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Mathematics 9 Online
OpenStudy (anonymous):

Test tomorrow, need an overview-type review for chapter 6 of the calculus eight edition book by Larson, Hostetler, and edwards (the brown one, 8e)~

OpenStudy (kainui):

Don't know where to find that, but I can definitely answer any of your calculus questions.

OpenStudy (anonymous):

okay, uhm what about dy/dx=[x^2+y^2]/[2xy]

OpenStudy (anonymous):

solve :)

OpenStudy (kainui):

Wait, do you mean take the derivative of the right hand side implicitly or solve this like a differential equation?

OpenStudy (kainui):

Either way, take your best guess at solving it, I'm here to help, not do it for you. =P

OpenStudy (anonymous):

the latter, I believe ^^

OpenStudy (anonymous):

yeah, I mean, I've tried separation of variables, and the 1st deriv linear standard form

OpenStudy (anonymous):

TToTT can't get the variables separated~

OpenStudy (kainui):

This is a special case where you make a substitution, like u=y/x. Try it out! =)

ganeshie8 (ganeshie8):

hint : homogeneous equation

ganeshie8 (ganeshie8):

you said, you dont need to learn homogeneous stuff... but u will need to knw that method also to solve this

OpenStudy (anonymous):

hmmm ok then ^^ sounds good then XD

OpenStudy (kainui):

So here if we try to separate it out we get: \[2\frac{ dy }{ dx }= \frac{x}{y} + \frac{y}{x}\] So we make a substitution with: \[u=\frac{y}{x}\] Then since u is a function of x, if we take the derivative after rearranging... \[y=ux\]\[y'=u'x+u\] (just the product rule) then when we plug it all in we get: \[2(\frac{du}{dx}x+u)=\frac{1}{u}+u\]

OpenStudy (kainui):

If any of the reasoning to getting this separable equation seems fuzzy I can explain it further if you like.

OpenStudy (anonymous):

there's this one vertical motion word problem: A falling object encounters air resistance that is proportional to its velocity. The acceleration due to gravity is -9.8 meters per second per second. The net change in velocity is dy/dt=kv-9.8. (a) find the velocity of the object as a function of time if the intial velocity is v(subscript "0"). (b)Use the result of part (a) to find the limit of the velocity as t approaches infinity. (c) Intergrate the velocity function found in part (a) to find the position function "s".

OpenStudy (anonymous):

@Kainui so then... why did you choose to substitute they/x rather than the x/y?

OpenStudy (kainui):

I don't think it matters, they should both work.

OpenStudy (anonymous):

okay, s then integrate both sides? although... how would you integrate the left side?

OpenStudy (kainui):

I just prefer to have it easier to plug in, since if I made it u=x/y then it would look like: y=x/u and I hate to do the quotient rule.

OpenStudy (kainui):

Show me what you got and your attempt at it.

OpenStudy (anonymous):

the integration?

OpenStudy (kainui):

Yep! What does the integral look like?

OpenStudy (anonymous):

wait hold on ...

OpenStudy (anonymous):

would the du/dx x come out as a clean x out of the integration?

OpenStudy (kainui):

No, you'll have to rearrange it so that x and dx are completely separate from the u's and du.

OpenStudy (anonymous):

OoO

OpenStudy (anonymous):

before integrating?

OpenStudy (kainui):

Of course!

OpenStudy (anonymous):

x/dx (du) +u

OpenStudy (kainui):

We're still talking about this equation right?\[2(\frac{ du}{ dx }x+u)=\frac{ 1 }{ u }+u\]

OpenStudy (anonymous):

yes~

OpenStudy (anonymous):

:|

OpenStudy (kainui):

So what does it look like when completely separated to both sides?

OpenStudy (kainui):

\[2\frac{du}{dx}x=\frac{1}{u}-u\] That's after a couple steps, maybe it's easier now?

OpenStudy (anonymous):

{1/2x} dx= [1/{(1/u)-u}]

OpenStudy (kainui):

Perfect. Can you simplify the fraction part of the u's any more? It might make the integration a little easier.

OpenStudy (anonymous):

the only thing I can think of are conjugates, but I don't remember how to do those anymore...

OpenStudy (kainui):

\[\frac{ du }{ \frac{ 1 }{u }+u }=\frac{ du }{ \frac{ 1 }{u }+\frac{ u^2 }{u } }=\frac{ du }{ \frac{ 1+u^2 }{u } }=\frac{ udu }{ 1+u^2 }\]

OpenStudy (anonymous):

O-O" I'm stupid. hahaha thnx ^^

OpenStudy (kainui):

sorry, those should be minus signs, but it's the same haha.

OpenStudy (anonymous):

ok so I got (1/2) ln(x)=(-1/2) ln|1-u^2|+c

OpenStudy (kainui):

Looks right to me. I suggest picking your constant to be in a logarithm so that you can easily use log rules to completely get rid of all the logarithms!

