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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

once again an arc length problem leaves me stumped.

Recall the formula for arclength is the integral of sqrt( 1 + (dy/dx)^2 ) dx from x = a to x = b.

**Find the length of x^3 /12 + 1/x from x = 1 to x = 2**

first we have to take the derivative to find dy/dx and square it:

dy/dx = x^2 /4 - 1 / x^2

(dy/dx)^2 = [x^2 /4 - 1 / x^2]^2

now we apply the formula:

∫ sqrt ( 1 + [x^2 /4 - 1 / x^2]^2 ) dx from x = 1 to x = 2

I cannot for the life of me figure out how to integrate this. Not even with a trig sub. Typicaly, in a trig sub you have a x^2 term and x is some expression who's derivative does not contain x. For instance ∫ 1/( 1 + 4x^2) dx use the trig sub: tan u = 2x so now you can replace 4x^2 with tan^2. Now we have to replace dx. Differentiating tan u = 2x we find dx = ( sec^2 u) / 2 du notice this expression for dx does not contain the variable x.

But in this case, we have

∫ sqrt ( 1 + [x^2 /4 - 1 / x^2]^2 ) dx

we cannot let u = x^2 /4 - 1 / x^2 and rewrite it in the form:

∫ sqrt ( 1 +u^2) du

because we cannot get an expression for du that does not contain the variable x. :-(

Any idea's?

*Last edited by mikau (2006-02-13 12:58:32)*

A logarithm is just a misspelled algorithm.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

Well technically, you could let u = x^2 /4 - 1 / x^2. The problem would be in solving that in terms of x. If you could then you would be left with;

∫f(u)√(1+u²) du

Then you could do integration by parts. Alas, I couldn't solve for x in this problem. Why don't you just use a calculator? Mine spit out an answer instantly. I probably mentioned this before, but my calculus text makes it a point to say that finding arc lengths with this formula is very rarely possible without technology.

edit*

If you do use your calculator, notice how simple the answer is. Why do these seemingly simple solutions come from such difficult problems?

*Last edited by irspow (2006-02-13 13:39:48)*

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

if its a simple solution then it can probably be expresed as a simple algebraic expression and is manually integratable. (is that even a word?)

Usually my book says when to use a graphing calculator. So I don't think I'm supposed to.

*Last edited by mikau (2006-02-13 13:46:01)*

A logarithm is just a misspelled algorithm.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

God bless you mikau, I wish you the best. My experience tells me that you are fighting an uphill battle here though. At least get yourself a large book of integrals. If you are using the little list of 50 to 100 or so in a standard calculus text, they only cover the most standard of integrals that you will encounter in calculus.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Thanks for the advice. I didn't know they had books of nothing but integral problems. That sounds awsome! I'll have to get my hands on one.

Still... this integral appears in my math text so it should be doable using at least one of the methods they've taught me.

A logarithm is just a misspelled algorithm.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

I'm sure that there **is** a do-able solution. I could easily have forgotten something from college. It was (blush) years ago.

Oh yeah, they have whole books of integrals. Your text has only a handful. There are a mind boggling amount of known u substitutions out there to be had. A decent university bookstore should have at least one version for sale. Obviously, you won't come across anything like that in the "Borders" or "Barnes and Noble" part of the world being what the average reader is.

Have fun.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

The problem is the possibility exists that I am supposed to use the calculator even though they didn't tell me. In which case I'd be barking up the wrong tree. :-(

A logarithm is just a misspelled algorithm.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

You got it to here;

∫√( 1 + [x^2 /4 - 1 / x^2]^2 ) dx

If you actually square your derivative, you get;

∫√(1 + x^4/16 + 1/x^4 - 1/2) dx;

∫√(1/2 + x^4/16 + 1/x^4) dx;

∫√([8x^4 + x^8 + 16] / 16x^4) dx;

1/4∫1/x^2 √(8x^4 + x^8 + 16) dx;

1/4∫1/x^2 √([x^4 + 4]^2) dx;

u = x^4 + 4, x = (u - 4)^(1/4)

du = 4x^3 dx, dx = du/4x^3

x^3 = (u - 4)^(3/4);

1/4∫1/(4[u - 4]^(3/4)) √(u^2) du;

1/16∫(u - 4)^(-3/4) u du;

You should be able to do integration by parts from here, huh.

