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)~

4 years ago
OpenStudy (kainui):

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

4 years ago
OpenStudy (anonymous):

4 years ago
OpenStudy (anonymous):

solve :)

4 years ago
OpenStudy (kainui):

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

4 years ago
OpenStudy (kainui):

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

4 years ago
OpenStudy (anonymous):

the latter, I believe ^^

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

TToTT can't get the variables separated~

4 years ago
OpenStudy (kainui):

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

4 years ago
ganeshie8 (ganeshie8):

hint : homogeneous equation

4 years ago
ganeshie8 (ganeshie8):

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

4 years ago
OpenStudy (anonymous):

hmmm ok then ^^ sounds good then XD

4 years ago
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: $2(\frac{du}{dx}x+u)=\frac{1}{u}+u$

4 years ago
OpenStudy (kainui):

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

4 years ago
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".

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (kainui):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
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.

4 years ago
OpenStudy (kainui):

Show me what you got and your attempt at it.

4 years ago
OpenStudy (anonymous):

the integration?

4 years ago
OpenStudy (kainui):

Yep! What does the integral look like?

4 years ago
OpenStudy (anonymous):

wait hold on ...

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (kainui):

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

4 years ago
OpenStudy (anonymous):

OoO

4 years ago
OpenStudy (anonymous):

before integrating?

4 years ago
OpenStudy (kainui):

Of course!

4 years ago
OpenStudy (anonymous):

x/dx (du) +u

4 years ago
OpenStudy (kainui):

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

4 years ago
OpenStudy (anonymous):

yes~

4 years ago
OpenStudy (anonymous):

:|

4 years ago
OpenStudy (kainui):

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

4 years ago
OpenStudy (kainui):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (kainui):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (kainui):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (kainui):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
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!

4 years ago
OpenStudy (anonymous):

y=x-c?

4 years ago
OpenStudy (anonymous):

^ that's what I got for my final answer

4 years ago
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!

4 years ago
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

4 years ago
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})$

4 years ago
OpenStudy (anonymous):

you forgot the negative sign on the lnc ^^

4 years ago
OpenStudy (kainui):

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

4 years ago
OpenStudy (anonymous):

hmmmm ok~

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

Or did you get a positive c?

4 years ago
OpenStudy (anonymous):

@Kainui ???

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

@wio did you work this out?

4 years ago
OpenStudy (anonymous):

which one?

4 years ago
OpenStudy (anonymous):

the one that was first mentioned ^^

4 years ago
OpenStudy (anonymous):

what is the equation?

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

but what was the original equation?

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

Oh ok.

4 years ago
OpenStudy (anonymous):

well, then what did you let $$u$$ be again?

4 years ago
OpenStudy (anonymous):

y/x

4 years ago
OpenStudy (anonymous):

well, undo the substitution I guess.

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

4 years ago
OpenStudy (anonymous):

@ganeshie8 please come and help with this last portion~

4 years ago
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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

You need initial conditions to solve for C.

4 years ago
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,...

4 years ago
ganeshie8 (ganeshie8):

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

4 years ago
OpenStudy (anonymous):

oh! right~ the plus and minus~

4 years ago
ganeshie8 (ganeshie8):

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

4 years ago
ganeshie8 (ganeshie8):

it changes how the final equaiton wud *look* like

4 years ago
ganeshie8 (ganeshie8):

where do u wana put c ?

4 years ago
ganeshie8 (ganeshie8):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
ganeshie8 (ganeshie8):

looks you're fine, till this point ?

4 years ago
ganeshie8 (ganeshie8):

yup, it doesnt matter

4 years ago
OpenStudy (anonymous):

okay~ sounds good~

4 years ago
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)

4 years ago
ganeshie8 (ganeshie8):

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

4 years ago
ganeshie8 (ganeshie8):

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

4 years ago
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

4 years ago
OpenStudy (anonymous):

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

4 years ago