Welcome 2013-14 BC Calculus Students! You should have been given a summer work packet to prepare you for our journey into the study of calculus this fall. If you have not received your packet, or want another copy, download it from the course folder on the website.
Use this forum to post questions and discuss solutions regarding the summer packet. Look to offer help to those who ask and post your questions when you have them. There is only one rule on this forum: ALWAYS RESPECT EACH OTHER!
Some reminders: 1. the packet is due the first day of class (9/3/13) for a significant grade, 2. your original work should be on separate paper and done without a calculator except when specified, 3. solutions (without work) will be posted to the course folder weekly over the summer, 4. the list of helpful links is also posted in the course folder.
Good luck! See you on September 3, 2013! ~Coach Wilson
Hi Everyone!
ReplyDeleteSome of you have already began working on the summer work, which is great! I have been working hard as well, getting ready for a terrific year ahead. As a reminder, please post your questions here to the blog instead of emailing me directly, that way we can all benefit from the question and other students can provide advice as well.
Solutions are now posted for Week 1 in the course folder.
***IMPORTANT***
I have altered Week 2's problems #6, 8, 13, 14, and 19. You will need to look at the updated handout in the course folder. I have also changed Week 5 problem #11 to be in QIV.
You can follow us on Twitter: @GrovesBCcalc
Hello everyone,
ReplyDeleteI hope you enjoyed a safe holiday week with family and friends. Week 2 solutions are now posted in the course folder. Please see my last post regarding changes to some of the problems. As always, please post your questions to this forum...still waiting for our first :)
Regards, Coach Wilson
Hello Everyone,
ReplyDeleteI hope you're finding ways to stay refreshed in this extreme heat. I also hope you're keeping up to date on the summer work. If you wait until the last week of August, you'll find it an insurmountable task. Please keep up with the suggested pace and post questions here.
Week 3 solutions are now up in the course folder via the BPS website.
I spent last week at an AP conference garnering some great information and ideas for the course this year. Just an FYI, we have 147 class meetings before the AP exam next May (excluding snow days), thus we will NOT spend in-class time reviewing any of the algebra or trig content contained in your review packets...that is why you have the packets!
Stay cool, Coach Wilson
Hello everyone,
ReplyDeleteWeek 4 solutions are now up in our course folder. This week you will be reviewing exponential and logarithmic equations and functions. YOU WILL USE THESE EXTENSIVELY IN CALCULUS, so be sure you understand them...particularly e^x and its inverse function ln x. You will also review sequences and series this week. Series are a very big topic in BC calculus. We will spend an entire chapter (chapter 9) on this in February-March. Each year at least one free response question on the AP exam is dedicated to the series topics we will study, so this review of how a series works will be helpful. I would keep this week's review handy for when we get to February!
Comment with your questions as they arise.
Regards, Coach Wilson
Greetings all,
ReplyDeleteWeek 5 solutions are now up. This week you will be doing a quick review of some geometric areas and volumes as these formulas are helpful in calculus. You will also be doing a review of trigonometric functions. You should have the major angles of the unit circle committed to memory (it is expected on the AP exam!) and their respective sine and cosine values.
Let me remind you that this summer packet is not optional and will be collected on the first day of class for a grade. This packet was put together at the request of prior BC students and other teachers, and it is designed to prepare you for day 1! We will NOT spend any class time reviewing topics in this packet.
I hope everyone, including varsity football players, are up to date on their summer work...don't fall behind.
Please post questions here if you have them...still waiting for our first discussion to ensue.
Regards, Coach Wilson
Hello everyone,
ReplyDeleteSolutions are now posted for weeks 6 and 7. Many of you will begin to have extra-curricular commitments soon, so the start of a great school year is just around the corner!
Post questions here, as always. Good luck!
Coach Wilson
I finished a few weeks ago, and now I've been going through and checking my answers and doing the problems I had skipped. There are some answers that no matter how many times I check are not the same as the ones on the key. I've checked these problems with my graphing calculator, with other students, and also with my older brother.
ReplyDeleteHere are my questions, organized by week.Sorry that there are so many! My questions kept accumulating but I never posted them... now that I'm going back through it, I've realized how many questions there are, so sorry I'm posting them all at once.
Week 1
#3: I have the same answer as the key except that in my final answer the part in the parenthesis is being divided by 3, instead of multiplied by 3.
For the “Function Domain” section, should we find the domain and range by finding the domain and range of the inverse function? Also, some of my answers are sort of intuitive and I don't know how to show my work.
