Turn in #3 is due Wednesday, Sept. 24, 2014!
This is for discussing assignments from WEEK 3 (this week will count as 1.5 weeks for your discussion board bonus), including homework,
turn-in, and in-class work or lessons, or anything else related to the class from this week. Please be sure to include your
name at the end of your post for credit. When answering a question, DO
NOT GIVE SOLUTIONS! Provide hints or explain a method that you used, but
do not give the final result. As always, RESPECT IS A MUST! Anyone
abusing this forum will be banned from future use (meaning, no extra
credit!!!).
How do you find the 11th derivative of a polynomial without finding the 3rd-10th derivative?
ReplyDeleteThanks,
Rachel Hersch
If the highest exponent of the polynomial is less than 11, the derivative there equals 0. If the highest exponent is equal to or greater than 11, there is no shortcut you have to take the derivative 11 times.
Delete-Eva A-L
Eva is right on. Think about a 4th degree polynomial. The power rule ensures that each derivative will be one degree less than its predecessor, so after 4 derivatives we will be at a constant function (degree 0). Then, the next derivative will be equal to zero. Hence, if you have an n-th degree polynomial, the (n + 1) derivative will be zero.
DeleteDid Wilson mention there's an answer sheet to Derivatives Worksheet we got in class today online? If there is, where can I find it? Who knows though, I could be making this up.
ReplyDeleteThanks,
Julia Trombley
Look in the folder "solutions" inside our course folder. It is up now.
DeleteFor #9 in the homework, how do i simplify (-csc^2x)/(1+cotx )^2?
ReplyDeleteMy hunch is to look for a Pythagorean identity relating cscx and cotx. It may not simplify....
DeleteThis comment has been removed by the author.
ReplyDeleteIn class, there were times when you simplified the answer you got via the chain rule and times when you just left the answer untouched. When do you usually simply your answer?
ReplyDelete-Rafey Rehman
If it involves expanding an expression to a power, don't waste the time. If it involves a simple distribution and collection of like terms, do it. As always, leave denominators from quotient rule unexpanded (is that a word?). We also ran short on time 3rd hour so I told you on the last two examples what should be further simplified.
DeleteAre negative exponents treated any differently than positive exponents when applying the chain rule?
ReplyDeleteThanks!
Allison Honet
I don't think so, just treat them the same as you would a positive exponent when applying the chain rule.
ReplyDelete-Daniel Honet
Do you prefer us to write our answers using negative exponents or fractions?
ReplyDeleteThe AP has no preference, nor do I. The key is that it be correct :-)
DeleteWill we need to have the derivatives of trig functions memorized for the quiz on Monday?
ReplyDeleteHe said in class that we will. Everything we have learned this week (including the chain rule) will be on the test.
Delete~Katie Weitzel
You should know the derivative rules we have covered thus far by Monday, yes.
DeleteFor the homework tonight, we have to do #21-37 odds only, but on the Turn-In, we need to do problem 28. Is this a typo or do we need to do both 28 and 29?
ReplyDelete~Katie Weitzel
Please include the problems that I listed on the turn in. #28 (not hashtag 28) will be assigned tomorrow.
DeleteFor 1d of the turn in i set f(x) equal to f'(x) and when i simplified it i got x=n...so to get f(0) to equal 0, i did 0^0/k and it seems like there is a controversy as to what 0^0 is, so what is it really? 0, 1, or undefined? If it isn't equal to 0 then what am i doing wrong in approaching this problem???
ReplyDelete-Eva A-L
Rather than setting the two functions equal, I'd get rid of the variable x entirely by substituting 4 in for x, since we know f and f' are equal at x=4. Once you have f(4)=f'(4), you can solve for n. You can then use that value of n to solve for k using either f or f', since we know that f(4) and f'(4) are both equal to 1.
DeleteFor number 2 on the turn-in I was wondering if the point at 0 was an open or closed circle. This would effect the answer to 2(b)
ReplyDeleteThanks,
Rachel Hersch
It did not copy too well, but yes, v(0)=0.
DeleteFor number 2 on the turn-in, should we assume that the line stops at 6 or continues to infinity?
ReplyDeleteThanks,
Allison Honet
I think it's better to assume that the line will continue to infinity because there is no reason that the particle will stop suddenly after 6 seconds.
