Solutions for study problems.

Hello, calculusians. Click on the images below to see solutions to the study problems. I hope that you have first tried them yourself.

If you have disagreements and/or questions, post them to the blog. You should answer each other's questions as I cannot guarantee how often I will check the blog. Feel free to stop by after your other finals if you are still struggling.











Don't forget to study related rates, implicit differentiation, and extrema. Practice some chain rule problems from your text or of your own creation.

Good Luck.

QOD:

Each one of these represents position, velocity and acceleration.

Acceleration is the derivative of velocity, and velocity is the derivative of position.

We will therefore note acceleration as the second derivative, velocity as the derivative and position as the function.

Key point on the graph:

We not that the slope of the green line approaches it's lowest quantity here, and that the y value of the red one reaches it's lowest value. Following that, the blue line reaches a y value of 0 when the red line's slope approaches zero.

Taking this point as representative of the function and its derivatives as a whole, then we assume that the green line is the function, the red line is the derivative, and the blue line is the second derivative.

They match up at all points. At the beginning, green (position)'s slope is zero, and red (velocity) is at zero, while it's slope is at it's lowest value -- and the green point is at it's lowest y value.

We then went over the function:

c(x) = 5000 + 10x + .05x^2

Where c is cost and x is designer glasses.

Asking for a) The average cost of designer glasses of 100 pairs, b) The marginal cost of 100 and c) The change in cost of 101 and 100.

a) asks for the plugging in of 100 as x and then dividing it by 100, finding the cost of producing each pair, at a production level of 100. Doing so:

(5000 + 10(100) + .05(100)^2)/100 = 65

It costs 65 to produce each pair of glasses at 100 glasses being produced.

b) asks for the change in cost at 100. We find this deriving. I used the power rule.

c(x) = 5000 + 10x + .05x^2

c'(x) = 5000(0)(x^-1) + 10(1)(x^0) + .05(2)(x^1)

c'(x) = 10 + .1x

We then substitute x for 100 to find the slope at that particular point.

c'(100) = 10 + .1(100)

c'(100) = 20

So the cost is increasing by 20 at that particular point.

c) asks for the difference between the cost at 101 and 100.

c(101) - c(100)

5000 + 10(101) + .05(101)^2 - 5000 - 10(100) - .05(100)^2

20.05

To our athletic compadre

Hey Max, whether you check this during your trip or when you get back I hope you guys did well and came back with your heads held high. I am going to try my hardest to post as nice as you do it but I am not sure if i am quite that blog savvy yet. Thank goodness there was no graphs! I really don't know how to write powers on this so I just put carrot ... sorry about that. Homework: Read- Pgs.128-129, Do questions #1, 3, 9, 11, 23, 25.

QOD

A bacteride was added to a growing population of bacteria. The size of the population in time t, in hours was given by: P(t) = (10^6)+(10^4)x-(10^3)x^2. This is 10 to the power of 6 plus 10 to the power of 4, x plus 10 to the power of 3, x squared. Find:

a)the rate of growth at 3 hours, 5 hours and 8 hours.

b)is the reate of growth increasing or decreasing at t = 7 hours?

a) In order to find the rate of growth we took the derivative of P(t) which is (10^4)-2(10^3)x which come out to: P'(t) = 10000-2000x We then plugged in each number... P'(3) = 10000-2000(3) = 4000 bacteria/hour , P'(5) = 10000-2000(5) = 0 bacteria/hour , P'(8) = 10000-2000(8) = -6000 bacteria/hour

b) In order to find out if the rate of growth was increasing or decreasing we had to (this might be a little confusing but you will understand it hopefully) find the rate of growth or the rate of growth. This would show if it is positive or negative, thus if it is increasing or decreasing. This is the same as d^2y / dx^2. This is finding the derivative of the derivative.

P''(t) = -2000 (beacause the derivative of P'(t) = 10000-2000x = -2000) P''(7) = -2000 (because there is no x so it is constant) bacteria/hour/hour Therefore it is decreasing (because it is negative) by 2000.

