Related Rates Exam Review

SUMMARY

• Exam Review Intro
• Related Rates: Area, Perimeter, and Diagonal of a Rectangle
• Related Rates: Shadows Including Similar Triangles
• Related Rates: Distance and Velocity and Spongebob!

Exam Review Intro

We do our exam review in three ways:

1. Doing mini-exams pre-test style
2. Doing questions in class related to rusty topics
3. Doing old exam free-response questions and study how each question evolved throughout the years and do 3 questions a night

Today's class we decided to do #2, reviewing related rates.


Related Rates: Area, Perimeter, and Diagonal of a Rectangle



The length (L), the width (W), and the rates of which the lengths (dL/dt) and widths (dW/dt) are changing.

By convention, we're going to designate an increasing rate as a positive rate and a decreasing rate as a negative rate.

a) We know the formula for the area of a rectangle as A = L * W, where A is area, L is length, and W is width. Since we're looking for the rates of change (That's the definition of a derivative!) we differentiate the formula, with respect to time.

Since L * W is a product, we use The Product Rule to differentiate.

Since area, length, and width are all with respect to time--meaning that area, length, and width are functions of time--we must use The Chain Rule to differentiate.

Answer: dA/dt = 32 cm/s^2; increasing

b) We know the formula for the perimeter of a rectangle as P = 2L + 2W. Differentiate the formula using The Chain Rule, since P, L, and W are with respect to time--a function within a function! Then plug and chug.

Answer: dP/dt = -2 cm/s; decreasing

c) We know that the diagonal (D), the length (L), and the width (W) are related in The Pythagorean Theorem: the square of two sides (in this case, L and W) equals the square of the hypotenuse (D), so D^2 = L^2 + W^2. Differentiate the formula using The Power Rule and The Chain Rule.


Answer: dD/dt = -33 cm/13 s; decreasing


Related Rates: Shadows Including Similar Triangles



We know the height of the lamppost (L = 16), the height of the man (M = 6), the rate at which the man walks toward the streetlight (db/dt = -5), and the length of the man's shadow from the base of the lamppost (b = 10).

Since the man is walking toward the lamppost, by convention, the rate is negative.

b) Using similar triangles, we can see that the ratio of L to M equals the ratio of s to b+s. We simplify our proportions and differentiate to determine the rates of change (That's the definition of a derivative!). To differentiate, we use The Product Rule and The Chain Rule--Refer to Related Rates: Area, Perimeter, and Diagonal of a Rectangle for reference. We plug in the numbers to obtain ds/dt.

Answer: -3 ft/sec

a) Note this question is underlined in blue. The exam would never ask you to do part B because you need to do part B anyways to answer part A. The rate of the tip of the man's shadow is dP/dt. To obtain dP/dt, we differentiate P = b + s.

Answer: -8 ft/sec


Related Rates: Distance and Velocity and Spongebob!

HOMEWORK!



HOUSEKEEPING

• Next scribe is bench.
• Wiki constructive modification due Sunday midnight.
• Developing Expert Voices projects due soon.
• AP calculus exam is in two weeks.
• Three AP calculus exam free-response per night.
• Homework: Olympic Spongebob!
• We will be reviewing applications of derivatives (related rates and optimization), applications of integrals (density and volume), and techniques of antidifferentiating (integration by parts).
  

BOB v.9: Differential Equations

The intro exercises were simple at first since it's a review of chapter four, including those questions where we had to antidifferentiate the acceleration due to gravity twice to get displacement, until they started asking questions that I haven't learned until later on in the chapter. Luckily, I caught on, but, unfortunately, I still forget the +C when antidifferentiating...and it becomes more of a problem when there are more than two +C's for those questions where we antidifferentiate both sides after separating the variables. And antidifferentiating e^-t...I'll be prepared the next time I'm asked to antidifferentiate e^-t.

Solving differential equations graphically via slope fields, as I remember, was the easiest out of the three ways to solve a differential equation, since all there is to do is draw lines at each (lattice) point on the graph OR use prgmSLOPEFLD to draw the lines for me.

Solving differential equations numerically via Euler's method was a bit too much to absorb at first, but reviewing Newton's linear approximation technique was helpful since those two methods are similar.