OpenStudy (anonymous):

y=x-c?

OpenStudy (anonymous):

^ that's what I got for my final answer

OpenStudy (kainui):

\[\frac{ 1 }{ 2} lnx=-\frac{ 1 }{ 2}\ln(1-u^2)+\frac{ 1 }{ 2}lnC \] picked that constant since it's completely arbitrary. Nope, that's not how logs work!

OpenStudy (anonymous):

yeah... I guess, but wouldn't putting the ln on the c be making the proceess longer? I just raised everything to the e after mulitplying both sides by 2, and got x=C(1-u^2)^-1

OpenStudy (kainui):

Actually I'll put the constant on the x side. \[\frac{ 1 }{ 2}lnx+\frac{ 1 }{ 2}lnC =\frac{ 1 }{ 2}\ln(\frac{1}{1-u^2})\]\[ lnx+ lnC =\ln(\frac{1}{1-u^2})\]\[ \ln(xC) =\ln(\frac{1}{1-u^2})\]

OpenStudy (anonymous):

you forgot the negative sign on the lnc ^^

OpenStudy (kainui):

Nah, I just completely decided since it was arbitrary that I would start over with it on the other side.

OpenStudy (anonymous):

hmmmm ok~

OpenStudy (anonymous):

didi you get y=sqrt(x^2-cx) ?

OpenStudy (anonymous):

Or did you get a positive c?

OpenStudy (anonymous):

@Kainui ???

OpenStudy (anonymous):

uhoh... left before we finished... QQ

OpenStudy (anonymous):

@wio did you work this out?

OpenStudy (anonymous):

which one?

OpenStudy (anonymous):

the one that was first mentioned ^^

OpenStudy (anonymous):

what is the equation?

OpenStudy (anonymous):

we got to here:\[\ln(xC)=\ln(1\div(1−u2))\]

OpenStudy (anonymous):

but what was the original equation?

OpenStudy (anonymous):

dy/dx=[x^2+y^2]/[2xy]

OpenStudy (anonymous):

Oh ok.

OpenStudy (anonymous):

well, then what did you let \(u\) be again?

OpenStudy (anonymous):

y/x

OpenStudy (anonymous):

well, undo the substitution I guess.

OpenStudy (anonymous):

I'm not sure that you're going to find an explicit formula.

OpenStudy (anonymous):

no, you're solving the derivative, so you're integrating~

OpenStudy (anonymous):

did you already integrate?

OpenStudy (anonymous):

@ganeshie8 please come and help with this last portion~

OpenStudy (anonymous):

yeah, I integrated, and I raised everything to the e, and then I played around witht the numbers, but I'm not quite sure what to do with the c, and if it's negative or positive because it matters if I'm asked to solve for it

OpenStudy (anonymous):

you have \[ Cx = \frac{1}{1-u^2} = \frac{1}{1-\left(\frac{y}{x}\right)^2} \]

OpenStudy (anonymous):

You need initial conditions to solve for C.

OpenStudy (anonymous):

oh, we're not solving for c, it's just that I'm worried that if I was to be asked to obtain the value of c, then I would get mixed up on what side to put it on... apparently, it doesn't matter as long as you keep it constant while solving, but Kainui was putting it somewhere else, and I was wondering if that made too much difference on the final answer,...

ganeshie8 (ganeshie8):

y= +- sqrt(x^2-cx)

OpenStudy (anonymous):

oh! right~ the plus and minus~

ganeshie8 (ganeshie8):

placement of c wud change the final equation, but it will not change the numerical answers

ganeshie8 (ganeshie8):

it changes how the final equaiton wud *look* like

ganeshie8 (ganeshie8):

where do u wana put c ?

ganeshie8 (ganeshie8):

(1/2) ln(x)=(-1/2) ln|1-u^2|+c

OpenStudy (anonymous):

hmmm, since the numerical values won't change, it doesn't matter, right?

ganeshie8 (ganeshie8):

looks you're fine, till this point ?

ganeshie8 (ganeshie8):

yup, it doesnt matter

OpenStudy (anonymous):

okay~ sounds good~

ganeshie8 (ganeshie8):

(1/2) ln(x)=(-1/2) ln|1-u^2|+c say, c = -1/2lnk then u wud get : (1/2) ln(x)=(-1/2) ln|1-u^2| - 1/2 ln(k) ot, ln(x) = ln(1-u^2) + ln(k) x = k/(1-u^2)

ganeshie8 (ganeshie8):

or, x/k = 1/(1-u^2)

ganeshie8 (ganeshie8):

the relation between ur constant and the earlier kainui's constant is : C = 1/k

ganeshie8 (ganeshie8):

it doesnt matter, k is as good a constant as 1/C or p or q or s or c2 or c^2

OpenStudy (anonymous):

ok~ and could you help me with that word problem I typed earlier? Actually I'll open another question and tag you ^^

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