Check out the discussion that's already happened here: Class Discussion Due 20120207
If you didn't already participate in that discussion, please do. Participating is easier than you think.
A new thing you can comment on is what you like or don't like about section.
Last week I had you look at Appendix A and Reference page 1 Algebra.
This week, look at the left-hand side of Reference page 2.
Try to finish reading Appendix A and do some problems (every 8 is a good suggestion).
If you tried to post, but couldn't feel free to send your comment by e-mail. Hopefully you'll eventually be able to post on your own. Sometimes it's a matter of web browser compatibility. Personally I use Google Chrome when I browse on www.blogger.com.
I think working on problems as a group then as a class worked really well today. It gave me a chance to work out problems that I otherwise would have ended up just copying off the board without necessarily understanding the process.
ReplyDeleteIt was interesting to work as we did, taking care of every possibility to deal with a unique problem.
ReplyDeleteIt allowed me to learn a new Trig Identity: sin2a=2sina*cosa
Excellent!
Delete$$\sin (2a)=2\sin(a)\cos(a)$$
I was doing 7.3.30 when I realized that I could only solve the integral down to its indefinite form, but when I tried to substitute the bounds in, I got non-existent answers. What was the method to fix the bounds so that doesn't happen?
ReplyDeleteI suppose it depends if you're changing the bounds as you go, or substituting back your original substitutions. What I like to do when making substitutions is keep track of the variable for my bounds separately.
DeleteLet's look at 7.3.4:
$$\int_0^1 x^3 \sqrt {1-x^2}dx$$
Let $x=\sin(\theta)$. Then $dx=\cos(\theta)d\theta$
I could either write the next step as
$$\int_{0=x}^{1=x} \sin^3 (\theta)\sqrt {1-\sin^2(\theta)} \cos(\theta)d\theta$$
or
$$\int_{0=\theta}^{\pi/2=\theta} \sin^3 (\theta)\sqrt {1-\sin^2(\theta)} \cos(\theta)d\theta$$
During the course of the problem, we may choose to revert to old bounds or change them, depending on what the problem calls for. But the KEY is to keep track of which variable the bounds are in.
For instructive purposes, let's see how the rest of the problem unravels.
$$\sqrt {1-\sin^2(\theta)}= \left| \cos(\theta) \right|$$. But since $\cos(\theta)$ is positive on the domain of integration $\theta=0$ to $\theta=\pi/2$, we can drop the absolute value sign.
Thus the integrand simplifies to $\sin^3(\theta)\cos^2(\theta)$
We use tools learned in section 7.2 to do this integral and our answer is a function of $\theta$. If our bounds were still in the variable $x$, we could choose to change those bounds in terms of $\theta$ or solve for $\theta$ in terms of $x$ and use our original bounds. It happens that for this problem, it would be easier to change the bounds in terms of $\theta$.
Remark: Earlier in the problem, we needed to determine the bounds for $\theta$ anyways, to simplify $\sqrt{\cos^2(\theta)}$
Ah there it is. I accidentally never switched back from theta to t. Thanks!
DeleteWhen I was doing my homework I was having some problems doing trig subsitutions and I was wondering if you could go over some more examples in class?
ReplyDeleteYes, in fact we should quickly go through all the odd problems (5 to 29) on page 483 and identify which function to substitute. You should be able to recognize the proper substitutions for all those problems in under 2 minutes. Shoot for 1 minute. I'll give you guys 1 minute to do 5-15 odd. We'll discuss and answer. Then another 1 minute to do 17-29 odd. We'll discuss and answer. Again, just choice of substitution.
DeleteThen I'll work out a couple in full. Perhaps 18, 25, and a modification of 30.
Please see this link. 7.3.18, 7.3.25, and help on 7.3.30
DeleteIt was really interesting to finally work through some problems about partial fraction decomposition and finally learn this technique. But would it be possible to try some more of these kinds of problems in class?
ReplyDeleteMichelle Bohrson
Note that "you" in the following refers to the reader.
DeleteI will set up the partial fractions for the fractions below. You are encouraged to attempt to setting them up yourself before coming to class. You are welcome to make up examples which I may consider setting up for you in class.
$$\frac{1}{(x-1)^2(x-2)^3}$$
$$\frac{1}{(y^2+2y-3)^2(y^2+2y+3)^2}$$
$$\frac{1}{(x+10)(x-5)^2(x^2+x+1)}$$
$$\frac{1}{(x^3+27)}$$
On a related note, if you did as I recommended last week, then you know how to factor:
$x^2-y^2$, $x^3+y^3$, and $x^3-y^3$
You would also know the quadratic formula.