Solving differential equations symbolically via separation of variables was a doozy, although I personally prefer this method over the numerical method, just because. Comparing the separation of variables questions in our book to the questions on slides, to me the slides were harder (and more helpful) since the slides had just more.

I'll admit the pretest wasn't my best pretest though, but I did keep track of where I went wrong! The main thing I need to watch out for are those +C's!

Slope Fields: Graphically

SUMMARY

• AP Exam Practice Non-Calculator Quiz #3
• Introduction to Differential Equations (Cont'd)
• Slope Fields: Graphically

AP EXAM PRACTICE NON-CALCULATOR QUIZ #3


Since f'(1)=0, that means that when we integrate starting at x=1, we should integrate up to the point where the integration yields 0, to satisfy the condition that f'(x)=0. (Where else would f'(x)=0 besides at x=1?) We can see through symmetry that occurs at x=3.


To find P'(x), we differentiate P(x) via chain rule. Plug the numbers in to get P'(3) = 12. Straight-up application of chain rule.


It's given in the question that H(x)=f^-1(x). If we insert both sides of the equation into f(x), we get f(H(x)) = x, which is the key idea to solving the question. Use implicit differentiation to find H'(x). Solve for H'(x). Plug and chug.


Let's visualize the graph. sqrt(1-x^2) is the equation of a semi-circle, derived from the equation of a circle, x^2 + y^2 = 1, and x is a factor that distorts the perfect semi-circle when x is multiplied by sqrt(1-x^2).

Because the function given contains a square root function, we know that the function has a limited domain. (The radicand can't be negative.) So, the interval of the function is [-1,1].

Note that the question is asking for the total area, which means we make sure that the top area (above the x-axis) doesn't cancel the bottom area (below the x-axis). That can be avoided by taking the absolute value of the equation. But, in this case, if we graph x(1-x^2)^0.5, we see that the graph is symmetrical about the y-axis. So, we can instead find twice the area of one-half the semi-circle-like figure. Either way yields the same result.

To find the area underneath a function (integration), we antidifferentiate the function, in this case by integration by parts, and apply the fundamental theorem of calculus to get 2/3. This is shown on the slide in blue font.

INTRODUCTION TO DIFFERENTIAL EQUATIONS (CONT'D)


Continuing from our previous class' introduction to differential equations...

a) We know that a(t)=-32 ft/s^2 is the acceleration due to gravity. Antidifferentiating a(t) yields velocity. We know v(0)=-16. So, we know from the family of velocity functions (-32t+C) that we're pinpointing the -32t-16 function. Antidifferentiating v(t) yields s(t), the position function. We know s(0)=96. So, we know from the family of position functions (-16t^2-16t+C) that we're pinpointing the -16t^2-16t+96 function.

b) When the rock strikes the ground, the stone would stop on the ground, so the stone's velocity would be zero. v(t)=0 at t =-3,2, which is determined by finding the roots of the velocity function. We reject t=-3 because we designated t=0 when we dropped the stone. At t=2, s(t)=0, which is determined by plugging t=2 into the s(t) function.

c) Plug t=2 into v(t) to get v(2)=-80. (Refer to part b.)


SLOPE FIELDS: GRAPHICALLY

dy/dx = y, meaning that the derivative function is the parent function itself and vice-versa, That's e^x. So, the slope is 0 at all the points on y=0. The slope is 1 at all the points on y=1. The slope is -2 at all the points on y=-2. The slope is 5/pi at all the points on y=5/pi. Etc. We drew the tangent lines at lattice points to easily visualize this field of slopes.

So, if a bird were to be at coordinate (0,1) and the slope fields were the direction the wind is blowing, then we can see the bird flying in the same direction as the wind in such a way that its trail can be seen as the graph of a function, which is shown in green.


HOUSEKEEPING

The rest of the slides are for homework.

Solution to 9.1 #9 is on the last slide.

The human rights assembly for The International Day for the Elimination of Racial Discrimination, hosted by The DMCI Human Rights Group, which we all enjoyed watching, is an assembly that occurs prior to Spirit Week. When Monday comes, Daniel McIntyre will enter the era of Spirit Week, a week of festivities hosted by Student Council, including Monday and Thursday period 2. So, Tuesday's scribe is .:. J + ME .:..

