Write stuff. Try to reply to each other's questions.
There's always time for learning material that has already been covered. Professors occasionally like to put a question missed by students on the first midterm back on the second. Or from midterm to final.
I make a due date more of an incentive and reminder. In general, I don't want you to try to only post when you need to make a due date. The goal is that you post without worrying about the artificial due date. Similarly, the goal is that you learn without worrying about your grade. Theoretically, when you meet your goal, the other follows.
Similar philosophy would be to consider, "You shouldn't study to improve your grade; you should improve your grade, because you study."
These are closely related to, "It's not the destination that's important, it's the journey."
[20120319]
Reiterate Some Goals with the Commenting System:
(1) I want students to comment consistently and abundantly.
(2) Students wait until its too late to ask questions.
(3) It's good to ask questions, but sometimes we need to ask the right questions. Something I'm still learning to do myself.
(4) Realize that it's okay to ask the TA and professor questions.
(5) Realize that it helps to reflect on what you've learned.
(6) Helping each other.
I'm confused about something from a lecture last week. Brown said that the integral of 1/X^P from 0 to 1 diverges if P is greater than or equal to 1 and it converges if P is less than 1 while the same integral but from 1 to infinity converges if P is greater than one and diverges if P is less than or equal to 1. I don't see why that is.
ReplyDelete(1) Just compute the improper integral.
Delete(2a) In a more intuitive sense, if $01$, then $X^P\frac{1}{X}$. While for $P<1$, then $X^P>X$ and $\frac{1}{X^P}<\frac{1}{X}$. This fact doesn't imply the convergence on the interval $(0,1)$, but rather it helps convince yourself why the result is true.
(2b) if $X>1$ and $P>1$, then $X^P>X$ and $\frac{1}{X^P}<\frac{1}{X}$. While for $P<1$, then $X^P>X$ and $\frac{1}{X^P}>\frac{1}{X}$. This fact doesn't imply the convergence on the interval $(1,\infty)$, but rather it helps convince yourself why the result is true.
(3a) The divergence of $\int _1 ^\infty \frac {1}{x} dx$ is similar to the divergence of $\sum _{n=1}^\infty \frac{1}{n}$. Just because the terms are going to zero, doesn't mean their sum goes to zero.
(3b) You get convergence of a series if somehow the terms go to zero "fast" enough. $\sum _{n=1}^\infty \frac{1}{2^n}$ for example converges. Similarly, the improper integral on $(1,\infty)$ converges if somehow the function goes to zero "fast" enough.
(3c) The functions $\frac{1}{X^P}$ go to $\infty$ as $X$ goes to $0$ from above. Let's consider breaking up the interval $(0,1)$ into several intervals, $I_n = [\frac{1}{n+1},\frac{1}{n}]$ for $n=1,2,3,\dots$. The width of $I_n$ is $\frac{1}{n(n+1)}$. Consider the average value on that interval $A_n$. Then the area over that interval is $B_n =A_n\cdot I_n $ and improper integral is $\sum _{n=1}^\infty B_n$. So if these terms, $B_n$, go to zero fast enough then the integral converges, but if not, then it diverges.
Thanks Tim! This thorough explanation was really helpful for clarification!
DeleteI'm completely lost on how to find the bounds when finding the area of polar coordinates. Although the hw for this is due tomorrow can we still go over it next week just to clarify?
ReplyDeleteNext week is spring break! Also, we mostly have to move on. We'll try to do more examples. But I recommend that you try to go to office hours (Dr. Brown's and/or mine) and help room to figure it all out.
DeleteMostly you just have to figure things out on the fly. So practice practice practice. Do as many problems as you can and if you run out, do the same ones again.
See Dr. Brown's latest post on the Facebook page (20120313).
Does anyone know how much work we are supposed to show when "sketching" the graph of a polar function? I've just been finding critical points and going from there.
