Calc help

[Trajan]

My coefficients are wrong when I expand (2x + 3y)^3, I just noticed that, hoping that is why lol...

Edit: So yea, (2x + 3y)^3 = 8x^3 + 36yx^2 + 54 xy^2 + 27y^3. Divided by 6 it is: (4/3)x^3 + 6yx^2 + 9xy^2 + (9/2)y^3.

Which is slightly different than what I got when I integrated, hoping I integrated wrong..

Edit #2: God damn do I hate doing math at night

[Kaylia]

Well, with one method, you will necessarily end up with a y³, while the other can't give you more than y² . It was bothering me earlier when I wrote the integral, but I didn't put much though to it.

It's probably because I'm tired as well, but I seriously don't see why we wouldnt be allowed to split the problem like this. All 3 constants should dissapears when you evaluate it, and x/y are independant.


[edit]

nvm, this part was wrong


[edit2]

Oh well, I give up for now. I need sleep, and the mathematics fairy Woozie will solve the problem while I'm sleeping.

[edit3]

Edit #2: God damn do I hate doing math at night

I was going to bed when you posted this. Now i can't sleep because it's bugging me, and I can't understand what's going on because my braincells are on strike. (and thanks for pointing out that mistake. I hate admitting that I don't understand something, but it's for the best)

[Trajan]

Actually, now that I think about it, getting (1/6)( 2x + 3y )^3 after integrating with respect to x isn't even possible.

Like we said, expanded this is equal to (4/3)x^3 + 6yx^2 + 9xy^2 + (9/2)y^3.

That (9/2)y^3 term doesn't even have an x in it, after you just integrated with respect to x lol.

Edit: For the same reason, you couldn't get this after integrating with respect to y first, because of the (4/3)x^3 term, or after integrating with respect to both x and y as every term would have an x and a y in it.

Edit #2: Yea I know. I hate not being able to solve a problem lol, I can't stand it.

Edit #3: In regards to op's 2nd post, is this supposed to be a definite double integral, with x going from 0 to 1 and y the same? It kind of seems like it from this post. Also, I don't recall ever using U substitution in multivariable calc... I don't think you can.

[Bugpop]

The original function is quadratic. Expanding a quadratic function is not time consuming at all and it's very unlikely that he'd make a mistake. For me, it would be the least likely way to make a mistake for a quadratic function. With u substitution I have a tendency to forget constants.

I've also noticed that people have a tendency to be really good at recognizing and integrating stuff that looks like polynomials. When I'm tutoring, if I give someone 9x^4-24x^2+16, they're like "that's easy", but if I give them (3x^2-4)^2, they're usually like "wtf???"

You get cubic functions when you integrate quadradic functions. Expand the quadradic function and integrate. The OP asked why the denominator was 6. You're not gonna see it if you expand and integrate the function without factoring it back.

The substitution method is a much simpler way to explain where the 6 in the denominator came from.

Expanding the function tends to invite error and factoring it in eats up time. There is no easy way to factor cubic functions.

Woozie, what method do you uses to factor quadratic function (serious question)? It's still giving me a lot of trouble when I'm trying to find an answer with rational number.

I agree u substitution is more confusing than expanding a quadratic (minus the factorization for me!), but I have a hunch he will have to use this method sooner or later to solve different type of integral (trigo). It's a good idea to use it a couple of time on simple problem, and get used to the method.

The u substitution makes the problem easier since you don't need to factor the cubic function.

[Woozie]

You get cubic functions when you integrate quadradic functions. Expand the quadradic function and integrate. The OP asked why the denominator was 6. You're not gonna see it if you expand and integrate the function without factoring it back.

If you expand the (quadratic) function first and then integrate, you're going to get a polynomial, which most people are very comfortable with integrating regardless of it's power. You said factoring cubic functions isn't an easy task. I was pointing out that there was no need to factor or expand any cubic functions. The only thing that needs to be expanded is the quadratic function. Anyone at his level is extremely used to expanding quadratic functions and integrating polynomials.

I didn't understand what he meant in the original post, so I didn't address the issue of why the first step resulted in the expression he posted. After I realized what he meant, I didn't bother addressing this part of the question because Kaylia had already done so. I explained the entire method of solving the problem to him, and Kaylia explained how to evaluate the integrals. There's really nothing left to say.

Now that I realize that the function is to be integrated over a certain region, I'd probably go the u substitution route too at this point. So I'm not even disagreeing with you when you say he should do the u substitution. I just don't know where you keep getting this idea that the cubic function needs to be expanded or factored at all if we choose to go the other route. Maybe I'm missing something.