Spirit Week Themes

1. Monday: Sports Day - Sports Obstacle Course
2. Tuesday: Superhero Day - Save the Day Relay Race (you could dress up in a costume or wear a t-shirt that has a superhero logo on it)
3. Wednesday: Twin Day - Talent Show
4. Thursday: PJ/Crazy Hair Day - The Chris Frolic Comedy Hypnosis Show ($2/ticket)
5. Friday: Colour Code Day - Grade War Gym Riot (gr.12s wear green; gr.11s wear blue)

In preparation for the comedy hypnosis show and to continue the tradition of a YouTube vid per scribe post, here's a YouTube video of THE INCREDIBLE BORIS! But, keep in mind that he's a Las Vegas performer, and Student Council has enough money to hire a Comedy Network performer.



Fortunately, this year's Student Council is carrying on the Daniel McIntyre tradition of having a hypnotist perform at our school. This only happens once every two years, and our graduating year just so happens to be that year! It's $2/ticket for Thursday, March 26 to see your friends get hypnotized. Proceeds may go to Grad Committee if enough fundraising is made.

This will also be Daniel McIntyre's last traditional 5-day Spirit Week. Let's make the best of it. =)

BOB Version 8: Applications of Integrals

The first section of the chapter, integrating velocity/speed to determine displacement/distance, was a breeze. Then rotating graphs came along... We learned how to take a function, integrate it, and then revolve it around the x-axis, revolve it around the y-axis, revolve around y=-1, and revolve it up the wazoo! Hopefully, no one's going to ask me how to revolve it around a parabola because I don't know how--a possible DEV question?

The visualizing of the washer and our quest in search of the hole took a lot of time for me to wrap my head around. Determining the integral of a function that's rotated around a line besides the x- and y-axis kinda messed me over in the pretest, but I managed to re-learn some of it thanks to the collaboration.

But when The Mean Value Theorem of Integrals came along, it was like a break from all this revolving and rotating, which made my head revolve and rotate too, not in the literal sense.

I realize I have to emphasize working more on the density problems since when Mr.K said "this should be a gimme!" it wasn't really a gimme as of yet.

You have reached the end of my BOB.

AP Exam Practice Quiz 1 and Intro to the Intermediate Value Theorem of Integrals

Introducing the first of a series of AP exam practice quizzes, the intermediate value theorem of integrals, the day when the AP Calculus 2008: Without Bound blog is unblocked at Daniel McIntyre Collegiate Institute, the series of YouTube videos in tandem with Flickr pics, and kristina's powerpoint presentations!

AP EXAM PRACTICE QUIZ 1


Let h be a function defined for all x does not equal 0 such that h(4) = -3 and the derivative of h is given by h'(x) = (x^2-2)/x for all x does not equal 0.


a) Find all values of x for which the graph of h has a horizontal tangent, and determine whether h has a local maximum, a local minimum, or neither at each of these values.

To determine if there are local extrema at the critical numbers, we must first determine the critical numbers. Remember that there are critical numbers when a function has an asymptote or is undefined. h' = 0 when x = sqrt(2) and -sqrt(2). h' is undefined at x = 0 since 0 isn't in the domain (as stated in the question). So critical numbers are x = sqrt(2), -sqrt(2), and 0.

We use the first derivative test on h' to see where h' is positive or negative. Why? Because when h' is positive, h is increasing; when h' is negative, h' is decreasing. So a change in sign in h' would indicate a slope of zero at that point and that's where there are local extrema. According to the line analysis, we see that to the left of -sqrt(2), h' is negative; between -sqrt(2) and 0, h' is positive; between 0 and sqrt(2), h' is negative; and to the right of sqrt(2), h' is positive. Wherever h' changes sign from negative to positive, h has a local minimum; wherever h' changes sign from positive to negative, h has a local maximum. By the first derivative test (line analysis), there are local minimums at x = -sqrt(2) and x = sqrt(2). We don't look at 0 because it's not part of the domain of the function.


b) On what intervals, if any, is the graph of h concave up?