Part A tells us: If our starting point if 0, from 0 to 5 hours the population of bacteria is increasing. At t>5 the population of bacteria is decreasing. The maximum population is at t = 5 hours.

Part B tells us: As the population of bacteria decreases the rate of growth increases (in an absolute value sense).
_________________________________________________________________

What is the meaning of f '(a)?

• the slope of a tangent to f at x=a
• the limit of the slopes of the secants to f at x=a
• instantaneous rate of change of f at x=a
  

What is the meaning of f ''(a)?

• the derivative of f '(a)
• rate of change of the slope of f at x=a
• instantaneous rate of change of the instantaneous rate of change of f at x=a

-the value of f ''(a) tells you whether f '(a) is increasing, decreasing or constant

Higher Order Derivatives
1st derivative f '(x) or dy/dx
2nd derivative f ''(x) or (d^2)y/dx^2
3rd derivative f''''(x) or (d^3)y/dx^3
nth derivative f^n(x) or (d^n)y/dx^n

Pg. 133 #29
Cost of Production: a function that gives the cost associated with varying levels of production

Marginal Cost of Production:
"The change in cost when production in increased by 1 unit." - Definition from Mr. A

*Marginal cost of production always varies depending on where you are starting.
For example: If you are starting at zero prodution and want to change your unit to 1000 production, the change in cost is going to be very large because you need to pay for startup costs such as machinery. As opposed to if you are starting at 1000 to 2000 production the change in cost is not going to be as much because you already have payed the startup costs.

The equation for the average rate of change is your classic:

C2-C1 /X2-X1

Where c=cost, x=# of items produced. The numerator is the 2 costs compared (C1 being the starting production cost and C2 being your new production cost). The denomonator is the jump in production (X1 being your starting amount of items being produced and X2 being your new amount of items you wish to produce)

This is an average rate but in Calculus, we want instantaneous. How do you find this...... THE LIMIT!

The equation for the instantaneous rate of change is:

wow, that was super hard to write that and post it. haha. anyways...

This is the marginal cost when x=x1.

So what is marginal cost?

It is setting the level of production to maximize prodution.

Here is an example and some questions to try out:

Total cost in dollars of producing x designer eyeglass frames is given by:
c(x)=5000+10x=0.05x^2

*The cost of the function should be continually increasing.
*You should also see the fixed cost vs. variable cost. A fixed cost may be the cost of a machine to shape plastic, a variable cost may be say cost of screws for the frames because this depends on the amount of frames you are producing.

a)find the average cost of producing 100 frames
b)find the marginal cost at a production level of 100 frames
c)find the actual cost of increasing production from 100-101 frames
-actual cost is found by taking the cost of 100 and 101 and then you will know the cost of this increase.


Good Luck!

Scribe Post Evaluation Form

Here is a copy of the form that I will fill out and return to you after you make your scribe post. Use this post and the scribe post instructions (see sidebar) to guide you in writing your scribe post.

For Max/ Emily

QOD: A car starts at rest, rapidly accelerates at a constant rate for 2 sec, coasts for 2 sec, slowly decelerates at a constant rate for 4 sec, coasts 5 more sec, and then brakes to a stop with a constant acceleration.









d = accelerates with a curve for 2 sec
maintains slope for 2 secs
decreases with a curve for 4 secs
maintains slope for 5 secs
momuntarily maintains slope, but changes to a horizontal line

d' = accelerates at a constant rate of change, a line, for 2 secs
continues unto a horizontal line for 2 secs
decreases at a constant rate of change, another line, for 4 secs
continues unto a horizontal line for 5 secs
decreases into a horizontal line

d'' = horizontal line for 2 secs
hole, due to corner, turns to a horizontal line at y=0 for 2 secs
hole, due to corner, turns to a horizontal line at y=-.5 for 4 secs
hole, due to corner, turns to a horizontal line at y=0 for 5 secs
hole, due to corner, turns to a horizontal line at 7=-.25

Test Review Questions

Pg 173 #s 7, 55, 57, 58

Good Luck Everyone!!!