Edit: You all HAVE to watch this video

http://www.youtube.com/watch?v=BipvGD-LCjU

[Kaylia]

^
Rofl...the lyrics are great




And woozie, can you explain why evaluating the expanded term give a different value than evaluating (x+y)². I seriously don't see where the y³ order dissapears

[Woozie]

I don't know. I just did it on my calculator (every single step, so there's no chance I made an error). Integrating these in different ways is definitely giving different answers x_x

bugpop was right, apparently expanding it wasn't a good idea, though I can't see why that gives a different answer.

Edit: Direction from my calculator, I'm getting

http://latex.codecogs.com/gif.latex?...(3y+2x)^4}{72} if I integrate without expanding first

but I get

http://latex.codecogs.com/gif.latex?...frac{4yx^3}{3} if I expand first. This equation only goes up to powers of three whereas the above goes up to powers of four.

(this is after taking both integrations, though the errors are already there even after the first integration)

Edit: Wait a minute, I'm starting to see what we're doing wrong. Give me like 20 minutes and I'll post what's going on.

[Julian]

Hijacking this topic!

Kind of the same topic (well actually it's the previous chapter, meh), I have a practice exam question for my exam tomorrow, it's on Lagrange multipliers. I was absent that day, so I don't really fully understand them

The problem is to get the min/max using Lagrange multipliers, f(x,y)=1/x+1/y; constraint 1/x^2 + 1/y^2 = 1

The way I did this problem I didn't really use lagrange multipliers so idk if I did it right it's also an even problem so I can't check it at all

[Woozie]

Okay, this is what's going on.

Suppose we take the first form of the integral

http://latex.codecogs.com/gif.latex?\int(2x+3y)^2dx

We get

http://latex.codecogs.com/gif.latex?...\frac{9y^3}{2}

Now if we expand first and then integrate, we get

http://latex.codecogs.com/gif.latex?...3}+6x^2y+9xy^2

Remember how in calc 1, our teachers kept marking off all our points because we kept forgetting to add "+c", the arbitrary constant, when we evaluated integrals? Well we should have listened to her.

The difference between the two terms is 9y^3/2. That's just the arbitrary constant (constant because we integrated with respect to x and that doesn't depend on x). When we expand first, we should add an "+f(y)" term for us to be correct. Then note that since any equation of that form is correct, and both of our answers are of this form, then both answers are correct (just like how "x^2+3" and "x^2+5" are both the correct answer to the integral of 2x even though they're two different functions).

Note that since his problem had boundaries, the arbitrary constant drops out. So he could solve the problem by expanding first or by not expanding and he would still get the same answer. So expanding the quadratic term is a legitimate strategy afterall.

But the reason we're gettting different answers when we do it different ways is because of how we're not taking into account arbitrary constants. Both answers we get returns the original function when you differentiate them. Both final answers return the original function when we differentiate with respect to y and then with respect to x. So essentially we got the multivariate equivalent of two antiderivatives being correct because they only differ by an arbitrary constant. And there's infinitely other possible correct answers too.

@Julian: I haven't done lagranged multipliers in so long, so I can't really help you here. I'd have to look it up first, but your exam is tomorrow and I probably wont get around to looking it up today. Sorry x_x I noticed that every time you ask a question I'm always unable to help you for some reason lol

[Julian]

@Julian: I haven't done lagranged multipliers in so long, so I can't really help you here. I'd have to look it up first, but your exam is tomorrow and I probably wont get around to looking it up today. Sorry x_x I noticed that every time you ask a question I'm always unable to help you for some reason lol

I just looked over my problem and realized that I did it wrong, and that I probably have to use them lagrange multipliers for it <_< If anyone can explain the process behind how to do that problem, I'd really appreciate it.

I can do lagrange problems, but this one just messes with my head (why do they all have to be 1/x, 1/y, etc, fuck) I'm probably just failing to "see the creativity" in solving this one

edit: Actually, I may have done it right <_<

Doing the whole fx=Lgx, fy=Lgy, g=1 system of equations, using L for lambda, I get the following system of equations:

1/x^2 = 2L/x^3
1/y^2 = 2L/y^3
1/x^2 + 1/y^2 = 1

I figured.. x=y, so 2/x^2 = 1, so x^2=2, so I got x=+/-root 2, and since x=y, the two anwers are (-root2,-root2) and (root2,root2), plugging into the original 1/x+1/y, the negative one is the min and positive one is the max. Did I do this right? >.>;

[Kaylia]

Okay, this is what's going on.

Suppose we take the first form of the integral

http://latex.codecogs.com/gif.latex?\int(2x+3y)^2dx

We get

http://latex.codecogs.com/gif.latex?...\frac{9y^3}{2}

Now if we expand first and then integrate, we get

http://latex.codecogs.com/gif.latex?...3}+6x^2y+9xy^2

Remember how in calc 1, our teachers kept marking off all our points because we kept forgetting to add "+c", the arbitrary constant, when we evaluated integrals? Well we should have listened to her.

