Do some method recognition on your own. Perhaps we'll do odd in class and you can do evens on your own. If you do evens now and if we exhaust the odd ones in class, then by the time we do evens as a class, it'll be half-new. Note that there's also the review section of chapter seven for which we can go through method recognition.
Any feedback on things you find distracting or that can be improved on?
What is something you learned from reading the book?
I can't be sure I can cover all the bases, but at the minimum, try testing your knowledge here (20120209) and here (20120221).
9.3.18- How do you treat the K variable? I've looked through the book and couldn't apply the method they used, and just can't seem to see a solution.
ReplyDeleteRyan Schneider
Sorry for replying late. The $k$ in 9.3.18 is just a constant. As a remark, we note that $t$ must be greater than $0$. So you have $$\int \frac{1}{L^2}dL =k\int \ln(t) dt$$. From there you get $$-\frac{1}{L}=k \left[t\ln(t)-t\right]+C$$. Then use the given initial condition to determine that $C=1+k$. Thus $$L=-\frac{1}{k\left[ t\ln(t)-t+1\right]+1}$$.
DeleteThe $k$ that appears in problems 29-32 is also a constant, but due to the nature of the problem, you have to deal with it as explained in Example 5 on page 597.
DeleteI was wondering if there is a good method of checking to see if I got the correct answer for a linear differential equation. Is there any easy way to do this, other than going back over my work?
ReplyDelete-Ashleigh Thomas
Depending on the original equation and the complexity of the answer, it is sometimes reasonable to check your answer with the original equation. Using the question and answer for Example 3 on page 595, $$\frac{dy}{dx}=x^2 y$$ $$y=Ae^{x^3/3}$$. Then to check our answer, we differentiate to get $$\frac{dy}{dx}=Ae^{x^3/3}x^2$$. That's the left-hand side. Plugging $y$ into the right-hand side of the original equation, we also get $Ae^{x^3/3}x^2$, so we know we found a correct solution.
DeleteFor problems where there is a constant C, is it ok to just leave it as "+C" even if you later manipulate the equation so that the actual value of C changes. for example if you have a C then you divide it by 2 (or drop in an ln or e, etc.), do you have to write C/2 or are we allowed to just leave it as C?
ReplyDeleteIn general, you can replace a manipulated constant by another constant, but you should have the habit of giving it a new name.
DeleteFor example, $2y=2x+C$ becomes $y=x+C_1$, where $C_1=C/2$.
But there are other minute details to pay attention to. For example, $\ln\left|y\right|)=x+C$ becomes $\left|y\right|=e^{x+C}=e^x e^C=Ae^x$ where $A=e^C$. Thus, $A$ isn't just any constant, $A$ is a positive constant. But since $\left| y \right|$ is either $y$ for $y>0$ or $-y$ for $y<0$, then we can have $y=Be^x$ where $B=A$ or $B=-A$. This issue came up in one of the graded homework problems, which I will eventually write a solution to.
After I wrote the above, I glanced down at the book and something very much like what I wrote was sitting right there on the page. Then end of example 3 on page 596. I wish all students would supplement their learning by reading the book more.
DeleteYesterday in class we learned about parameterization, and I did not understand how you find your bounds. Can we go over this at some point?
ReplyDeleteFinding bounds is quite a general question. If you are going from Cartesian coordinates to a parametrization, then you start with points $(x_1,y_1)$ and $(x_2,y_2)$. Then you have to solve for the times $t_1$ and $t_2$ so that $x(t_1)=x_1$, $y(t_1)=y_1$, $x(t_2)=x_2$, and $y(t_2)=y_2$.
DeleteA related reply is about 8 posts down. Under "Chris Skoff."
If you're using CTRL+F, then search for "Chris.Skoff".
DeleteA student had a related question for polar coordinates at Parametric Equations; CTRL+F "Julia Deutsch".
Can we go over graphing parametric equations?
ReplyDeleteIs there a specific or convenient method such as DETAIL to know what to replace our new equations for parametric equations. Professor Brown uses sin and cos, but how do we know which goes with x and y?
ReplyDeleteI'm not sure of the context for this question. Would you mind elaborating or giving a specific problem for which this substitution comes into play?
DeleteI would like to go over how we choose the parametric functions for graphs more difficult than circles.
ReplyDeletePlease supply me with some specific problems. Otherwise, the best I can suggest is keep practicing and you'll get a good idea of how to choose the functions. Often the geometry of the graph should suggest the choice of functions.
DeleteTIm,
ReplyDeleteI don't really understand the direction, or how to draw the arrows on the curves, when dealing with parametrization problems. Going over one of them would be very helpful.
Thanks,
Marcelo
To summarize, in class I informed you that the direction of the curve can be determined by the first derivative. Thus, if the curve is $(x(t),y(t))$ then the curve will go in the direction of $x^\prime (t),y^\prime (t)$. Many students find plotting points for various values of $t$ useful.
DeleteI agree with most of the posts here. I really would like more detail about the parametrization problems. Thank you!
ReplyDeleteI've been doing the homework due March 2. On 10.2.54 I can set up the integral, but I don't know where to go from there, since a is a constant, but unknown. any suggestions?
ReplyDeleteI don't have the book with me right now, but I glanced at the problem earlier today and if you were able to set up the integral, I don't see why the constant $a$ should give you any difficulty.
DeleteFor example, how would you do the problem $$\int \cos (\beta x)dx$$?
That one is considerably less complex. The integral is like integrating sqrrt((-3acos^2(x)sin(x))^2+(3asin^2(x)cos(x))^2) It's really messy and multiplying out still looks pretty bad, from a stand point. I considered what it looks like graphically, and I think the integral would just be what it would be without the a, then you can multiply through by a and it would be right. But that seems like bad technique
DeleteOh and it would integrate to Bsin(Bx) +c
DeleteActually, multiplying that out and factoring out the common terms ends up being really nice. I'll multiply it out and you try to factor the common terms. Good luck. $\sqrt{9a^2 \cos^4(x)\sin^2(x)+9a^2 \cos^2(x)\sin^4(x) }$.
DeleteIn section today, i learned the proper time to use chain rule in an integral.
ReplyDeleteI feel like I would like to do some more examples of parameterization so that I feel confident about the subject. I think it is somewhat challenging.
ReplyDeleteWhen finding the length of a parametric curve, how do you determine the bounds on your integral?
ReplyDeleteThe question is quite general. In class, we worked on a problem where we had to find where the curve intersects itself. This means we needed $t_1 \neq t_2$ such that $x(t_1)=x(t_2)$ and $y(t_1)=y(t_2)$. We used these three pieces of information to solve for where the curve intersects itself.
DeleteUsing the picture of the curve can help as well. Let me know if there's a more specific problem you have in mind.
There is a related reply about 8 posts up. Under "Stacey Hall."
DeleteA student had a related question for polar coordinates at Parametric Equations; CTRL+F "Julia Deutsch".
DeleteI was integrating the final homework problem and I found that my answer for the length of the astroid was zero. Should I change the bounds of integration from (0 to 2pi) to four times (0 to pi/2) so that I end up with a positive distance?
ReplyDeleteYes and no. The "right" way is to realize that $\int _0 ^{2\pi} \left| 3a\cos\theta \sin\theta \right|d\theta$ requires you to break up the integral from $0$ to $\pi/2$, $\pi/2$ to $\pi$, $\pi$ to $3\pi/2$, and $3\pi/2$ to $2\pi$. The reason for this is the absolutely value signs. Ultimately, by the symmetry, this is the same as four times the integral from $0$ to $\pi/2$.
Delete