Hello Rachel and Connor

10/22
QOD

Suppose that f and g are both differenciable at x = 3 and that f(3) = 7, g(3)= 2, the derivative of (f)= -1 and the derivative of (g)=4. Using x=3 find:

a) d/dx (g x f)
b) d/dx (g/f)
c) d/dx (-2f-gf+3g)


*You should try and answer these and i will post the work below*
























Ok now that you have tried these i will now walk you throught them and give you the answers.

a) d/dx (gf) (gf)' = f'g + g'f

= -1(2) +4(7)

= -2+28

=26

b) d/dx (g/f) (g/f)'= g'f-gf'/f(squared)

= 4(7)-(-1)2/7(squared)

= 30/49

c) d/dx (-2f-gf+3g) (-2f-gf+3g)' = -2f'-(gf)'+3g'

= -2(-1)-26+3(4)

= -12

I am sure you guys got those pretty easy. On to todays lesson.


Today's Lesson


Find all domains where the function is indifferencaible.

The function below is a piecewise function.


absolute value of x where x is less than 2
x(squared) - 2 where x is greater than or equal to 2 and less than or equal to 5
f(x) x-7/x(squared)-49 where x is less than 5


What are the first basic things that we need to check?

1. Check problems that may occur in the three individual problems
2. One sided limit boundaries
3. One sided derivatives at boundaries


Work for the listed above

1. Absolute value of x has a cusp or a corner at x = 0 therefore we must consider that because the function takes on this identity everywhere less than 2 and because zero is less than 2 the function is indifferenciable at x = 0. Now we must look at x(squared) - 2. Because that function is a polynomial it is continuous everywhere and thus differentiable at every domain. 1/x+7 which is the simplified version of the third element of our function has a hole at x = 7, which makes it discontinuous and thus it also differentiable at x = 7.

2. Now lets check the one sided limit values at the boundaries.

l 2 l = 2(squared) - 2
2 = 2

5(squared)-2 does not equal 5-7/5(squared) -49

Therefore there is a corner at x=5

3. Ok now we must check one sided derivatives at boundaries. So...

d/dx ( lxl ) = d/dx (x) for x greater than 0 so d/dx (x) = 1 .

d/dx (x*squared*-2) = 2x

because we are observing the boundary x=2 approaching from the right, we must subsitute 2 in for x and that yields 4. This slope is not equal to the answer of one which is the slope at 2 when approaching from the left, thus we may concur that there is a corner there. The same may be done for the slopes at x = 5 but we already know it is indifferencialbe there.

Another piece of quiz review in todays lesson was the application of the power rool....i just gave you a hint for this one. Try it and the answer will be posted below! :]


3(x*squared*) / x*power of 3/4





























Answer: DAILY DOUBLE!!!.....




































JK GUYS! THE ANSWER IS... F' (X) = 15/4 *FOURTH ROOT OF (X)

Tuesday 10/16

Today, I tried to prove the product rule but I made a mistake. The mistake was thinking that I had made a mistake, when I had not. I just needed to finish the proof.

I stopped when I had g(x+h)f '(x) and thought that it was wrong. I knew that I wanted g(x)f '(x) instead but one gets that by continuing on and evaluating the limit.

See the image below for the finished proof. Sorry for the brain block.

Wed 10/10

Rachel,

It's the least I could do for all of your effort and interest.






HW: R pg 105 -106, pg. 111 # 1-21 odd

I hope you feel better.

Lesson from September 17

Question of the Day:

Find the equation of the horizontal asymptote for the given function:

Factor the x^4 out of the numerator.

Multiply the entire equation by 1 by mutliplying the numerator and denominator each by 1 divided by the highest degree of x in the denominator.

