Remember that you may comment on pretty much anything related to the course. What you like about section/lecture, what you don't like about section/lecture, a problem you would like done, a question about homework, and something you learned in section/lecture are some of the choices you have to comment on.
Perhaps not enough people have been reflecting on what they've learned, and overlook how useful a learning tool this is.
[20120301, Updated 20120307]
Homework Bonus:
Because I can be a tough grader, I will implement a bonus system.
0) Have completed the homework.
1) Check your work with a friend (they must be in section 1 or 2, i.e., either of my two sections).
2) Get the same answers. After checking your work, and still disagree on which of you is correct, you may agree to disagree. However, in most cases, one or both of you is wrong, so the mistake should be found and corrected.
3) Underneath your name, write HWBuddy: (Their name). You may have multiple buddies, write HWBuddies: (Name1), (Name2), ... .
I will award between -2 and 5 points. See my comments below. If I get a higher response, I will increase the possible bonuses even more, so encourage your classmates to participate in this bonus program! (For HW4 the max was 4)
Staple Penalty:
Lose 1 point for not stapling your paper.
Comment Bonus:
Tier 1: At least ten posts. Six of which belong to different weeks of the year. A week is Sunday to Saturday. Result: Approximately fifteen points.
Tier 2: At least twenty posts. Eight of which belong to different weeks of the year. A week is Sunday to Saturday. Result: Double Tier 2.
Tier 3: At least thirty posts. Ten of which belong to different weeks of the year. A week is Sunday to Saturday. Result: Triple Tier 3.
[20120307]
Regarding Comment Bonus:
The goal should be to post consistently. This is best displayed by posting once every week. Beyond that, the distribution of how many a week you want to post is up to you.
Regarding Homework Bonus:
So it's Thursday night and your friend hasn't even done their homework. Unfortunately, it doesn't really make sense to check your work with their incomplete work. The purpose of the bonus is to go through your completed homework, preferably on/before Wednesday, so that you have time to figure out and correct your mistakes. In fact, a great time is to check before or after the Wednesday lectures. Grab a trustworthy classmate or two and agree to have your homework completed by then. If he/she doesn't complete it as agreed, don't keep a grudge, but just move on and choose another classmate next time.
Because of the effort I put into seeing if you are checking your answers correctly, I will deduct 2 points for not adhering to the spirit of the Homework Bonus, and I will award 5 points for doing it correctly. There might be points in between.
Homework Bonus is recorded separately. So it doesn't directly affect your homework score. I will write "HB#" to the right of your homework score. It'll be added at the end of the year, after a homework score is dropped. This is good, because theoretically you can get a low score like 5 on the homework, and a homework bonus of 5 (though if you're checking your homework, you should probably be averaging 25). If I kept the scores together, and your lowest combined homework was a 10, then that homework would be dropped and you essentially lose the bonus. By keeping the score recorded separately, your bonus is preserved. Of course by the previous paragraph, you might have a negative total number Homework Bonus points, and in that case I'd treat it like a zero. Though honestly I only expect you to gain points.
Part of the spirit of the Homework Bonus should be that your buddy should theoretically get the same score you get. Ideally you should either both get a problem wrong or both get a problem right.
Another part of the Homework Bonus is to complete all the problems, covering for cases such as accidentally skipping a problem and doing the wrong problem.
And last but not least, the spirit of the Homework Bonus definitely says don't just copy, that's cheating.
Today in lecture I like how Professor Brown explained the process of quantifying the area of polar coordinates in comparison to Reimann summations for continuous functions. It made the integrad so easy to understand and I instantly understood the method!
ReplyDeleteCan we please go over how to graph polar equations in section tomorrow? Thanks!
ReplyDeleteAs polar equations have been moved to the next exam, I will review this topic in section after spring break. I will be gone next Tuesday (20120313) and the other Calc II TA will fill in for me. Maybe you can get her viewpoint, in addition to mine.
DeleteI would suggest you come to section prepared with a specific problem for the TA to answer. If she does not get to your question, feel free to inquire on this blog.
DeleteI was wondering if for the calculus with parametric curves section if you could go over how to know the limits of the integrals are for finding the areas and arc lengths?