These are found on Reference Page 1.
Is making a trig substitution regardless of which one it is (either sin,cos or tangent) assure the same answer at the end?
ReplyDeleteYes and no. $\sin$ and $\cos$ are interchangeable. Either can be used in the situation of integrating an integrand that contains an expression of the form $\sqrt{a^2-x^2}$. I believe you should get the same answer to both indefinite and definite integrals, but I'm not absolutely positive. My doubt lies in the domain and ranges of the inverse functions. In any case, I think $\sin$ is a little nicer than $\cos$ because it's derivative doesn't involve a sign change. The change of bounds will also be nicer. $\sin$ on its restricted domain is an increasing function, while $\cos$ on its restricted domain is a decreasing function. See page 478 for a table of when to use which substitution. Observe that $\tan$ and $\sec$ play different roles, from each other and from $\sin$. Why? See also page 67 to 69 for a review of inverse trig functions.
DeleteI always struggle to recognize when a trigonometric identity is appropriate for substitution, any tips? I so often do a substitution and it ends up taking me in the completely wrong direction.
ReplyDelete-Max Aserlind
Are you referring to section 7.2 or section 7.3?
DeleteReading section 7.5 should help.
DeleteHi, I'm also having trouble with trig substitution, especially with these weeks problems. I was wondering if you could go over it in class on how to approach challenging problems such as in 7.3 similar to the ones that are due this week. examples: 16,24,30
ReplyDeleteThank you
- Nahyr S. Lugo-Fagundo
Please see this link. 7.3.18, 7.3.25, and help on 7.3.30
DeleteThere are so many ways to integrated. There are partial fractions, integration by parts, substitution, and then regular integration. How do you recognize which to do? I'm worried that during a test I won't be able to see the obvious.
ReplyDeleteYes, thanks for reminding me. I should do quick recognition tests. But for the time being, you can read the following, which the book does in a more structured manner on page 495 to 497.
Delete1) Simplify
2) If you see a composition of functions, then consider substitution. If you see the derivative of another function sitting around, then strongly consider substitution. For example $\sin(\ln(x))$ or $\cos(x^2)$.
3a) The product of trigonometric functions is a special type of integral you'll probably end up seeing on the test. Should be simple to recognize, yes?
3b) If you see a rational function (a quotient of polynomials), then use partial fractions.
3c) If you see a product of functions, one of the functions being $e^x$ or $\sin(x)$ or $\cos(x)$, then try using integration by parts.
4b) Other uses for integration by parts can be found by studying section 7.1 examples 1 through 5.
3d) If you see an expression of the form $\sqrt{a^2-x^2}$, $\sqrt{a^2+x^2}$, or $\sqrt{x^2-a^2}$ then you should probably use a trigonometric substitution. See the table on page 478.
Besides learning theoretically which methods you should try first, simply doing a lot of problems builds up a database in your mind that will improve your intuition when it comes to tackling an integral blindly.
You are presented with a list of integrals without the prescribed method on page 499 (section 7.5) and page 530 (review).
For number 7.3 #34 I can simplify the integral down using trig substituition to 3tan(theta) but for some reason if i integrate tan(theta)then draw a right triangle to get back to original x variable and original bounds the answer doesn't come out right. However, if I switch back to x varaible before integrating tan(theta)then integrate the function with respect to x, I get the right answer. Is there a reason for this
ReplyDeleteYou made an error. No, I'm kidding. I'd have to see your work to see what's up. Since it's a definite integral, your final answer should be the same.
DeleteWhen using trig substitution, when the trig function on top is equal to the du of the substitution except it is raised to a power is there a way to integrate it or does that just mean there was an error somewhere earlier working out the problem?
ReplyDeleteFor example if it is the integral of cos^2theta/sin^3theta where sintheta is the u and costheta is the du
Justin Drechsel
I'm currently unsatisfied with the previous answer I gave, and I'm not entirely satisfied with this new one. What motivated your original question?
DeleteIn this situation, it happens that we can convert your expression to $\cot^2\theta \csc\theta$. From here we can try using integration by parts. We know the part that has an antiderivative is $\cot \theta \csc \theta$. Then the remaining function is $\cot \theta$. Let $u=\cot \theta$ and $dv=\cot \theta \csc \theta d\theta$. Remember the extra signs when integrating or differentiating cotangent and cosecant related functions. Anyhow, we obtain $-\cot \theta \csc \theta-\int \csc\theta \csc^2 \theta$. Then it would seem that with a trig identity, our original integrand is $(\csc^2\theta-1) \csc\theta$. We add and subtract $\csc\theta$ to the right-hand side. So we get $2 \int \cot^2 \theta \csc \theta d\theta = -\cot \theta \csc\theta - \int \csc \theta d\theta$. The integral of cosecant is analogous to that of secant. See a table of integrals.
