Integration by Trig Substitution

Blog Post on google docs because I am not a fan of the Blogger's formatting. 

https://docs.google.com/document/d/1E5TqLG3DPgaYTAACS_1_mfq2iioFC9CYd3c6V6NFvpc/edit?usp=sharing

31-1-13 Definite Integrals, Riemann Sums, and Trapezoids

31st January, 2013

Class began today with 2 Explorations: "1-3 Introduction to Definite Integrals" and "1-4 Definite Integrals by Trapezoidal Rule," how fun.  If you want another copy, they can be found here Explorations 1-3/1-4 After doing our best to complete these explorations on our own, we went over the answers.

The 1st problem :

We basically just reviewed what we knew about Riemann Sums from the last class. To find how far you travel in the time interval [30, 50], it’s just a matter of finding the areas of rectangles. So, we simply multiply 20 (rate) by 60 (time) to find an area of 1,200 ft, which is how many feet traveled. Or, you could simply count up all the squares, and multiply that number (24) by the area of 1 of the squares (50) to get the same answer. With Riemann sums all we’re doing is adding up rectangles where the height of the rectangle is the velocity, v(t), and the base is change in time, d(t).

We then touched upon how if there is an unvarying rate of change, we can use base x height to get the area, but if there is a varying rate of change, we have to estimate to get the distance traveled. This is where definite integrals come in. We will use those later on in the football problem later.

First we discussed what exactly we need for a definite integral sum:
1. An a & b which are called the limits of integration
2. A  f(x)dx which is basically v(t)
3. A height/base

Or if you want something a little simpler : 

“In words: The definite integral sum is the limit of the Riemann sums as the number of subdivisions gets larger and larger.”

Here's a little visual representation :

Now onto the football problem :
For this problem it can be helpful to think about the football like this : [insert image here]. We have to add up all the squares, which ends up being around 230-ish. Then we have to sum up cross sectional areas to find out how much area is inside the football. Basically, the definite integral represents the volume of the football in inches cubed . It allows us to multiply the base with a variable height, how handy since footballs are not shaped like rectangular prisms. 

We then had some questions pop up:

Weston asked, “Why is it d(t) and not t ?” O’Brien said to consult his handy applet (which can be found here applet) to show that it isn’t height x time, it’s height x the change in time (velocity x change in time).

Alex also had a question concerning the units for the previous problem, something along the lines of “How is it ft?” O’Brien then showed how since velocity (ft/sec) x time (sec), it cancels out to be just ft. (Reminder that the football problem is in INCHES not FEET, otherwise that would be a pretty massive football).

O’Brien then took the time to tell us to calm down about the definite integrals.
“Just breathe with our definite integral!” He said gleefully.
The class was not amused.

...moving on...


Besides those pesky Definite Integrals we also were introduced to something even more ridiculous:  Trapezoids.

More specifically, this mess (let's try to ignore the terrible attempt at a Star Wars pun) :

Anywhoozies...trapezoids, why? Well, a trapezoid has fewer error than our Riemann sum rectangles because rectangles have 2 of the same points. To find the area of a trapezoid we have to average the bases, then multiply by height : 




So by adding up all our trapezoids we are going to get a more exact definite integral. Exactly what we always wanted! What makes this that much better is that we can use Geogebra to model this for us. Using the velocity function from the problem :

We simply graph that function like normal, then type “trap” and fill in all the areas. It should look like this:

Then, if you wanna get really fancy, you can set n to be a slider, so you can adjust the number of trapezoids to get a more or less accurate area. You should get something like this :

Oodles more accurate than those rectangles.

Some more tech tips were then brought up.
To do definite integrals on your calculator : fnInt
For Geogebra : type “integral” and plug in your function starting/ending points and you get your area integral. yay.

We then had a brief intermission from all this madness when OB went about pulling his awkward potted tree across the room on a string. Yes, this is our AP Calculus teacher.