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

You're homeschooled, right? Are you accountable to anyone for these problems? If so, ask them. If not, I say screw it, use a calculator.

Look, sane people use calculators. (I think I'll make that my sig.) Who wants to be puzzling over some bizarre integral when we've invented a tool to do it for us? Just working out the real-world problems that require integration is bothersome enough, and it's something no computer can do.

El que pega primero pega dos veces.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

ah! I tried squaring it once but I forgot to add -1/2 to 1 to get 1/2

I think your right. Let me give it a shot!

A logarithm is just a misspelled algorithm.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

lol ryos! Great sig.

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**mikau****Member**- Registered: 2005-08-22
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Yes thats it! The thing is we have [x^2 /4 - 1/x^2]^2 this yields x^4 / 4 - 1/2 + 1/ x^4, but when we add + 1 the middle term -1/2 changes to positive and just so happens to be the perfect square of [x^2 /4 **+** 1/x^2]^2 so we can eliminate radical. Hooray! Thanks irspow!

A logarithm is just a misspelled algorithm.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

I blew right past that realization, good job mikau. I told you that I was rusty.

*Last edited by irspow (2006-02-13 15:06:24)*

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

You're homeschooled, right?

More like self schooled

Are you accountable to anyone for these problems?

Yes. To myself.

If so, ask them.

ok.

"Hey mikau? What should I do?"

"uh... ask Ryos, he might know, in the meantime have a banana!"

If not, I say screw it, use a calculator.

I don't have a graphing calculator.

Look, sane people use calculators.

thats why I don't use one.

(I think I'll make that my sig.) Who wants to be puzzling over some bizarre integral when we've invented a tool to do it for us?

Because thats cheating.

Just working out the real-world problems that require integration is bothersome enough, and it's something no computer can do.

Then more power to me.

Anyways, I like practicing integration techniques. I figure they must be usefull if I'm being taught so much about it. We could teach kids in gradeschool to use calculators only and not how to do anything by hand. But first they wouldn't learn as much and they're mathskill would be entirely dependant on the calculator. They would be completely worthless without one.

Anyways, back to the original problem, the thing to remember is the only difference between (a + b)^2 and (a - b)^2 is the sign of the middle term 2ab. Adding 1 to -1/2 changed the sign so we were able to refactor what was (a + b)^2 into (a - b)^2 and take the square root to get (a - b).

Many thanks irspow.

*Last edited by mikau (2006-02-13 15:17:15)*

A logarithm is just a misspelled algorithm.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

Yes, I realized where I went wrong after scribbling through it again. My **real** problem was in assuming the problem was harder than it was. I can be a lunk head sometimes, I looked right at a perfect square and still didn't eliminate the radical sign. Anyway, the solution is the same as what the calculator gave, so yay to us. Thanks for nudging me to keep at it.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

well I didn't read your whole explanation, I just noticed the sign change of -1/2 to 1/2 and something clicked.

but if your really clever, maybe you can figure out how to solve this one:

∫ sqrt( 1 + 4/9 x^(-2/3) ) dx me and another guy were working on this one for hours and couldn't get it.

*Last edited by mikau (2006-02-13 15:39:01)*

A logarithm is just a misspelled algorithm.

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

mikau wrote:

I figure they must be usefull if I'm being taught so much about it. We could teach kids in gradeschool to use calculators only and not how to do anything by hand. But first they wouldn't learn as much and they're mathskill would be entirely dependant on the calculator. They would be completely worthless without one.

I thought you might say that (everyone does). The answer is that both of our arguments are valid, and neither stands on its own. It's a question of where to draw the line.