Rememberize these rules (from chapter 5 of your textbook):

If the second derivative is positive, the first derivative is increasing, and the parent function is concave up.

If the second derivative is negative, the first derivative is decreasing, and the parent function is concave down.

So wherever h" is positive, h is concave up.

We determine h" by differentiating h using the quotient rule.

We see that h" is positive, so h is concave up everywhere.


c) Write an equation for the line tangent to the graph of h at x = 4.

Pull out the point-slope formula: y-y1=m(x-x1)

The question gave us the x-coordinate: x = 4.
The question gave us the y-coordinate: y= -3.
m is the slope at x = 4, so plug x = 4 into h' which spits out 7/2.

Plug those numbers into the equation. BING! BANG! BOOM! We're done part c.

y+3=(7/2)(x-4)


d) Does the line tangent to the graph of h at x = 4 lie above or below the graph of h for x > 4?

If we draw a line tangent to the graph at x = 4, the line is below the graph at x > 4, because h is concave up everywhere.


INTERMEDIATE VALUE THEOREM OF INTEGRALS

Similar to the intermediate (or mean) value theorem of derivatives (that in a closed interval between a and b there exists a point on a continuous function which equals the average value), there exists a point on a continuous function which equals the average integral of the function.


We have b = 3. We have a = 0. We have f(x) = 1-2x. Plug the numbers into the equation. BING! BANG! BOOM! We get the average integral. But why does it work?

In this graph, pivot the yellow area found between c and b to the white area found between c and a. Notice that they fit together like a jigsaw puzzle to yield one big yellow rectangle. We can imagine that one big rectangle having the same area as the area under the graph, which we can see as the average value of the function.

HOUSEKEEPING

• Next scribe is benofschool.
• I'll be away on Monday to write the grade 12 English pilot exam.
• Don't forget to check Graeme's comment on Rence's post as it will prepare you for the exam!
  
• Don't forget to give Jamie your $1.25 donations so she can make her Betty Crocker style cheesecake!
• Pi Approximation Day is coming! Are we organizing the annual Coin Hunt for Pi Approximation Day too?

BOB Version 7: Finding Antiderivatives

I thought this unit easier than the previous unit because it's mechanical, but I was surprised on what I got on the antiderivative quiz. Substitution and integration by parts now makes sense to me, although antidifferentiating arctrig functions isn't second nature to me as of yet. I hope I'll meet the expectation of having these skills "fluid" by the time I write the test. It's also my hope that I can solve problems like the ones we had on our pretest without the use of a calculator, but using the calculator isn't time-consuming, if you understand, but, just like the pretest, I predict there will be one question on the test in the open response that will be challenging to answer. I hope I'm prepared for that when the time comes.

Antidifferentiating the Inverses of Trigonometric Functions

OVERVIEW:

• Formulas for the derivatives and the antiderivatives of the inverses of the trigonometric functions
• Antidifferentiating the inverses of trigonometric functions using techniques such as completing the square, the chain rule, substitution, and multiplying by one and using the constant multiple rule
• Proving an antiderivative of a derivative is equal to the derivative of an antiderivative

Slide 2

These are the derivative and antiderivative rules that we discovered in the previous class.


Slide 3

The denominator with a square root! Factoring won't do us any good here (I've tried already) because we want the fraction to look similar to arcsinx, so we complete the square. *zeph thinks back to grade 11 precal*

AND VOILA! We were lucky. It looks very similar to the arctanx function when, using the method of substitution, we let u = x - 3 and find the derivative of u (du), and also because the denomiator has 1 - u as its radicand. (But what if that 1 was another constant? We'll look at that later, but for now, let's solve this problem first!) Using the antderivative rules, we know that the antiderivative of 1/sqrt(1-u^2) is arcsinu + C. Resubsituting u into arcsinu + C, we get the answer!


Slide 4

Now what if that 1-(x-3)^2 underneath the radicand was 5-(x-3)^2? There's no 5 in the derivative of arcsinx, so we put it there by doing a neat trick. In the green on the right side of the slide, we factored out a five, rewrote (x-3)^2/5 as ((x-3)/sqrt(5))^2 (since 5 = sqrt(5)^2), and rewrote (x-3)/sqrt(5) as x/sqrt(5) - 3/sqrt(5) in blue. We let the blue stuff equal u and differentiated it to du. Rewriting the expression in terms of u, we get 1/sqrt(1-u^2) which we can resubstitute u back in to get the answer.


