume that this equation we have just found is an identity, meaning that it is true for all values of x in order to be a continuous function which can be integrated at all points. 

If we can plug in any x value and be given a correct result in this equation, we can use this to our advantage! Lets cheat the system! The following is a means of performing fractional decomposition which does not work in every circumstance - its a trick like the "drop the power" rule for differentiation, as it only works in simple cases like Case 1 - Distinct Linear Factors. Lets plug in x values which will result in answers for A and B. Try plugging in x=2

Thus, we know now that B = 2/5

Lets do the same strategy for A: say that x = -3

Likewise, we now know that A = 3/5

Now, lets use this knowledge to decompose our origional integral!

 and thus, plugging in our values, 

This is actually , as the now defined A and B are constants.

The new form of the original integral we've found is much easier to integrate now. By using the rule of the integral of 1/x, and remembering that this integral does not have set limits and thus requires a +C, we find the answer to be:

From that messy fraction integral, we have produced an answer! Nice job!

This trick of setting things equal to zero by exploiting the identity of the function we are using can also be applied to problems which belong to Case 2: repeated linear factors. When you see functions which you need to integrate which contain linear functions as the factors of their denominator, but which are repeated more than once, you approach the problem in much the same way. 

How would you go about attacking this? 

Use the same approach as before, making sure to take into account all possible factors of the denominator. This denominator has many more factors than our original easy example - let's see if we can't find them all.

The parts which go into this function's denominator are (x+1) and (x), yes, but don't forget that its also true that a root for this function is (x+1)^2 and (x+1)^3 as well. With this knowledge, we can set up our decomposed series of fractions. Again, we drop the integral sign for the sake of clarity, and will pick it back up later when we actually solve the integral.

Are you seeing a pattern so far? The fractions we use are comprised of the factors of the bottom of the original function, with letter placeholders over each one which represent constants we will solve for. Integration by partial fractions is a long and painstaking process, but its most important feature is that it yields results. You can probably see how we would go about answering this problem: multiplying by forms of one in order to create one single denominator, then substituting in values of x to isolate constants and solve. Then, we plug in values and perform integration just like the first time. 

Now, lets look at the other two cases which are possible in this method. So far, we've been dealing with distinct linear factors, but Case 3 - Distinct Quadratic Factors and Case 4 - Repeated Quadratic Factors are more difficult to solve as they are often at least semi-immune to this trick we've learned so far. So, let's learn the one true (yet painfully slow) method of reducing these equations down. It starts a lot like all the other forms of decomposition, by breaking down the denominator into fractions bearing the parts of said denominator. In Case 3, we see equations like 

Notice first off that the denominator is made of irreducible quadratics, aka quadratics which can't break down into linear functions without involving the use of imaginary numbers. Thus, they must remain in the form they are in. Why, you may ask, have the simple constants of A and B become so complicated in this new form, with x's and all? When broken apart in this manner, these equations containing quadratic factors must contain generic linear factors with generic terms like A and B and C and D on top instead of even simpler constants, since these denominators must be one degree greater in power than the numerator which they are underneath. When we were only dealing with linear factors, our numerators were one power less than them, having x's raised to the 0 instead of to the 1. Now, we've raised all the powers by one. Case 4 takes this idea to yet another level in much the same relation that case 1 had to case 2. A sample case 4 problem would look like this: 

Just like in case 2, we approach this problem by finding all possible factors and creating a decomposition which reflects them.

Now, this is much more complicated a setup than we've had to deal with so far. Let's learn the principle which will serve us better for these more complicated quadratic decompositions. Our trick works well for easier equations, but there is a longer and more time-consuming method which will work every time. 

We'll use an easy example: let's refer to our original original equation way back in case 1 which told us that the following relationship was valid.  

We will now put a name to this new process - it's called Equating Coefficients. First, we multiply our A's and B's in. 

Now, let's look at the invisible exponents in front of our x.