ReplyDeleteIt depends on what you mean by critical points, seeing as they can change for each graph depending on what modifiers are applied to the equation internally and externally. In general though, once you have translated the critical points for a particular equation (or determined what said equations critical points are)they are usually sufficient.
DeleteYou should sufficiently label the graph. Drawing the graph of $r$ as a function of $\theta$ can also be useful, though not necessary.
DeleteAs Stacey pointed out, it is very hard to determine bounds, because we need to evaluate a case by case scenario.I remember that you did a an strategy of integration recognition in class, can we do the same for polar bounds? That would be VERY helpful.
ReplyDeleteThere were different ways to try and integrate functions, but finding bounds is a single task. It's now up to you to do a lot of problems and figure them out. Then if one really stumps you, then let me know and I'll work it out in class or online.
DeleteSee Dr. Brown's Facebook post on 20120313.
I am also having trouble identifying the bounds and I would really appreciate a sort of "bound recognition" session. Also, I've been unsure about sketching polar graphs. If I memorize the basic shapes and practice with transformation is that enough preparation for the midterm? Maybe we could dedicate a short period of class to practicing transformations to help with recognition.
ReplyDeleteSee the above reply.
DeleteA key to finding bounds has to do with sketching the graphs. Memorizing basic shapes will definitely help with determining bounds.
Practice makes perfect.
I am forming some ideas about what we can try to do, but in the end I recommend you put a lot of your own effort into figuring out how to determine the bounds.
Personally, I attack such problems with the aid of the graph.
I sometimes have trouble remembering the ways of computing limits, which is relevant because these skills are needed in computing improper integrals. Is there an easy way to remember which techniques should be used, like L'Hospital's Rule, etc?
ReplyDeleteI believe the book has a summary of rules. In addition to l'Hospital's Rule, there are rules dealing with pulling out a scalar multiple, when the sum of the limits is the limits of the sum (and similar for multiplication and division). Then there's taking the limit inside a continuous function.
DeleteExample: $$\lim_{x\to a} e^{2x+5}= e^{\lim_{x\to a} (2x+5)}$$.
Then there was a specific limit you might have learned: $$\lim_{x\to 0} \frac{\sin(x)}{x}=1$$
Then sometimes you need to multiply by some sort of conjugate.
Example: In finding the limit of $\lim_{x\to \infty} (\sqrt{x}-\sqrt{2x})$, we multiply and divide by $\sqrt{x}+\sqrt{2x}$.
Similar examples exist.
And the last technique I can remember off the top of my head is computing the limit of a quotient of polynomials, or algebraic expressions. The official method is to divide the top and bottom by the greatest degree and use some of the above laws. The boils down to looking at the "dominating" degree of the numerator and denominator. When they are equal, the limit is the quotient of their "true" coefficients; if that of the numerator is greater, then the limit is $\infty$; and if that of the denominator is greater, then the limit is $0$.
Good luck reviewing!
In section this week could we do a challenge problem involving polar coordinates and finding the areas enclosed by two functions that do not have obvious endpoints? Thank you!
ReplyDeleteI like how this request was a little more focused than previous requests on the general topic. I will likely prepare such a problem. But as I said above, practice practice practice.
DeleteIt may be a question of just practicing, but how can you tell if a function is greater than another function on some interval. This just came up during comparison testing in 7.8.
ReplyDeleteThanks
It's a matter of practicing combined with knowing the basic rules regarding inequalities.
DeleteExample of Basic Rules:
DeleteIf $a < b$ and $c > 0$, then $ac < bc$.
If $a < b$ and $c < 0$, then $ac > bc$.
If $0<a<b$ and $0<c<d$, then $ac<bc$ and $bc<bd$, so that $ac<bd$.
If $0<a<b$, then $\frac{1}{b}<\frac{1}{a}$.
Example:
On the interval $(0,1)$, $0<\sin(x)<1$ and $x^2<x$. So then $\frac{1}{x}<\frac{1}{x^2}$ and $\frac{\sin(x)}{x}<\frac{1}{x^2}$.