Slide 5

To let the left-hand side equal the right-hand side, let's take the derivative of both sides since we're dealing with antiderivatives. Differentiating arcsin(x/a), via chain rule, and replacing the resulting denominator with an algebraic equivalent (look in box--we let the 1 = a^2/a^2 and factored out 1/a^2) to get 1/sqrt(a^2-x^2). Therefore, left-hand side equals right-hand side.

Slide 6

For the left question, we used what we generalized in the box in slide 5 to "simplify" the denominator. Then we factored out a 1/10 using the constant multiple rule. Using the method of substitution, we let u = 3x/10, find its derivative (du), and rewrite the expression in terms of u. We realize that the antiderivative of the blue is arcsinu+C. We can then resubstitute u back into the expression to get the answer.

The middle and right questions we haven't done yet.


HOUSEKEEPING:

• Breaking the chain, Hi I'm Justus, you're up next.
• Middle and right questions, do for homework, I assume.
• Today is Day 3 of the school day cycle. Today is an optional class.
  
• All calculus students will be writing the grade 12 contest on the 18th, which are able to win $1000, but .:. J + ME .:. and Hi I'm Justus may need to reschedule their plans for the DMCI Jr. High Tours on the 18th and 19th.
• You may achieve a high school credit for tutoring an EAL student in math for 100 hours.
• And don't forget to listen to the Dr. Love messages at the last ten minutes of class!

SINE! COSINE! TANGENT! INVERSE?!? ... OR An Introduction to the Inverses of the Trigonometric Functions

OUTLINE:

• Definitions of many-to-one and one-to-one functions
• Unit circle
• Sine, Cap-Sine, Arcsin!
• Cosine, Cap-Cosine, Arccos!
• Tangent, Cap-Tangent, Arctan!
• Using Right Triangles

SLIDE 2: BLAST FROM THE PAST! GRADE 12 UNIT CIRCLE UNIT!

• Review of grade 12 precalculus (unit circle unit)
• Ask yourself what angle you're looking for and imagine the unit circle
• We gave the answers to the above questions, but Mr.K then says that one of the solutions in each answer is wrong, but why?
• Introducing the inverses of the trigonometric functions!

SLIDE 3: VOILA! THE UNIT CIRCLE!


SLIDE 4: SINE! CAP SINE! ARCSIN!

• We have the sine function. We want to know its inverse, but we can't since it's a many-to-one function.
• If we restrict the domain of the sine function to [-pi/2, pi/2], we have a function that has the same range as the sine function. This restricted sine function is the Sine (pronounced "Cap-Sine") function.
• Take the inverse of Sine (switch the x- and y-coordinates). We get the arcsine function.
• Note that Sine takes up Quadrants I and IV of sine. Thus, only values of sine found in QI and QIV are in the domain of Sine.

SLIDE 5: COSINE! CAP COSINE! ARCCOS!

• We have the cosine function. We want to know its inverse, but we can't since it's a many-to-one function.
• If we restrict the domain of the cosine function to [0, pi], we have a function that has the same range as the cosine function. This restricted cosine function is the Cosine (pronounced "Cap-Cosine") function.
• Take the inverse of cosine (switch the x- and y-coordinates). We get the arcsine function.
• Note that Cosine takes up Quadrants I and II of cosine. Thus, only values of cosine found in QI and QII are in the domain of Cosine.

SLIDE 6: TANGENT! CAP TAN! ARCTAN!

• We have the tangent function. We want to know its inverse, but we can't since it's a many-to-one function.
• If we restrict the domain of the tangent function to [-pi/2, pi/2], we have a function that has the same range as the tangent function. This restricted cosine function is the Tangent (pronounced "Cap-Tangent") function.
• Take the inverse of tangent (switch the x- and y-coordinates). We get the arctan function.
• Note that Cosine takes up Quadrants I and IV of tangent. Thus, only values of tangent found in QI and QIV are in the domain of Tangent.