This is where the equating of the coefficients comes into play: since these two identities are equal to each other, the coefficients of each x^1 should equal out, as well as the coefficients of the x^0. This way, we know that 1 = A+B, and that 0 = 2A+B

We now have two true equations, which can be solved any number of ways - the chief two being either equating of equations, or substitution. We'll demonstrate both.

     1 = A+B                                                 1-A = B

-    0 = 2A+B                                               0 = 2A + (1-A)

____________

     1 = -A

A=-1, substitute back in to find B = 2

Are these answers familiar? They should be, they're the exact same answers we found for this example when we did it via our little trick! Equating coefficients is longer and more time-consuming, but will never fail you. Combine it with the trick (for finding of easily-identified coefficients) in order to get to the values you're looking for, and decompose your function into a more easily-integrated form. If you want any practice, UC Davis has 10 problems which they explain well, which will help you practice the basics.

Surfaces of Revolution

Surfaces of Revolution
Measuring the Enigma

Surface area really is tricky. Volume, as we know, can be measured using liquid. An interesting aside, Archimedes originally discovered this principle while taking a bath, and he subsequently ran through the streets naked shouting about it (you can learn all about that story here). Measuring length is a similarly simple process: take a string, match the curve, and the measure the string. Easy. Surface area, however, is different. Certainly, surface area is simple enough to find on prisms, and they have flat sides, but once curves are involved the process becomes more difficult. (the video I showed a part of during the presentation)

Take this torus for example. The surface area of this object could be measured by calculating the surface area of each individual square on the surface on this torus. However, it would not be all that accurate. But what if we called the surface area of these squares as representative rectangles? What if their size was our change in x? If that was the case, we could set up a limit as the size decreases towards 1/∞ that acts on a summation of all the surface areas. This limit would give us an approximation of the surface area (seen below).

However, this equation is only useful as a model of an idea. It is not really feasible, at least at our level of mathematics. For this reason, we will think about surface area in terms of circumferences. 

Just as each representative radius led us to the area when finding the volume of a surface of revolution (below), that same radius begs us to use circumference when finding surface area.

VOLUME                                         SURFACE AREA

  

The surface area method sums up the surface area of the sides of our beloved representative rectangles. For this reason, it is important to pay attention to the shape of these surfaces of revolution as each limit is approached. If the surface of revolution animation above, there are two small disks at either end. Those disks both have their own surface area, so their area must be individually calculated. If we say that r(x) is the radius function, then the Surface Area integral evolves to look more like this:

But this isn't all. Consider, for instance, the function below. This function has both an exterior surface (y=√x) and an interior surface (y=x/10).

Due to this, the integral must change again, along with some of it's notation. Let r_big(x) be the exterior radius, and r_small(x) be the interior radius. I will leave r(a) in the equation, even though r(a)=0 for this particular problem set. If we think about it, we must then sum both the interior and exterior surfaces, along with the area of the two washer at each end. 

***NOTE: there are now washers instead of disks are each end. This change occurred because of the switch from one function being revolved to two functions being revolved. Now we must use washers to find the difference between the larger and smaller circles.***

       LEFT WASHER              RIGHT WASHER              EXTERIOR SA              INTERIOR SA

Of course, this notation makes everything look a bit more complicated that it needs to be, but it gets the point across. So now let's try this method out for the last video.

We are revolving the area between f(x)=√x and g(x)=x/10 on the interval x=[0,4] about the x-axis. We know that there will be no washer on the left side, as both f and g are equal to zero when x=0. So our initial equation looks like this:

But wait, the second surface [with g(x)] is just a cone, isn't it? An angled line? Yes, it is! So we can use the SA formula for the side of the cone (SA=πrl) instead of the crazy integral mess.

Now that is better. One integral, a couple functions appraised at single values. No problem. Slap that into the calculation (good old fnInt), and … SHAZAM!

That whole mess is equal to ~50.600587.

You are a genius. Go do your happy dance, then try the quiz.