The difference between the two terms is 9y^3/2. That's just the arbitrary constant (constant because we integrated with respect to x and that doesn't depend on x). When we expand first, we should add an "+f(y)" term for us to be correct. Then note that since any equation of that form is correct, and both of our answers are of this form, then both answers are correct (just like how "x^2+3" and "x^2+5" are both the correct answer to the integral of 2x even though they're two different functions).

Note that since his problem had boundaries, the arbitrary constant drops out. So he could solve the problem by expanding first or by not expanding and he would still get the same answer. So expanding the quadratic term is a legitimate strategy afterall.

But the reason we're gettting different answers when we do it different ways is because of how we're not taking into account arbitrary constants. Both answers we get returns the original function when you differentiate them. Both final answers return the original function when we differentiate with respect to y and then with respect to x. So essentially we got the multivariate equivalent of two antiderivatives being correct because they only differ by an arbitrary constant. And there's infinitely other possible correct answers too.

@Julian: I haven't done lagranged multipliers in so long, so I can't really help you here. I'd have to look it up first, but your exam is tomorrow and I probably wont get around to looking it up today. Sorry x_x I noticed that every time you ask a question I'm always unable to help you for some reason lol

Thank, make a lot of sense, and dammit, that was stupid.

I realized last night the y^4 term would be missing because of the constant, but I couldn't see why the answer in the end would be the same.

[Trajan]

I just looked over my problem and realized that I did it wrong, and that I probably have to use them lagrange multipliers for it <_< If anyone can explain the process behind how to do that problem, I'd really appreciate it.

I can do lagrange problems, but this one just messes with my head (why do they all have to be 1/x, 1/y, etc, fuck) I'm probably just failing to "see the creativity" in solving this one

edit: Actually, I may have done it right <_<

Doing the whole fx=Lgx, fy=Lgy, g=1 system of equations, using L for lambda, I get the following system of equations:

1/x^2 = 2L/x^3
1/y^2 = 2L/y^3
1/x^2 + 1/y^2 = 1

I figured.. x=y, so 2/x^2 = 1, so x^2=2, so I got x=+/-root 2, and since x=y, the two anwers are (-root2,-root2) and (root2,root2), plugging into the original 1/x+1/y, the negative one is the min and positive one is the max. Did I do this right? >.>;

It looks like you did it correct to me... LM are a pain in the ass though, so I could be wrong.

[Julian]

It looks like you did it correct to me... LM are a pain in the ass though, so I could be wrong.

Yeah.. the TA that grades my quizzes is harsh One tiny mistake and it's like an automatic -5 (out of 20 pts/question), and I make a ton of mistakes on quizzes cause there's only 20minutes to do 5 rather hard problems, I hate being rushed when doing math, cause I always need to think out what I'm doing, and think it rationally to make sure it makes sense, and I never have time to do that. :/

Hopefully the exam is different, 2h and only 10 problems, so I have all the time I need to do it right. But then again, each question is only 10 points, and if I get -5 for some stupid thing, ugh. Most of the time my stupid thing is just reading what I wrote wrong, like in one step I'll write x^2, and suddenly in the next step its x^4 cause I'm retarded, and BAM, -5. Or, the question will say pi/3, and for some stupid reason I write down pi/6 when I'm rewriting what's given. With calculus it's usually either that or me fucking up some simple differentiation or integration, while still doing every single other step correctly with my wrong answer >_<

[Eliseos]

Going to ask my question in this thread instead of making a new one.

Continuing the theme of vector integration, my problem is with surface integrals. We're using the equation
http://latex.codecogs.com/gif.latex?...0\textbf{n} dS

I don't understand how to choose a good dS here. Some examples in my book use dxdy and then divide that by the magnitude of n dot j, then other examples say that a projection on the xy-plane isn't possible, but it doesn't explain why.

I'll add a specific example to make things more clear. Say A= yi + 2xj -zk and S is the surface of the plane 2x+y = 6 in the first octant cut off by the plane z=4.

So far I have del(2x+y) = http://latex.codecogs.com/gif.latex?...i+j}{\sqrt{5}}

This is where I get stuck, since I have to choose where my projection is. I chose http://latex.codecogs.com/gif.latex?...c{1}{\sqrt{5}}

So http://latex.codecogs.com/gif.latex?...%20\right%20|} = http://latex.codecogs.com/gif.latex?\sqrt{5}%20dxdy which eventually leads to

http://latex.codecogs.com/gif.latex?...{4}(2y+2x)dxdy

But this obviously won't work since I would have a variable that doesn't get integrated. I'm kind of thinking the denominator of ds would have to dot n with the variable we aren't using in the denominator, but if I do that I'd get 0 for n dot whatever and the whole function would integrate to 0. The exam I have tomorrow covers gradients, divergence, curl, line integrals and surface integrals, and this is the only thing I don't really understand. Any help is appreciative.