#28 – 30: are we supposed to estimate or is there a method to this? Like, do we have to find the equation for each line and work with those, or just work visually with the graph?
Week 2
#1-4, I graphed them so that the x value is from [-10, 10] but I changed the y interval so I wouldn't have to draw really tiny. Is that okay? For #1 I had it go from [-100, 100] in intervals of 10. For #2 I did [-6, 15] in intervals of 3. For #3 I had [0, 2] in intervals of 1/5.
#8: I have (x+4)(x-4)(x^2+5) and you have (x+4)(x-4)(x^2-5). This is taking into consideration the change made in the original problem.
#11: I have (x-3)^2(3x-2)^2(4x-5). I checked it with a graphing calculator and I think it's right, but the answer key has (x-3)^2(3x-2)^2(2x+1)
There is no #14 on the answer key.
#19: I can't get your answer, even when taking the change in the original packet into consideration.
Week 3
#6: In the answer key, there is supposed to be an equal sign after “r”, right?
#13: I have 4^x, isn't that the same as 2^(2x)?
#15: I have (x-2)/(5x+1), but the key has (x+2)/(5x-1).
#21: I think this is just a typo, I have 1/3 and the key has 31/3.
Week 4:
#2: When I checked my work, I found that only the 2 of the answers worked, the (-3/2+...) and the (-5/2-...). I didn't get that all four answers worked.
#24: I think there's a typo in the packet?
Week 5
#33: For the 2nd equation, I got the equation 1/6 +1/3n instead of the two equations on the answer key, but I think all three work...
Week 6
#7: In a hyperbola, isn't the operation supposed to be subtraction? I got [(x+1)^2]/9-[(y+1)^2]/25=1. Also, for the covertices I got (-1, 4) and (-1, -6) because since the y values are being subtracted, then the covertices would be on the conjugate axis, and the square root of 25 is 5, so I got the covertices were (-1, -1 + or – 5). Also I got my foci as (-1 + or – square root of 34, -1) not the square root of 33... because:
a^2+b^2=c^2
9 + 25 = c^2
34 = c^2
c = + or – square root of 34
And that also changes the eccentricity.
#16: I got 0=-x^2+4 – 4y, because
r = 2/(1 + sin@) ← sorry I am using @ as theta
r+rsin@=2
r+y=2
r=2-y
r^2=4-4y+y^2 ← I'm allowed to square each side like that, right?
x^2+y^2=4-4y+y^2
0= -x^2 – y^2 + y^2 +4 – 4y
0=-x^2 + 4 – 4y
Week 7
#3: I got -1/(x(x+h))
#5: I got 23, not 21
#6: I got 6, not 5
Did anyone else get those? Or am I just bad at arithmetic...?
For the “shapes of parent graphs,” I knew most of the graphs just by looking at them, and a for few of them I just made a table on some scrap paper and drew little sketch graphs... how should I show my work?
Once again, sorry about posting all of these at once!
I hope everyone's had an awesome summer!!!
oh by the way this is Erica G... just ignore the "Heather" part, when I made my email address I was afraid of internet predators and my name is Latin for heather.
DeleteHello everyone,
ReplyDeleteI have been away at Groves football camp the past few days and that can only mean one thing - school is coming! I hope you are all excited about a new year, and for many of you your final year at Groves! :)
I have changed the settings to allow for anonymous comments, that way you do not need to sign in with an email to post a question. Sorry about that, "Heather" (Erica).
Regarding the solutions, some of the problems came from other sources and the solutions might not be correct for those. I will check all of the questions you posed, Erica, and get back to you/everyone as soon as possible (probably Friday).
When it comes to showing your work, some problems may not have any work to show at all, such as the parent graph matching in week 7. It is fine on that part to simply print off the graph page and write your answers next to each graph.
Hope this helps for now, Coach Wilson
I had some time to look over your questions. Thank you for posting them, as many of them turned out to be typos in the solutions. I have updated the documents in the course folder to reflect the below:
ReplyDeleteWeek 1: #3, yes it should be 1/3, not 3 out front. There is not work to show for the domain and range pieces. #28-30 you should visually estimate from the given graphs.
Week 2: Choose any window for your graphs as long as they show adequate detail. #8 should be x2 + 5, the answer key is a typo. #11 should be (x-3)^2 (3x-2)^2 (4x-5), also a typo on the answer key. The problem for #19 should read x^4-4x^2-12=0.
Week 3: yes, #6 is “r =”. #13, yes 4x = 22x because 4 = 22. #15, the answer key is correct. Interchange x and y, then solve for y. #21, yes the answer key has a typo, it is 1/3 not 31/3.