The Quiz

Thanks!
Cole

p.s. Duncan values SATs more than a day in O'Brien's classroom. Let us collectively dishonor him.

More math animations here

Quiz answers (be honest, no one is grading you)

Polar Coordinates

POLAR COORDINATES!

An Introduction to Polar Coordinates

     The polar coordinate plane is a two-dimensional coordinate system in which each point on a plane is determined by a distance from a fixed point and an angle from a fixed direction. We've been exposed to this in math classes before, but we'll include a quick refresher on some of the basics of the coordinate system to bring everyone up to speed.
     A point on the polar plane is expressed as (r, θ), where r is the distance from the fixed point (generally the origin) and θ is the angle from the fixed direction (again, generally the origin). It is fairly easy to convert between polar and Cartesian coordinates with simple trig: to find the x Cartesian coordinate from a polar coordinate, you just need to do rcos(θ): x=rcos(θ). To find the y Cartesian coordinate, the process is very similar: y=rsin(θ). This beautiful image may help:

     Now, what about changing from Cartesian to polar? From the diagram above, it's fairly easy to tell that r can be found through the Pythagorean Theorem: r=√(x²+y²). This image also makes clear how to find θ, again through the use of trig. If given x and y, you can solve for θ through the following: θ=tan^-1(y/x). There are at least five PatrickJMT videos on this topic, but here is one that is very brief and straightforward. Let's move on to polar coordinates in calculus now!

Finding Areas with Polar Coordinates

Finding areas in the polar plane is similar to finding areas in the cartesian plane in that integrals are used in both cases. However, the limits of integration and the equation are slightly different.

Cartesian coordinates:                                                                               Polar Coordinates:

As you can see, we are no longer working with values of x and y, but instead with values of theta.

Also, the integral looks slightly different because there is a 1/2 out front and an r² inside the integral. This is because the area of a circle is given by πr² and we’re summing little tiny segments of different circles. However, because each part is not an entire circle, we need to put it into proportion, and we do this by multiplying πr² by dθ/(2π), because dθ/(2π) is the ratio of the little tiny change in θ to the total number of radians in a circle. This gives us r²dθ/2, or (1/2)r²dθ. Then, since we’re summing all the little areas, we throw in a definite integral sign with limits from the initial angle θ to the final angle θ and we’re done.

In terms of finding areas once you know the equation, the hardest part is visualizing the graph. PatrickJMT does a wonderful job explaining how to find an area with polar coordinates here:

https://www.youtube.com/watch?v=RrVEuywFhjg

Even more complicated is finding areas between two polar curves. PatrickJMT does it again in this thrilling sequel:

https://www.youtube.com/watch?v=obo7QAvaxSU

and this: https://www.youtube.com/watch?v=AkoR2eIGvQc

and this: https://www.youtube.com/watch?v=4I3K_-UK6u4

Derivatives in Polar Coordinates/GeoGebra

One crucial element of understanding the mystical overlap between polar coordinates and Calculus is the process of finding dy/dx in polar coordinates. First, you must realize that x=rcos(theta) and y=rsin(theta).  Just like y is often seen as a function of x, r in this case can be viewed as a function of theta.  Therefore,  y can really equal f(theta)sin(theta) and x can really equal f(theta)cos(theta). Taking the derivative of that, you must use product rule to get dy/d(theta) and dx/d(theta). For example, dy/d(theta) would equal f prime(theta)sin(theta)+f(theta)cos(theta). Then you replace f prime (theta) with dr/dtheta since r is the function and you put dy over dx and you get:

Another interesting and vital skill is knowing how to use Geogebra to plot polar points and functions. First, you’ll want to go into grid options and change grid type to polar.  You can show the grid and you’ll start to see the magical unit circle-like grid.  

You can always plot a point in polar by just entering coordinates in style (r;theta) with a semicolon separating the coordinates.  To graph a polar function simply create a slider with value t.  Then create point a with polar coordinates (r(t);t) where r(t) is your function of t.  To make the function visible, find the locus of all points by creating a locus with point A.  