Distribute the 1/(x^2). Also, recognize that the (2/x^4) in the numerator has been replaced by 0 because it approaches 0 as x gets closer to infinity.

Recognize that (x^2/x^2) in both the numerator and the denominator are equal to 1. Also, recognize that (4/x^2) has been replaced by 0 because it approaches 0 as x gets closer to infinity.

Simplify.

Therefore, the horizontal asymptote is y= -1.

Now onto the lesson. We learned the Intermediate Value Theorem and it's Corollary, and to do so, we performed the following instructions.

-On a graph, draw a function between two endpoints

-On that line, choose two points 'a' and 'b', and label them.
-Find f(a) and f(b).

-Define any point W between f(a) and f(b).

Intermediate Value Theorem:

Hypothesis:
-Suppose f is continuous on the closed inverval [a,b].

-W is any number between f(a) and f(b).

Conclusion:

-There exists as solution within [a,b] such that f(c)=W.

-There is at least one x value for W, your chosen y value.

-If f(x) is continuous, it can't skip any values.

Corollary to the Intermediate Value Theorem:

Hypothesis:

-Suppose f is continuous on a closed interval [a,b].

-f(a) and f(b) have opposite signs.

-[This is true of the above graph].

Conclusion:

-There is at least one number within [a,b] such that f(c)=0.

-The function must cross the x axis at least once, so there is at least one zero.

So, that's about it. Hurrah for finally figuring out how to work this thing. If anybody has any questions, come talk to me and I'll try to explain it better. Good luck studying for the quiz.

The Work of the Day (September 13, 2007)

Hey guys we sure missed you today! I hope that you both are feeling better and i hope you get this blog in time to make use of it. I know you both are dying to get this info... Love you guys and see you tomorrow! = )
-Max Chellemi

Here is the Question of the day!

True or false...sorry Rachel ; ) hehe

The function f(x) = cos(x) - sin(x) is increasing in


Lets analyze the question. We need to be able to visualize the graphs of sin(x) and cos(x). We also need to know that we are looking to see that the value of the output {f(x)} is increasing in value between the given parameters. Remember when the function is said to be increasing there may be no decreasing values in the output of the function between the given parameters.

Ok we know that the graph of sin(x) looks like

We also know that the graph of cos(x) looks like








Ok sorry for the layout issues but you don't know how hard this is...or how slow i am haha but i am trying. Ok so now that we have the graphs we can start tho think about this.

alright. We don't care about any part of the function that has an x less than zero or greater than π/2. We observe that the two functions cross at pi/4 and are the farthest apart as we approach π/2.

Graphically we can already see that the function doesn't increase the entire time but now we must examine it numerically to solidify our argument. Lets pick two easily identifiable points and check our hypothesis with a table.

x	f(x) = cos(x) - sin(x)
0	1
π/4	0

We need go no further ladies. As clearly indicated in the table above we increased the "x" value and the "y" value decreased, ergo the function is not solely increasing between

Today’s lesson talked about the Corollary to the Theorems that we received yesterday concerning continuity (the three things). The theorem reads: suppose “g” is continuous at “a” and “f” is continuous at g(a), then the function is continuous.

I am having some real trouble conjuring up the graph right now but i will try and get it up in a little while along with an explanation.

I hope this was helpful for you guys! Get well! I mean it!
Night
-Max Chellemi

Limits: Are They Your Friends?

One of the purposes of the blog is to provide a medium through which you might express your concerns, anxieties, and triumphs, free from the time pressure of our 55-minute slot. After ample time for reflection, you may find questions or understandings crystallizing in a manner that they often would not do during class.

We are now well under way in our opening foray into calculus. I am interested to hear how you have been coming to grips with limits and our four-fold analysis of them. Please respond to the following prompts but also feel free to add other thoughts that may be of help or interest to your classmates.

In regard to limits:
With what ideas, if any, have you been struggling? Describe any confusion and how you have been making sense of it.