DeleteLet's assume that we are looking at the half -open interval for t, [0,6).
DeleteOn the Turn-In, for number 1, it is is hard to tell exactly at which intervals of time the particle/ acceleration is doing what, so should we approximate?
ReplyDeleteI assume you mean #2. We will cover this in class tomorrow, so I won't say much now. I will say that you can assume that v(t) has horizontal tangents at t=1 and t=4.
DeleteI don't think the graph is necessary to solve the problem. I did it algebraically.
Delete-Rachel Hersch
yes, i did mean 2, and thank you
DeleteFor the quiz tomorrow is it required to simplify negative exponents?
ReplyDeleteThanks,
Rachel Hersch
If you look at Mr. Wilson's answer keys for the worksheets, he has the final answers in both ways so I would assume that you don't have to simplify them on the quiz.
DeleteAngela Satullo
We do not need to simplify negative, or even rational exponents. This is also acceptable on the AP exam.
DeleteFor number 57 on the Turn-In, there is no d and e, it only goes through d
ReplyDeleteMake sure you are checking page 148. It is the first problem on the page. The problem goes through g, but we are only doing d and e.
Delete-Allison Honet
oh thanks, apparently I don't know the alphabet ;-)
DeleteI know this is a little late but anyone still working on chain rule, this website helped me a lot! https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainrulesoldirectory/ChainRuleSol.html#SOLUTION 5
ReplyDelete-Kai Selwa
This is great! I will add it to our references document as well. Thanks, Kai!
DeleteIf there is a trig identity, should we simplify? i.e. 2sinxcosx = sin2x
ReplyDelete--Paul Tremonti
It is usually helpful, but not always necessary. In most cases it is not necessary, but will help make some steps a bit easier, such as finding a critical point.
DeleteFor number 20 on the turn in for the textbook section can we use our graphing calculator or do we have to use the quadratic formula?
ReplyDelete-Laura G.
Calculator is fine.
DeleteIs instantaneous velocity just the derivative of average velocity?
ReplyDeleteThanks,
Rachel Hersch
The derivative of velocity is acceleration. To find instantaneous velocity, solve for the velocity at a specific x value.
Delete-Allison Honet
Keep in mind what average rate of change (velocity) is. We must examine an interval and use the position function to calculate an average rate of change.
DeleteFor the turn in #2a would the answer have to be on an open interval?
ReplyDelete-Laura G.
Yes, because you don't include points where the particle is at rest.
Delete-Lexi Kizy
So the same idea applies to 2c for acceleration?
Delete-Laura G.
Yes, you also don't include points where the acceleration is zero
Delete-Lexi Kizy
For 2d on the Turn-In, you take the absolute value of the velocity graph to get the speed graph that you use to determine whether the particle is increasing or decreasing, right? Just wanted to clarify.
ReplyDeleteThanks!
Julia Berthel
Yep! Since speed cannot be negative, you take the absolute value. The graph of speed will be all above the x-axis so if you sketch the graph of speed, it might be easier to answer the question.
Delete~Katie Weitzel
I think you can do that and then look at if velocity is approaching 0(slowing down) or velocity is increasing away from the x-axis(speeding up).That's what I did.
ReplyDelete-Laura G.
For number 20 part a, I'm a little confused on how we are supposed to describe the motion of the particle.
ReplyDeleteThanks!
Allison Honet
I would look to use terms such as " moved left", "moved right", "increased speed", etc.
DeleteIf you find the derivative of the graph given in the textbook it will be easier to recognize the motion.
ReplyDelete-Rachel Hersch
In #2 of the turn in, how can you tell if you should use closed on open time intervals?
ReplyDeleteScroll up and there is a good discussion on this. They key to ask yourself is "what is the particle doing at the endpoint of the interval?"
DeleteDo you find the position of the particle in 20f by plugging 5 into the x(t) equation?
ReplyDeleteYes, plug 5 in for x(t) then solve for t.
Delete-Allison Honet
Yep! Then subtract the 5 to the other side and solve for the x-intercepts. There should be three answers I think...
ReplyDelete-Laura G
Or you could use your graphing calculator to find the zeroes too...you don't actually have to solve it.
ReplyDelete-Laura G
Great participation this week! Any further comments on this board will not be counted for your bonus. Please begin using Week 4.5-5 board.
ReplyDelete