On last weeks homework I was a little confused as to how to relate the answers for the five different methods of solving the integral. While they should all be equal, I was not sure how to show that for the less obvious ones.
ReplyDeleteRyan Schneider
Let's first list the four answers:
Delete56a) $-\cos^2 (x)/2+A$
56b) $\sin^2 (x)/2+B$
56c) $-\cos(2x)/4+C$
56d) Should give you 56a or 56b depending on which trig function you choose for $u$.
A, B, C are constants and you should keep in mind that each answer is a family of functions.
Explicitly:
$-\cos^2(x)/2+A=1/2-\cos^2(x)/2-1/2+A$
$=\sin^2(x)/2+B$
Thus 56a) and 56b) are the same family.
$-\cos^2(x)/2+A=-\frac{1}{4}-\frac{1}{4}\cos(2x)+A$
$=-\cos(2x)/4+C$
Thus 56a) and 56c) are the same family.
By transitivity, 56b) and 56c) are the same family.
Abstractly:
Ask yourself, "What makes two family of antiderivatives the same?" Or perhaps, "How can I describe a family of antiderivatives?"
Well, we can choose a representative, and any other antiderivative can be obtained by adding a constant.
$A={f(x)+C\mid C\in\mathbb{R}}$
$B={g(x)+C\mid C\in\mathbb{R}}$
Let's show that if two representatives differ by a constant, their families are equal. That is if $f(x)-g(x)=C$ for some constant $C\in \mathbb{R}$, then $A=B$.
Proof: Let $h(x)\in A$. Then $h(x)=f(x)+C_1$ for some $C_1\in \mathbb{R}$. Then $h(x)=f(x)-g(x)+g(x)+C_1=C+g(x)+C_1$. Let $C_2=C+C_1\in \mathbb{R}$. Then $h(x)=g(x)+C_2\in B$. We conclude $A\subset B$. We can repeat the argument to show that $B\subset A$ and therefore $A=B$.
Using this result, we just need to show $f(x)-g(x)=C$ for some constant $C$. Well $\sin^2(x)/2-(-\cos^2(x)/2)=1$. Hooray!
I find that doing problems on the board is helpful and I liked working in groups during section.
ReplyDeleteAlso could the assigned HW problems on the website change as late as Thursday or are the problems posted now finalized? --Tatiana
The syllabus officially says: "Homework based on the week’s lectures will be posted as official on the course web site sometime on Friday (Homework may be posted earlier, but may change as the lectures evolve for the week)." That means the homework 2 posted last Friday was official as of Friday.
DeleteI still, however, recommend you check you've done all the problems listed, and that you've done all of them correctly. How should you go about checking you've done them correctly? The easiest way is to go through with a friend and see if your answers match. If you're unlucky then you both got it wrong, but at least you tried to check your answer.
For 7.3.2, I'm having trouble visualizing the triangle. They give the substitution x=2tanT (say T=theta). This means that T=tan^-1(x/2).
ReplyDeleteSince tanT=(sinT/cosT), would arctan(x/2) mean that the side opposite T would be 2, the adjacent side would be x, and the hypotenuse would be (x^2+4)^1/2?
Or would the opposite and the adjacent be flipped?
Ashleigh Thomas
The easiest visualization is probably to write $\tan (\theta) = \frac{x}{2}$. Since $\tan(\theta)=\frac{\mathrm{opp}}{\mathrm{adj}}$ then choose the side opposite $\theta$ to be $x$ and the side adjacent to be $2$. The hypotenuse would still be $\sqrt{x^2+4}$ as you computed.
DeleteIn the last lecture Prof. Brown mentioned direction fields and I thought this seemed like an interesting concept. Are there any applicational uses of this concept or is it just a concept to illustrate the possible behaviors of a differential equation?
ReplyDeleteI'd say that they definitely can help you analyze the solutions to a differential equations. Perhaps prior to seeing the picture, you wouldn't know what you're looking for, or if the solution you arrived at was the correct one. However, with the direction field in hand, you have some idea. I'd say that's pretty useful.