Week 4: For #2, you are correct, only 2 of the solutions work. #24, the 4th term should be -4, not -2.
Week 5: #32, yes your answer is equivalent to mine. Also note that I have renumbered the last 3 problems to omit the duplicate #30.
Week 6: #7, you are correct. My typo of “+” threw off the entire problem. I have updated the answer key. #16, the answer key should read “4y” instead of “2y”, making your solution equivalent.
Week 7: #3 is “-1”, a typo. #5 is 23. #6 is 6.
Hope this helps! Coach Wilson
Thanks Coach Wilson!
DeleteHi Mr. Wilson,
ReplyDeletehow do you find the answer to week 6 # 10? I had the equations right for both the circle and the parabola. I substituted in and got an equation with only y terms, and no x terms, but it went up to y^4 and it doesn't factor. Can I use my calculator to solve for y and then substitute back in to get the x values? or do I have to show work to solve for y, because I don't know how to do that.
Also, I can't get the same answer as you for week 6 # 9. When I did (a-c)/b I got cot(2θ)=-1/4 and then when I solved for θ I got .5cot^-1(-1/4) not (-3/4)
ReplyDeleteWeek 6 #10 you should use your graphing calculator's "intersect" function to find the coordinates of intersection. For #9 the A coeffecient is -2 and the C coeffecient is -1, so A-C = -1. Your answer is correct, the key has a typo there. I have updated the answer document to reflect this. Thanks! Coach Wilson
ReplyDeleteHi Coach Wilson,
ReplyDeleteI have a few questions. I've listed them by week.
Week #1-#7- I keep getting x^(3/2)+ x^3 - x^(5/2) through distribution. How did you get the exponent of (7/2) and the absolute value of x?
Week 5-#1- I got (4-pi)/4. Can that be reduced to pi/4? If so, how?
#29-I used the half-angle trig identity, but I'm not sure where to go from there. Do I need to keep the plus/minus in there?
Week 7-#17- I've done this one several times, and I just get 3. not 3x. Could be a typo, or a mistake on my part.
Thanks for your input!
Hi!
ReplyDeleteFor week 1 #7, this definitely appears to be a typo. As I stated previously, many of these problems were taken from other sources, this one included. The answer should be: |x^3| - x^5/2 + x^3/2, the absolute value is necessary due to the square root of x^6 (x^(6/2)).
Week 5 #1, you are correct, it should be 1 - pi/4. The given answer is the ratio of the area of the circle to the square. #29 you are finding the zeros of the function h(x), so I would subtract the sin x to the LHS (left hand side), and use the half-angle identity for cosine. Then, you can square both sides, eliminating the +/- and the square root. You should be able to solve from there using factoring and a Pythagorean identity.
For week 7 #17, my answer key has 3 also, so you are correct. I will upload it again, perhaps it was an old version?
I have changed the answer keys to reflect the above as well. Thanks to everyone for posting your questions. As I expected, we are finding some typos in the answer key. Consider this year a "beta version" and we are the guinea pigs :) Please keep your questions coming as you find them!
One week left...Coach Wilson
Thanks so much! I'm still having trouble with 29 for week 5. Once I got rid of the root, I changed sin^2(x) into 1-cos^2(x) so all the terms would be cosines. However, after I multiplied each side by 2, the expression I got wasn't factor-able, and if I use the quadratic formula, I end up with imaginary solutions.
DeleteHi!
DeleteUpon multiplying both sides by 2 you should obtain:
2 - 2cos^2x = 1 + cosx
Setting equal to zero you get:
0 = 2cos^2 x + cos x - 1
Which factors to:
0 = (2 cos x - 1)(cos x + 1)
:)
Hi Mr. Wilson. Once I got (2cosx-1)(cosx+1) then I solved for cosx and got cosx=1/2 (so on the unit circle x=pi/3 and 5pi/3) and cos x=-1 (so x=3pi/4 and 7pi/4)
DeleteSo I understand where the equation x=(pi/3)+2(pi)n comes from but I don't understand where the other 2 equations on the answer key come from...
Sorry & Thanks!
Hi!
DeleteWell, you should start with a review of the unit circle :) Your solutions for cos x = -1 are incorrect.
Haha thanks I was thinking tanx=-1!!
DeleteOK wait now I understand the pi/3, and the pi, but I don't understand to get the 2pi/3. I keep getting 5pi/3. 2pi/3 has a cos value of -1/2, not +1/2.