If this looks complicated at first, don't worry, it is. Good luck!

Final Project: Multiple Integrals

MULTIPLE INTEGRALS

By Scotty and Bex

Multimedia


Click Here To Go To The Quiz

Review: This year we have only done single integrals, both indefinite and definite. To review, an integral indicates that we must find the area beneath and curve. To do this, we have to find the antiderivative of a given function. If we are working with an indefinite integral, we will end up with a function in terms of x or y (depending on whether we are taking the integral with respect to dx or dy) + C, a constant of integration. However, if we are using a definite integral, we will have certain values that we evaluate the antiderivative at. See below for a visual:

But what happens if we are no longer looking at flat surfaces?

Double Integrals:

When you graph a curve on the x-y plane, you use a single integral to find he area beneath that curve. But what happens when you add a third axis, (let’s call it z) and that 2D plane becomes a 3D space? And what does something like that even look like? See below...

So that’s kinda wonky. But let’s say z is a function of whatever x and y value you happen to be at on the 2D plane. In other words, z is a function in terms of variables x and y. Mathematically, we’ll refer to it as f(x,y). But what does that look like? Well, you know that 3D shapes have length, width, and height. So if you think of x and y as your length and width, z must be your height. What does that form? It forms a shape, and it can be any shape, depending on what your specific f(x,y) function looks like. But to get started understanding this, let’s draw an arbitrary f(x,y) function on our triple-axes below...


Hmm, nifty. Now what if we want to find the volume of the space enclosed between that shape and the x-y plane. Well, as you already know, a single integral is what you use to find the area under a curve. Therefore, a double integral can be used to find the volume under a shape! Let’s start by setting boundaries. On what intervals of x and y would we like to find the volume of this space? Let’s set a constant x and a constant y, as shown below...

Wow, that looks super great! We’ll call our constant x endpoint “a”, and our constant y endpoint “b”. Finding the volume of that whole space probably seems pretty intimidating, but we’re super-duper AP Calc students! We’ve conquered over 200 O’Brien opportunity days, literally! A real, full-blown AP test! A million online applets... we can handle a double integral. Let’s first think about how we find a single integral. We take a bunch of tiny slivers and add them up, the length being the y value (which changes depending on which one you chose, making it a function), and the width being an infinitely small change in x, otherwise known as dx. For a 3D shape, it’s the same idea. We want to find the 2D areas of a bunch of little slivers of our 3D shape, and add them together. See the illustration below! The sliver we’d like to find the area of is highlighted in purple, and its infinitely small thickness is highlighted in green...

Represented algebraically, that sliver looks like this: 


So what is the area of that 2D sliver? Well, it’s the length times height. We know the length is going to be x = a, and z is a function of x and y, remember? f(x,y). And remember to multiply by that infinitely small change in x for the width, dx.

Now you want to add a bunch of those slivers up--every sliver on the y interval from 0 to b. This is where the second integral comes in. You want to integrate that first integral, which was in terms of dx, again, in terms of dy. See below...



One very important thing to remember is that when you integrate your first function, f(x,y), you integrate it in terms of dx. This means you treat all y values as constants and only apply the power rule (in reverse, of course) to x terms. Then, the second time you integrate, you are integrating in terms of dy and therefore can apply the power rule to the remaining y terms, as all x terms have become constant through the FTC. Here, let’s do a practice problem:

So in this case, f(x,y) = x^2 + y^2. First, we need to integrate in terms of dx, so we treat the y term as a constant:


Now, just use the FTC to turn the x values into constants.


Ok, now it’s pretty simple: just integrate from 0 to b in terms of y and plug in your endpoints with the FTC for the final answer.



TA-DAAA!

Triple Integrals: Triple integrals involve functions that are in terms of x, y, and z.

One way a triple integral can be used is to find the volume of a box.