ReplyDeleteMichelle Bohrson
I hope some of the problems I did helped a little with this question. In some sense it's not an easy skill to teach, because it depends so much on the specific problem. It helps to go back and forth between different pieces of information. Sometimes you use the x-y plot and sometimes instead you use the time variable. Sometimes you use the plot of x over t and sometimes you use the plot of y over t. Sometimes you need to combine various functions to solve for the final answer. Good luck.
DeleteGoing over the homework, I've realized that I don;t even know how to begin on problem 50 in section 7.4. If we could possibly go over it in section or perhaps a quick hint on how to start off would be helpful.
ReplyDeleteYou should be able to go through problems 39 through 52 and know which substitution you should make in order to obtain a rational function. For number 50, the substitution should be $u=e^x$. See page 492 for an example.
DeleteI was wondering if we could go over some parametric problems that might be on the test. I don't know which parametric to use when breaking the original into components.
ReplyDeleteWhat do you mean by the original?
DeleteProfessor Brown mentioned that we would just have basic parametric stuff on the exam and the rest of the material we've covered will be included on the final. What constitutes "basic parametric stuff"? Thanks!
ReplyDeleteYeah, that's a good question. Does he mean basic for him or basic for you? Joking aside, I flipped through sections 10.1 and 10.2 and thought the following skills are definitely "basic."
DeleteSketch the curve.
$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ if $\frac{dx}{dt}\neq 0$.
Finding a tangent line using the above formula.
Finding where the tangent is horizontal or vertical.
Areas (pg 647)
Arc Length (pg 648)
I would definitely read/review examples 1 to 6 from section 10.2. You might as well look at the formula for $\frac{d^2 y}{dx^2}$ while you're at it, the book uses it in Example 1.
Thank you so much Tim! This helped me narrow what I needed to study before the exam!
Delete-Erica
I am having problems with differential equations, because I get lost when trying to decide when they are or are not separable, and therefore when I should or should not use the P(x) method. If you could go over it quickly, I think it would be helpful for tomorrow.
ReplyDeleteThanks a lot,
Marcelo
If you can put the differential equation into the form $y^\prime+P(x)y=Q(x)$ then use an integrating factor. One reason you wouldn't be able to do this is if you had a non-linear function of $y$ somewhere in the equation. For example, if you saw $y^2$ or $\sin(y)$. Since you basically know two methods of solving differential equations, if it has a non-linear term, then it's probably a separable equation.
DeleteThanks Tim.
DeleteOn one of my Hw papers you wrote 5b4 or sb4 at the top? What does this mean? The only thing I can think of is "staple before" but I always staple my papers..
ReplyDeleteHaha. I just realized I should have wrote "HB" instead of "SB," for Homework Bonus. For this assignment, those who I felt adhered to the purpose of the Homework Bonus system got 4 points. HB4. Those who sort of did it how I wanted got 2 points. HB2. And those who tried, but didn't really do what I wanted got 1 point. HB1. I am recording the homework bonus separate from the main homework score. That way you get the points, but I can still monitor your unaltered homework performance. (With the exception of -1 for homework which isn't stapled. That's been a policy from day 1).
DeleteDo you have a concret method to match parametric equations with the corresponding graph when there is no values, like in ex 28p642 (section 10.1) ?
ReplyDeleteThanks a lot,
Vincent.
There is no concrete method. But the general idea is to match properties of x and y as functions of t with the x-y plot. For example, 28a) $x=t^4-t+1$ and $y=t^2$. Thus $y\geq 0$. Two plots satisfy this description, I and V. So let's take a closer look at $x(t)$. What is the minimum of $x(t)$? Do basic calculus to arrive at the conclusion that $x(t)>0$. Thus it must be V. Similarly, we can determine 28b)is I. One could have also argued that 28b) is $(0,0)$ at $t=0$ and so must be I, not V.
DeleteThe x and y for 28f) decay to 0 as $t\to \infty$. The only graph with this property is III.
Aside from looking at the behavior and/or graphs of x and y as functions of t, then looking at the derivative $(x^\prime(t),y^\prime(t))$ can also help. 28b) and 28c) both contain the origin, but the derivative for 28b) is $(2t-2,\frac{1}{2\sqrt{t}})$ and at $t=0$ this is $(-2,\infty)$. This must be I. Or for 28c) $(2\cos 2t,\cos(t+\sin 2t)(1+2\cos 2t))$ at $t=0$ is $(2,3)$.