Once the tree had been returned to its rightful place in the corner of the room we continued with class as usual. Jumping back to those Riemann Sums (rectangles, not trapezoids) we learned that when we’re approximating definite integrals with Riemann Sums, we get 5 different choices of Rectangular Approximation Method (RAM). They are :
LRAM — “L” for left [0]
RRAM — “R” for right [1]
MRAM — “M” for middle [.5]
Upper RAM
Lower RAM

The way we use these on the ever-powerful Geogebra (for the first 3): type in rectangular sum and plug in the all the values except for “Position for rectangle start”. Then, depending on which RAM you’re doing you plug in either 0, .5, or 1. LRAM is 0, RRAM is 1, MRAM is .5. However, if you’re trying to use Upper/Lower RAM you simply type in upper sum or lower sum. Easy peasy.

Now let's have some visuals :

LRAM

MRAM

RRAM

LOWER

UPPER

And lastly we got dealt this crazy-looking thing :

[The interval [2,1] is partitioned (yes, I spelled partitioned wrong in the above photo, I'm tired) into "n' subintervals of depth ∆x = 3/n let x sub 3 denote a point in the k^(th) subinterval]


Apparently though, if the change in x is 3/n then 3 is coming from the length of the partition interval. You can separate ∆x as the base and the stuff in parentheses as the height, and from there you can just plug that into your calculator as

That's all for now. Final thoughts from the class :

Autumn : why can’t this class be over?

Noah : O'Brien hates us all.


IW #7
The supercorrections for our Free Response Question Quiz and...
p. 274/5, 23, 31, 35
p. 286/3, 7, 13, 19, 43, 45
p. 318/31, 35



[STILL IN THE PROCESS OF EDITING. DON'T JUDGE ME]

Scribe Post 10/24

We began class by taking our quiz on IW’s 6 & 7. Once that was done we reviewed and praised Noah’s scribe post;  “That’s what an A+ scribe post looks like” said O'Brien with much zeal.

The main topic of today’s class was ...

Derivative of the Inverse Function

Since we already know how to find the derivative of practically every other type of function, why not inverse functions too? (I CAN THINK OF PLENTY OF REASONS TO NEVER LEARN THIS, but my opinions are not influential in the slightest)

However before we could delve into derivatives of inverse functions OB reminded us that he had posted the solutions/answers to IW#8 : https://dl.dropbox.com/u/3243156/CHRHS/apcalc/U2%20ap%20IW%20%238%20solutions.pdf

We then encountered a problem with the solution to “Negative Integers” on Question 2 of IW #8, OB said he would change it.

* It was also hinted that #3 on IW #8 could possibly be a finely crafted opportunity day question *

OKAY SO , The majestic Inverse Function

We already know that to find the inverse of a function we replace f(x) with y and solve for . Easy peasy.

(If you want a more in-depth refresher of finding the Inverse of a function patrickJMT is your BFF)

“The Basics” : http://www.youtube.com/watch?v=nSmFzOpxhbY&feature=relmfu

“Finding the Inverse of a Function” http://www.youtube.com/watch?v=Ec5YYVxyq44

We also know that if you graph a function and its inverse, they have y = x symmetry. These are created by interchanging the   and values. For example :

If you were to graph these functions it would look something like

with the diagonal y = x symmetry. At any point on the function its inverse point on the inverse is its reflection across the y = x line. Example : if you have a point (1, 2) inverse is (2,1).

Algebraically : If f and g are inverses then :

and if we take the derivative with respect to we get :  

which then can be rewritten as: 

 
Next, we looked at an example. Find:

Finding the derivative of this inverse by using the Chain Rule:

where

1 is the derivative of (x -1) in the Chain Rule.

Then we plug in 2 for x, since that was the original goal (finding the derivative of the inverse function) at 

x = 2 :

and if we then solve we get a lovely small answer of :

Now, let's compare this answer to the answer we would get if we used the reciprocal relationship we found above. 

If (2, 1) is a point on the inverse of f , then (1, 2) is a point on f. 

So, 

 

And if we plug in 1 for ,we get something like 

 

and if we go back to the our first answer of

we can see that it was the reciprocal.

The key to this approach was comparing points on f and points on f'.

Next class we will learn how to find the derivative of the inverse function of at x = 3

How exciting.

And as always continue to post questions! 

 

To do for next class : IW #9

Keep working on IW #8 and older IW

p. 169/1, 2, 4

p. 171/Exploration 1

p. 175/29b & c

#APCalculusproblems
#Whydowedothistoourselves
#O'Brienlikestomakeuscry

UPDATE:

Work from class on how to find derivatives of inverse functions :