Which representation most appeals to your understanding, numerical, graphical, or analytical? Explain.

Share any "Aha!" that you have had or connections that you have made with your previous mathematical understanding.

Ask a question, if you have one, about something for which you still need clarification.


Please comment before Tuesday's class.

I am taking AP calculus for a medley of reasons. First and foremost, I am taking this course to prepare myself for my major in college which i am almost certain will be in a field of engineering. I am not quite sure what kind of engineering, but never the less whether it be chemical, nuclear, aero, or civil, i will require the tool of calculus. I am talking the AP course not only because that is the only course that is offered at our school, but also because i am not from the most affluent family around and like most people i need help to pay for college. I intend on passing the exam, which will take care of an approximately $300 for $95 bucks. I also would like colleges to see that i aspire to learn as much as possible by not slacking off senior year.

I would like to learn as much as my resources (the text book) and/or my teacher (Mr. Alcantara) has to offer. Particularly to be better suited and if not to have a head start as i dive into college level courses.

I would like my classmates and friends to be as supportive I as I will be for them. I believe that we should all rise and carry each other through this possibly difficult time. I think encouragement and respect should be mutually expected as well as granted amongst ourselves.

I am content with the class structure. I really enjoy having digitally prepared notes along with the support of a computer and all of the facilitations that it brings to the table of learning. The visuals are something that definitively provide the largest advantage for our learning. I am looking forward to a great year!

Try Your Hand at This

Once you finish your assignment, go here

http://oos.moxiecode.com/examples/cubeoban/

to play a fun and deceptively challenging game. Level 1 is automatic. Level 2 is a quick hello. It's not until level 3 that you will appreciate the game. Remember, this is for AFTER you finish your assignment!

Happy bloggin'.

Response/Answers to questions in first blog.

I am taking AP calculus because I really like math. I am still very unsure about what I want to take in college but I think no matter what I take this class will benefit me in the future.

From this class I want to learn and I want to have a challenge but at the same time I don't want to feel like I have hit a brick wall as far as math goes and what I am able to understand. I really just want to have a class in which I feel proud of myself when I finish it because I know I worked hard to get there.

I want support from my classmates when I need it and hopefully I am able to return the favor. I want to work in the class so I am hoping we can all stay on task and I won't be distracted. I want to be able to go to my classmates for help when I need it and I want them to be able to approach me as well.

I want the teacher to help me out when I am asking questions but I don't want the work to be done for me. I am really bad at asking for help when I am too lazy to figure something out myself even though I know I can do it on my own (you will see that soon Mr. A haha). Hopefully we can work together to help me get a 5 on the A.P.

So far I think the class has been good. At first I felt a little intimidated but on the third day I felt a lot better about where we were going and what we were doing. I really like the class dynamic we have going on and I really like the way the class is organized. Hopefully the year will continue like this.

By: Emily

Welcome

You found it!!!

Welcome to our mutual venture into the land of math blogging. This is the place to come when you did not quite understand that last topic from class, or you were too shy or confused to ask your question, or you want to share an interesting and useful math website or new problem-solving strategy, or maybe just to chat about your math struggles and/or successes.

As with everything, you will get out what you put in.


So let's get started. Here is your 1st blog assignment:

Why are you taking AP calculus?
What do you want to get out of it?
What do you want from your classmates so that you might achieve your goals and have a good experience?
What do you want and expect from me as the teacher?
Describe your feelings about the class or content so far. (Feel free to be critical but always do so in a manner designed to improve rather than as simple whining)


So, click on New Post and start typing. Please do not respond as a comment; respond as a new post. (Too late for Rachel) When creating posts you have more publishing options such as colors, text size, bullets, spell check, and inserting images, that are not available in comment mode.

The first 2 people to post adequate responses will receive a 105% for this assignment. (It will be your job, if needed, to help others figure out how to get here and post.) The deadline for posting is Friday at 2:30.