Though the period of $x$ in 28d) would suggest the plot for VI.
[Originally Posted Mar 6, 2012 9:06 PM]
Ok, thank you very much.
DeleteDoes anyone need a HW buddy? My email is mkelly48@jhu.edu if you're interested!
ReplyDeleteTim,
ReplyDeleteWhen/where can we come to pick up our exams?
See your e-mail. For those of you who don't know where my office is, it's listed on this blog.
DeleteHey,when we're missing with polar equations, is the substitution r=sqrrt(x^2+y^2) ok? because whenever I do that then check the graph compared to the polar form, it looks different. Also, any hints on 10.3.50?
ReplyDeleteYou have to be careful, because the relationship is $r^2=x^2+y^2$ and so $r$ can be $\sqrt{x^2+y^2}$ or $=\sqrt{x^2+y^2}$. Given the choice of the sign of $r$, we have to choose $\theta$ so that $(r,\theta)$ represents the correct $(x,y)$.
DeleteFor 10.3.50, for what $\theta$ does $r$ go to $\infty$? $-\infty$? It's also useful to know that $y=r\sin\theta$.
When going from polar to Cartesian, it's easiest to use $x=r\cos\theta$ and $y=r\sin\theta$. Then use the equation $r=r(\theta)$ to replace $r$ and obtain the parametric curve $x(\theta),y(\theta)$.
Ok thanks. I figured it out, I think. I'm not sure I proved it, but I got it to a point where it was obvious which I think counts as "showing" it
DeleteCan you go over how to find the bounds of integration for polar functions when finding area and arclength? Also, under what circumstances do I need to instead find where two functions intersect?
ReplyDeleteI would also appreciate some review on this topic!
DeleteI would suggest you come to section prepared with a specific problem for the TA to answer. If she does not get to your question, feel free to inquire on this blog.
DeleteWhile doing the homework for section 10.3 I'm not really understanding problem like 16 and 20 where you have to change r=(constant)(trig function) into a Cartesian equation. The book doesn't explain it very well and doesn't give very helpful examples. Any hints?
ReplyDeleteIt's mostly heuristic, but here are two steps you can consider.
DeleteStep 1:
Solve for either $r$ or $\theta$ and plug the information into the right-hand side of
$x=r\cos\theta$ and $y=r\sin \theta$.
(In 16 and 20, $r$ is conveniently already a function of $\theta$) (For $r^2$ see the second half of my reply)
Step 2:
You now have $x$ and $y$ as either functions of just $r$ or just $\theta$. (In 16 and 20, you have them as functions of $\theta$)
Eliminate that parameter. (In 16 and 20, you should be able to use the simpler of either $x$ or $y$ and solve for $\theta$ in terms of $x$ or $y$. Possibly using triangles in the process.)
Example:
$r=\tan\theta$
Then
$x=r\cos\theta=\tan\theta\cos\theta=\sin\theta$
and
$y=r\sin\theta=\tan\theta\sin\theta$
In this case, $x=x(\theta)$ is more informative. I see that I'll need $\tan\theta$ and $\sin\theta$. So I draw my triangle and get that $\tan\theta=\pm\frac{x}{\sqrt{1-x^2}}$. I conclude that $y=\pm\frac{x^2}{\sqrt{1-x^2}}$.
My Original Answer which Didn't Quite Address the Question, but Useful to Keep:
In problems 16, 17, and 20, you are given $r$ as a function of $\theta$. On the other hand, $x$ and $y$ are functions of $r$ and $\theta$. Putting the information together, you can get $x$ and $y$ as functions of $\theta$.
Example:
$r=\sin\theta \tan\theta+24$
Then
$x=r\cos\theta=(\sin\theta\tan\theta+24)\cos\theta$
and
$y=r\sin\theta=(\sin\theta\tan\theta+24)\sin\theta$
In problems 15 and 19, we are given $r^2$. I'd say a reasonable way to deal with those problems is to square the relationship for $x$ and $y$. That is
$x^2 = r^2 \cos^2 \theta$
and
$y^2=r^2 \sin^2 \theta$
Example:
If
$r^2=4\sin\theta$
then
$x^2=(4\sin\theta) \cos^2 \theta$
and
$y^2=(4\sin\theta) \sin^2\theta$
You might ask, for $r^2$, "Why don't I just solve for $r$ before plugging it into $x$ and $y$?" No problem, just remember that it should be $r=\pm \sqrt{5}$. Personally I think squaring the $x$ and the $y$ to deal with $r^2$ is more aesthetically pleasing.