DeleteNo problem, this is why we review. And yes, it should be 2/3 and 5/3 pi for the solutions of cos x = 1/2. Answer key has been updated. Thanks!
DeleteHi Mr. Wilson, for Week 7 # 20 I keep getting 12x^2-14x+2. Could you check that please?
ReplyDeleteYikes! I have no idea how they got their answer...not even by simply multiplying the binomials would one obtain that answer. Yes, your answer is correct. I have changed the answer key to reflect this. Thanks!
ReplyDeleteCoach Wilson
For week 3 #15 the answer key has the same exact thing as the problem.
ReplyDeleteHi!
DeletePlease see the above discussion regarding this problem. The key is correct (and not "exact" same, since it is the inverse).
How do you find the range for Week 1 number 8? If I do the inverse the domain of the inverse is not the same as the range of the original.
ReplyDeleteHi!
DeleteFirst, I hope you factored and reduced the function to simplest form noting any domain restrictions. Then, you should consider the shape of the graph (simple inverse shape with a single vertical asymptote at this point) to determine the domain and range. You should be able to do this from your prior knowledge of the shapes of basic graphs. If you're uncertain, use a graphing calculator to verify your ideas.
Hello All,
ReplyDeleteFor week 2 numbers 13 and 14, i'm not even sure I understand what the question is asking, but I am having a tremendous amount of difficulty doing what I think I'm supposed to do. Is there anyone who could post a step by step for either or both of these problems? I can't seem to factor them and I am probably completely misunderstanding the meaning of the question. Hopefully I am not alone in having no idea what these questions are asking because I've probably read it over 20+ times throughout the summer and still can't manage to pull meaning from it!
Sincerely,
Extremely Confused
I also have noticed that 14 is now gone from the course work, so please disregard that part of my question!
DeleteDear Extremely Confused! First, I don't know if you know this, but question 13 has changed from the original packet. The new version of #13 is online, but I'll post it here too.
Delete(-1/4)x+(5/4)x^(1/3)+(6/4)x^(-1/3)
Now, what you have to do is factor it to find the roots (or where the graph crosses the x axis. The first thing to do is find the GCF, or the Greatest Common Factor.
What I did to find it was to rewrite the equation like this so that all the x's have exponents. The first one used to have an exponent of 1, but now it is written as (3/3).
(-1/4)x^(3/3)+(5/4)x^(1/3)+(6/4)x^(-1/3)
Now you can see that all the coefficients of x are multiples of (1/4) and the exponents of x are all coefficients of (1/3)
So then I factored out the greatest common factor, which is (-1/4)x^(-1/3). Then I factored the rest of the equation so it ended up looking like this.
((-1/4)x^(-1/3)) * (x^(2/3)-6) * (x^(2/3)+1)
Sorry, that looks really confusing. The * means multiply.
Okay, now that this is split into 3 factors, you set each factor equal to 0 and solve for x.
((-1/4)x^(-1/3))=0
When you solve it, you'll find x=0. The answer key says this is a "discontinuity" but I'm not really sure what that means, sorry!
Next factor:
(x^(2/3)-6)=0
When you solve for x this time you get x=6root6 (6 times the square root of 6)
Next factor:
(x^(2/3)+1)=0
When you solve for x this time you get a nonreal answer! Which wouldn't be a root on a graph because you can't graph it, because it isn't real.
I hope this helped!
Hi Everyone,
DeleteI think it is awesome that you are answering each others' questions! Keep it up! I envision us using this throughout the year for the same purpose and I will set up a discussion for each chapter of the text soon.
To clarify, a discontinuity as you should currently know it from Algebra 2 and Pre-Calculus is a value that we restrict from the domain of a function because the function is undefined there (such as a vertical asymptote or a hole). You will learn next week that there are other types of discontinuities where the function is actually defined! But, for now, you should recognize that x = 0 is not a valid domain value because of the GCF term with x ^ (-1/3).
~Coach Wilson
Hi Everyone,
ReplyDeleteAs our summer winds down, I want to take this opportunity to commend you on your efforts and use of our discussion board. Also, I want to thank you for persevering through several issues with the summer work, from typos to wholesale changes of problems. It is not easy being the initial group, but your work is valuable to me as much as it is to you.
I will not be checking in over the weekend, as I am taking the opportunity to enjoy some time with family and friends before the start of the year. I encourage you to do the same. If you have questions, however, you can still post them and hopefully a classmate will help you out :)
See everyone on Tuesday in B-35!
Regards, Coach Wilson
P.S. Don't forget to follow our Twitter feed. An important tweet will come tomorrow!