SLIDE 7 TO 12: A BARRAGE OF QUESTIONS!

• We found the values of each using the unit circle. Note that some values of the unit circle aren't in the domain of the "Cap functions."

• 3a and 3b are undefined because the angle doesn't live in the domain of Sine and Cos.

• Sine and sine, Cosine and cosine, and Tangent and tangent are inverses of each other. Thus, they undo each other, like multiplying a number by 3 then dividing the number by 3.

SLIDE 13: USING RIGHT TRIANGLES!

• In 8a, we let sin^-1 2/3 = x to make a simplified expression cot x. Looks easier to solve, right? We used the definition of sine (the ratio of opposite side to the hypotenuse side) to determine what cot x is. We solved for cot x.

HOUSEKEEPING:

• Next scribe is Kristina.
• And don't forget to listen to the Dr. Love messages during the last ten minutes of class!

Integral Infomercial Introduction?

So, Mr.K, like the derivative commercials, we had our commercials as a reply to your introductory derivative commercial (the one where you said "I challenge you" etc.). Are you going to make one up for the integrals too? Because we're just doing the finishing touches to ours and we're not sure on how you want us to "submit" it.

Don't forget, you guys, Mr. Beaumont is interested in seeing our projects! =)

BOB Version 6: Integrals AGAIN!

The properties of integrals were straightforward, but those questions where they add a little twist and say, "Suppose f is an ODD FUNCTION and nonnegative on [0,2]..." perplexed me a little bit, but once I saw things visually, I saw what was happening in the problem and understood what was going on. (Hence, I am a visual learner and I like to see things visually.)

The Second Fundamental Theorem of Calculus, as pompous as it may sound, took a bit getting used to. I found the part where we had to prove (or was it argue?) that the derivative of the accumulation function is the parent function was easier to grasp than I anticipated it. Also, this is where I "lix up the metters," as Mr.K would say, talking about the variables of the accumulation function and the original function.

Visualizing the areas that's bounded by graphs helped a LOT. (Hence, I am a visual learner and I like to see things visually.) From the graphs, all I really need to do, basically, is to subtract or add the areas, however the case may be. Although the graphs dealing with the trigonometric function and integrating them or finding the areas, I don't consider them as my friends as of the moment. Sorry! (Yes, I apologized to the trig functions.) =)

Ben&Zeph DEV Timeline

(Click to enlarge.)

Properties of Integrals

OVERVIEW:

1. Developing Expert Voices (DEV) Project
2. StoryTools
3. The Fundamental Theorem of Calculus
4. Properties of Integrals - Introduction
5. Constant Multiple Rule
6. Sum and Difference Rule
7. Additive Interval Rule
8. Inequality Rule
9. End Notes

Developing Expert Voices (DEV) Project and StoryTools (Slide 1)

Timelines and sample questions were due today.

Continuing our discussion about our DEV projects, Mr.K provided us a link, http://cogdogroo.wikispaces.com/Dominoe+50+Ways, which is a list of web tools you can use for any project or assignment.


The Fundamental Theorem of Calculus (Slides 2 and 3)

Slides 2 and 3 is a summary of what we did in the previous class. Check the previous scribe post for an explanation.


Properties of Integrals (Slides 4)


Slide 4 summarizes the properties of integrals. (It's the most important slide in this scribe post, in my opinion.) Rules were given; then explained.

Constant Multiple Rule (Slide 5)

A function with a constant inside it can be factored out. Knowing this...

Integrating a function with a constant not factored out = integrating a function with a constant factored out.

For example, doubling the sum of areas equals the sum of all the areas doubled, if that makes better sense.


Sum and Difference Rule (Slide 6)

The integral of a sum is the sum of its integrals. In other words, when you are asked to integrate a polynomial function, you can integrate the terms inside the polynomial function, as seen in slide 6. However, the answer isn't complete in slide 6. Answer is 94/3.

Additive Interval Rule (Slide 7)

The area under a function of a graph in a given interval can be seen as the sum of smaller areas.