Let 0 ≤ x ≤ 5
Let 0 ≤ y ≤ 3

Let 0 ≤ z ≤ 2

When these limits are graphed on an x, y, z plane, a cube is formed. To find the volume of this cube, we can just use basic geometry : L x W x H = (5)(3)(2)=30m^3. However, we can also use a triple integral to find the volume. 

First, we take a very small cube from our large cube and call its dimensions dxdydz or dV for short.
We then make three integrals regarding the dimensions of the x axis, the y axis, and the z axis, respectively:

Note that there is a constant (1) in the triple integral above. This represents the density of the block which is 1 kg/m^3. Then we integrate as such: 

Although this gave us the volume of the block (we can check our answer with geometry), it also gave us the mass. #gotlucky We know that the formula for density is d=m/v. In our integral, we are taking density and multiplying it by volume. Therefore, our answer should be mass. AND IT IS!!! If the density is 1 kg/m^3, the mass and the volume will be the same! YAY!

But how would our answer change if the density was 2 kg/m^3? Mass and volume would no longer be the same. The answer to the triple integral will now be the mass:

However, if the mass is 60 kg and the density is 2 kg/m^3, with the equation d=m/v, we find that our volume is still the same (as we would expect it to be - the cube's dimensions did not change).

If density isn't given, and the question asks for volume, just assume the density is 1 kg/m^3 or whatever unit is being used. It's just a rule of thumb.  

This is just a mini intro....

One way we can use triple integrals is to find the volume bounded by certain curves. Let's say we have the curve x + 2y + z = 6, x=0, y=0, z=0. The surface itself looks like so:

(a semi-birds eye view; flat up-down is z-axis, flat left-right is x-axis, vertical up-down is z-axis)

How do we find the boundaries??? Well....

For the x-axis bounds, we know that x = 0. But for the upper bound? If we make y=0 and z=0 in the original equation x + 2y + z = 6, x=6. Therefore, for our graph, x=[0,6]. If we do this for y and z (making x and z equal 0 and making y and x equal 0, respectively), y = [0,3] and z = [0,6]. One thing that we can do is draw traces in order to see the what the three possible planes look like:

 Now we need to create a triple integral in order to find the volume. Because no density function is given, we can just assume the density of the pyramid is 1kg/m^3. We know with a triple integral, we will be summing up the small volume dxdydz within the proper bounds. It really doesn't matter what order we sum, so let's start with dz.

dz: We know that one of the planes will be z=0. The other one will be the surface given in terms of x and y: z = 6 - x - 2y. This means that the first part of our integral looks like such:

 dy: We know that one of our y-axis limits will be y=0. If we look at the graph of the x-y trace, we see that it's also bounded by that diagonal line. Because the x-y trace does not involve z, we merely eliminate z from the given equation and solve for an equation in terms of x. The result is another integral:

dx: The last limit is easy because we just sum up dx from 0 to 6. Straightforward. Our final product is: 

 

Then we integrate as we did with the double integrals, treating the variables which we are NOT integrating with respect to as constants. Here is a picture from the online solution:

Triple integrals seem to be commonly used to find the mass of a non-uniform massed object. The functions used are often in terms of x, y, and z and are representative of a variable density. We set these up the same way as before. Take the surface from above: z = 6 - x - 2y. Let's say they give us the variable density: p(x,y,z)=6-x. Then they ask us to find the mass. Because we already figured out the boundaries of the different axes, we can just use the triple integral we used earlier but instead of the density being 1, the density is now 6-x.

This is how far most problems will ask you to go because it becomes super complicated and shit. But we all know how to integrate so IN THEORY we could do it.

 #SENIORSUMMERBITCHEZZZ

Works Cited: 
"Triple Integral Problems." Triple Integral Problems. N.p., n.d. Web. 22 May 2013. <http://homepages.math.uic.edu/~dcabrera/practice_exams/topics/tripleintegrals.html>.

"Double and Triple Integrals." Khan Academy. N.p., n.d. Web. 22 May 2013. <https://www.khanacademy.org/math/calculus/double_triple_integrals>.