When I wrote $r=\pm\sqrt{5}$, I was referring to problem number 15.
DeleteNote that when using the triangle to determine expressions such as $\tan(\sin^{-1}(x))$, we are really only determining the absolute value of the expression. The true sign of the expression depends on the angle or original information.
Hi,
ReplyDeletecan tomorrow during section can we discuss teh area and length of polar coordinates. I was looking at some problems, and I have difficulties on when I should change my limits.
THANK YOU
nahyr
I would suggest you come to section prepared with a specific problem for the TA to answer. If she does not get to your question, feel free to inquire on this blog.
DeleteThe limits change when the variable of integration changes.
DeleteExample:
$\int _{x=1}^{x=4} x^2 dx$
Let $x=y^2$, $y>0$.
Then
$\int_ {y=1}^{y=2} y^4 2y dy$
Technically I can write:
$\int _{x=1}^{x=4} 2y^5 dy$
But after finding the antiderivative, which is $y^6/3$, I won't be able to evaluate unless I change the limits or solve back for the original limits.
Let me know if you're still confused. (As usual, references to "you" refer to the general reader.)
Not sure if this is the right place to put this, but here it goes.
ReplyDeleteIn class today we talked about limits of integrals as they approach infinity. I want to see if I'm understanding this right, so correct me if I'm wrong!
The purpose of finding a limit as a approaches infinity of an integral is to tell if the area converges. If the area converges, this means it approaches a finite amount of area, similar to when we take the limit of a function as a approaches infinity to see if it approaches a finite number.
True or not?
In Calculus I, we gave a certain meaning to integrating over a finite interval. Now we use this old definition to give a meaning to integrating over an infinite interval. The concept of integrating over an infinite interval is defined as the limit of a finite interval as the endpoint goes to infinity. Of course, this makes sense if the limit converges. This is why we are interested in finding the limit as $a$ approaches infinity.
DeleteI think I'd like to go over ways we find the bounds of integration when integrating polar and parametric equations. I think I have some trouble mostly with polar equations. Furthermore, should we always rely on the periodic nature of trigonometric functions when calculating bounds?
ReplyDeleteDr. Brown addresses this popular question in a recent (20120312 2:31pm) post on the Facebook page.
DeleteIn accordance to his post, I'd say that the periodic nature can both help and hinder. Proceed with caution, and use what's given to you to its fullest.
DeleteI was wondering if we could hold a practice test session for the next test that is timed. I know it would help me along with others to get into the habit of solving problems under a timed environmemt.
ReplyDeleteThank you!
If you can get at least 30 people from my section (that's half my students) to agree to come to such a practice test session, I will write problems and monitor such a session. Have them respond here: http://timothytran.blogspot.com/2012/03/practice-test-session.html
DeleteOkay I have a comment that is kind of from an old topic. When we were doing the linear differential equations, it never really occured to me how we are essentially un-doing the Product Rule. I went to Prof. Brown's office and he showed me. That just really cleared up alot for me because I didn't understand what we were really doing when we multiplied both sides by I(x) and then integrated both sides.
ReplyDeleteI'm glad it clicked for you! I'm also glad to hear you went to his office to go over material.
DeleteThus if you know you absolutely computed the integrating factor correctly, then you know that
$I(x)y^\prime +I(x)P(x)y=I(x)Q(x)$ becomes $[I(x)y]^\prime=I(x)Q(x)$. However, because mistakes happen, it's often useful to check that you correctly computed I(x), by differentiating $I(x)y$ with respect to $x$ and recover $I(x)y^\prime +I(x)P(x)$.
Hello, I was having trouble sort of understanding the more general point of polar coordinates as in why we would want to use polar coordinates instead of cartesian coordinates.
ReplyDeleteThanks
Michelle Bohrson
1) There are situations in which using polar coordinates makes calculations easier. If you go on to calc III, you will definitely learn how.