In this slide, using the Additive Interval Rule, the red area (an integral) plus the blue area (another integral) gives the green (total) area which is the integral for the function in the interval [-2, 2]. Remember that an integral is the area underneath the function of a graph.

Inequality Rule (Slide 9)
If a function is less than or equal to another function in the same interval, then the area underneath the former function is less than or equal to the area underneath the latter function.


End Notes:

• 6.1 The Definite Integral Again odd
• Post your DEV timeline on the blog by Sunday
• Next scribe is benofschool

Determining the Properties of the Parent Function Using the First and Second Derivatives

OVERVIEW:

1. Calculus Commercials: How Many Ways Can You Define the Derivative?
2. Mathematical and Computer Science Definitions of "AND" and "OR" Notes (slide 10)
   
3. Determine the Properties of the Parent Function Using the First and Second Derivatives (slide 2 to 9)
   

*************************************************************************************

1. Calculus Commercials: How Many Ways Can You Define the Derivative?

A continuation of last class' lesson, but first, we watched calculus commercials, which can be accessed here:

http://apcalc2008.blogspot.com/2008/12/gauntlet-is-thrown-down.html

The class agreed that the time restriction of 30 seconds to define the derivative was challenging, but there are our finished products! Note that people have commented on our projects.

*************************************************************************************

2. Mathematical and Computer Science Definitions of "AND" and "OR"

Let
A = go to the store
B = wear the hat

One day, you aren't sure if you should go to the store or wear the hat. So you think and think and think, but in terms of mathematical and computer science definitions, you notice that their definition of AND and OR are different from the AND and OR of English. The OR used in math is the inclusive OR. The OR used in English is the exclusive OR.

Here is a list of all the possibilities:

1. You go to the store and wear the hat. (A = true; B = true)
   
2. You go to the store but not wear the hat. (A = true; B = false)
   
3. You don't go to the store but wear the hat. (A = false; B = true)
   
4. You don't go to the store and you don't wear the hat. (A = false; B = false)
   

A U B = satisfy the conditions of A OR B
A ^ B = satisfy the conditions of A AND B


*************************************************************************************

3. Determine the Properties of the Parent Function Using the First and Second Derivatives

Using the derivative rules, we found the roots of the function.

Remember, k is a constant, not a variable!

We let k equal a negative number, like -1 000 000, in f' and discovered that to the left of 1/k, f is increasing, so f' there is positive.

We let k equal a positive number, like 1 000 000, in f' and discovered that to the right of 1/k, f is decreasing, so the f' is negative.

Since f is increasing to the left of 1/k and decreasing to the right of 1/k, we can imagine that at 1/k, there is a local max.

Remember, k is a constant, not a variable!

In the previous slide, we evaluated f' when 1/k>0.

In this slide, we evaluated f' when 1/k<0. style="font-style: italic;">

Since f is decreasing to the left of 1/k and increasing to the right of 1/k, we can imagine that at 1/k, there is a local min.

We have now obtained as much info as we can from the f'. On to f''!

Using the derivative rules, we determined f''.

We determined the roots of f'', so we can determine the sign of f'' on either side of the roots; this info can be used to determine f's concavity to either side of the roots.

We let k equal a negative number, like -1 000 000, in f'' and discovered that to the left of 2/k, f'' is positive, so f is concave up.

We let k equal a positive number, like 1 000 000, in f'' and discovered that to the right of 1/k, f'' is negative, so f is concave down.



Remember, k is a constant, not a variable!

In the previous slide, we evaluated f'' when 2/k>0.

In this slide, we evaluated f'' when 2/k<0. style="font-style: italic;">


Synthesizing all that we know, there is that list of properties f has.

• We know where f is increasing or decreasing at certain intervals.
• We know f's concavity.
• We know where f changes sign.
• We know f's local extema.

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END NOTES:

• Homework: the rest of 5.5 Antiderivatives
• Looking forward to the rest of the presentations tomorrow!
• Ending our scribe for 2008 is Francis.
• Reminder tomorrow's class will range from 10 to 25 minutes, with the Holiday Inn Gym Riot, starting at 10:30 AM, hosted by the DMCI Student Council. We will be called down to the gym between 10:15 AM to 10:30 AM. See you there! =)