Learning to integrate by hand is valuable as far as it goes, but there are some integrals that are just too tricky to be bothered with. This is why integral tables were invented, and later, calculators that integrate.

If I need to integrate for my job (Engineering, etc.), my employer will give me the tools I need to get the job quickly so I can focus on less mundane things. In this sense, an engineer *is* worthless without a calculator. Yet, engineering students are still taught how to do i.e. material balance problems by hand, even though there are advanced computer systems that do them for us. I think we both understand why that's valuable.

So, if an engineer is going to be using a computer to do tough integrals for him, does it make an ounce of difference whether he can do them by hand or not? Of course he should know how to do most integrals, but there are some...where do you draw the line?

Sorry to go OT on you. I'm better now. 8O)

*Last edited by ryos (2006-02-13 15:51:10)*

El que pega primero pega dos veces.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

lol. I do agree. I mean I use a calculator to add subtract multiply and divide when I can't do it in my head, rather then using long division or multiplication.

In the end I think it all comes down to this. A calculator is to avoid tedious time consuming tasks so your attention can be focused elsewhere when needed. A machine to do algorithms quickly and accurately. But never a replacement for understanding of a concept.

A logarithm is just a misspelled algorithm.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

Let u = x^(1/3)

du = 1/[3(x^(2/3))] dx

3x^(2/3)du = dx

3u^2 du = dx

∫√(1 + 4/[9(x^(2/3))] dx

3∫u^2√(1 + 4/(9u^2)) du

3∫(u^2)/3√((9u^2 + 4)/u^2) du

∫u^2√((9u^2 + 4)/u^2) du

∫u√(9u^2 + 4) du

∫u√(9(u^2 + 4/9)) du

3∫u√(u^2 + 4/9) du

3[1/3(u^2 + 4/9)^(3/2)]

[u^2 + 4/9]^(3/2)

u^2 = x^(2/3)

[x^(2/3) + 4/9]^(3/2) + C

*Last edited by irspow (2006-02-13 16:20:21)*

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

hmm... are you sure its legal to replace x with expressions containing u after differentiating? I don't see why it wouldn't be but there are a lot of things that I think should work but don't. :-/

A logarithm is just a misspelled algorithm.

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

It's just another substitution...

El que pega primero pega dos veces.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

wow! Many thanks irspow! I was stuck on that one for weeks!

A logarithm is just a misspelled algorithm.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

Yes, mikau, you can always replace the u with what you substituted it for after you are done integrating. I think it's just a matter of preference. You either do what I did or you change the limits of integration. I just think it is usually easier to do it the way I did because I already know what u equals. If you change the limits of integration then you have to solve for the values of u.

Oh, you won't get stuck much if you keep practicing. After a while, you will have a certain set of integrals that you remember best. What you find then is that you tend to manipulate equations to match the integrals that you do remember.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Very true. Nice observation.

as far as the u substitution, yeah I know you can replace u with the value it stood for after integrating, rather then changing the variable to evaluate it at (in terms of u).

Its just I don't recall ever having to do something you did and didn't know it was legal.

Typically if I made the substitution u = x^2 then du = 2x dx I would have figured u substitution wouldn't work, but you can replace x with sqrt u to get dx = 1/( 2sqrt u ) du ???

*Last edited by mikau (2006-02-14 14:21:48)*

A logarithm is just a misspelled algorithm.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

Yeah, mikau, I have never had any trouble with doing that. I don't recall ever being taught to do it specifically, but like I said, I haven't seen it ever getting me into trouble before.

In the example you gave;

u = x²

du = 2x dx

dx = du/2x

But x = √ u

So, yes,

dx = du/2√u

I do that kind of manipulation all of the time. Actually, I don't remember how else you would do it.

Oh, yeah!

Then 2x would have to actually appear in the integral for you to use the "standard" method. Or even x I guess would still leave you with du/2.

I can see how if you were "lucky enough to have the derivative of u in the integral it would make sense to use it, but how often does that happen?

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