Delete2) Polar coordinates are a particular change of coordinate system. Learning to change between Cartesian and polar coordinates helps gap the idea of a more abstract coordinate system.
3) Related to [2] is that polar coordinates are relatively simple as far as coordinate systems go.
4) At the beginning of the semester, Professor Brown proposed finding the value of $$\int _0 ^\infty e^-x^2 dx$$. One method of doing so makes use of polar coordinates.
5) Polar coordinates can be used to more simply describe curves or regions that would otherwise be more cumbersome in Cartesian coordinates. The circle of radius $A$ is ${(x,\sqrt{A^2 -x^2} )\mid x\in(-A,A)}\cup {(x,-\sqrt{A^2 -x^2} )\mid x\in(-A,A)}$ in Cartesian coordinates and ${(A,\theta)\mid \theta\in [0,2\pi]}$ in polar coordinates.
6) Related to [5], we can derive the area of a circle by $4 \int _0^A \sqrt {A^2 -x^2 } dx$ or by $\int_0^{2 \pi} \int _0^A r dr d\theta$. I find the first slightly harder to set up than the second, and the first slightly harder to compute than the second. What do you think?
Hey,
ReplyDeleteHow do we solve the following polar equation:
sin(3ø)=cos(3ø)
I know that there are more than the 2 obvious solutions ø=pi/12 and ø=5/12
But I don t see how I can get them.
Thanks!
We know sine and cosine are equal at $\pi/4$ and $5\pi/4$. Why? Of course we can also add any multiple of $2\pi$ to these. Thus $3\theta=\pi/4+\pi k$ for any integer $k$. Then all possible solutions are $\theta=\pi/12+\pi *k/3$.
DeleteI found Kalina's quick but thorough overview of polar coordinate systems very helpful today. I like that as she went through the sample problems, the material added layers to the previous problem. I especially found her transformation from Cartesian to polar coordinates helpful! I definitely needed a broad review of the current material!
ReplyDeleteTim, I am not sure where post once again, I didn't find the section that is due next week, but I just wanted to keep my postings up to date. So here it goes:
ReplyDeleteJust as a summary of today's lecture Professor Brown expanded on the ideas of improper integrals. He explained 3 cases, but focused on the issues (undefined or discontinuities) of the integral were in the middle, and not at the endpoints. He also explained a theorem similar to squeeze theorem, but applied to improper integrals.
If f(x) is greater than g(x) on the interval a to infinity and both functions converge, then the integral of f(x) from a to infinity is greater than the integral g(x) a to infinity. There are variations, which we also should look at.
Hope this is correct, bc that is what I got from the lecture
Marcelo
The next homework set is not posted on the website...is there any way it could get posted so I could get started on it? Also, am I posting in the wrong section (I am in the Tuesday section)?
ReplyDeleteI have both my sections post comments in the same area. I had mentioned getting homework posted during Spring Break, but I'm not sure if that happened. Hope you had a good Spring Break.
DeleteWhat is the purpose of Theorem 7 in section 11.1? because technically we've already determined the limited in the first part of the theorem.
ReplyDeleteThe assumption is we know the limit of $a_n$. Now we want to determine the limit of $f(a_n)$.
DeleteFor example, we might ask, what is the limit of $e^{1/n}$ as $n$ goes to $\infty$. We know that $\lim_{n\to \infty} \frac{1}{n} =0$ and that $e^x$ is continuous at $x=0$. Thus the theorem tells us that $\lim _{n\to \infty} e^{1/n}= e^0=1$.
Sometimes, books will write the result as $$\lim _{n\to \infty} f(a_n)=f\left(\lim _{n\to \infty} a_n \right)$$
IMPORTANT:
DeleteOf course, you still need to satisfy the conditions of the theorem.
Mathematical Induction
ReplyDeleteIn the book it goes from because
1/2(a_k+1 + 6) > 1/2(a_k + 6)
then
a_k+2 > a_k +1
I dont understand where the k+1 and the k+2 come from in the second part.
I believe you are referring to example 14 on page 699. The sequence is defined by a recurrence relation. That is, newer terms are defined by existing terms in the sequence.
DeleteAt the top, the author writes, $a_1=2$ and $a_{n+1}=\frac{1}{2}(a_n+6)$ for $n=1,2,3,\dots$.