DeleteOn a related note, direction fields are special types of vector fields, where at each point you attach the directional vector. But in general, you can assign a vector to a point and study the properties of the vector field. You could, theoretically speaking, at every point in a room, attach the direction and magnitude of the airflow at that point. Or you could attach the temperature at that point and how much its increasing or decreasing. You could ask yourself how would such a vector field change with respect to time.
Another answer is that direction fields show the solutions of a differential equation visually.
DeleteNow that we know all these new methods for solving Integrals (Integration by Parts, Identities, Trignometric substitution, Partial fraction decompisition, etc.), how do we know when to use which way? Like I know theres cetain patterns we should recognize but if Prof. Brown didn't tell us which approach to use on an exam I'm not sure I'd always start with the right method.
ReplyDeleteThe book provides a strategy in section 7.5. I briefly discussed it with you in class.
DeleteIn section I like working in groups first and the collaborating as a class to make sure everyonegetsit right. I also think it would be helpful to do a wide range of problems in each section to practice recognizing structures within integrals.
ReplyDeleteThe acronym DETAIL was really helpful.
ReplyDeleteDifferentiate
Exponential
Trig functions
Algebraic functions
Inverse functions
Logarithmic functions
just as a reminder
One of my biggest troubles in this part is the partial fractions. I understand the concept and am able to solve it when there are obvious factors, but how do you know when to use long division and what variable.
ReplyDeleteex:
1/(x^2+4x+8)^3
where do we begin?
You answered my question in section today... thanks!
DeleteNo problem. Though it does seem I left out when to use long division. Basically you want the degree of the polynomial in the numerator to be less than the degree of the polynomial in the denominator. If it isn't, then you use long division. (It also doesn't hurt to try and simplify by cancelling common factors whenever possible)
DeleteExample (of simplifying):
$\frac{(x-1)}{(x-1)(x-2)(x-3)}$
simplifies to
$\frac{1}{(x-2)(x-3)}$.
Thus instead of using partial fractions on the first expression, we use it on the new, simplified expression. This saves us time, as well as reducing errors made when expanding and regrouping.
As a counter to all of the "liked working in groups" comments, I did not feel that working as part of a group accomplished much. However, this may have just been based on my personal experience rather than reflecting on the concept as a whole (I guess we'll just have to see if it makes any difference today in section).
ReplyDeleteI thought the group exercise in which we just went through the problems and thought about how to approach them-u substitution, integration by parts, partial fractions, trigonometric substitutions, etc. was helpful. Maybe we can do more exercises like that and spend more time on explaining which method to use because sometimes it's not obvious which technique we should use to integrate the functions.
ReplyDeleteWhat's the deal with 7.3.16? I'm not sure if I substituted wrong or just couldn't integrate cos^4(x) correctly
ReplyDeleteSo we use the half-angle identity $$\cos^2(m)=\frac{1}{2} (1+\cos (2m))$$ to integrate $\cos^4 (x)$.
Delete$\int \cos^4(x)dx=$
$\frac {1}{4} \int (1+\cos(2x))^2 dx$
$=\frac{1}{4}\int 1+2\cos(2x)+\cos^2 (2x) dx$
We have to use the identity again with the $\cos^2 (2x)$ term.
Good luck!
What exactly makes a function an Ordinary differential equation? Does a linear diff eqn only differ in the way it is structured
ReplyDeleteAn ordinary differential equation is an equation where there is one independent variable and the equation involves derivatives in that variable. This contrasts with a partial differential equation where there is more than one independent variable and partial derivatives. A linear differential equation is one in which the differential operators are linear. Thus, there are linear ODEs and linear PDEs.
DeleteWhy is C often changed to another constant like K in some problems? Like 3C=K.
ReplyDeleteFor simplicity. As example 1 on page 595 explains, $C$ is an arbitrary constant, so there's no difference between $y=\sqrt[3]{x^3+3C}$ and $y=\sqrt[3]{x^3+K}$ where $C$ and $K$ are arbitrary constants. However, the second expression looks cleaner.
DeleteOn the other hand, there is a difference between $y=e^x e^C$ and $y=Ae^x$ where $C$ and $A$ are arbitrary constants. $e^C$ is always positive. Thus to make the above two the same, we'd have to specify that $A$ is an arbitrary positive constant.
Do we have to care about the constant C we get when we try to find I(x)=exp(integral(P(x)))?
ReplyDeleteOr can we just get rid of it?
Thank you.
Indeed, you only need a particular antiderivative. That's because of the role the integrating factor plays in solving for the solution.
Delete