Question

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)~

Answer 1

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

Answer 2

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

Answer 3

solve :)

Answer 4

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

Answer 5

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

Answer 6

the latter, I believe ^^

Answer 7

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

Answer 8

TToTT can't get the variables separated~

Answer 9

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

Answer 10

hint : homogeneous equation

Answer 11

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

Answer 12

hmmm ok then ^^ sounds good then XD

Answer 13

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:
\[2(\frac{du}{dx}x+u)=\frac{1}{u}+u\]

Answer 14

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

Answer 15

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".

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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.

Answer 20

Show me what you got and your attempt at it.

Answer 21

the integration?

Answer 22

Yep! What does the integral look like?

Answer 23

wait hold on ...

Answer 24

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

Answer 25

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

Answer 26

OoO

Answer 27

before integrating?

Answer 28

Of course!

Answer 29

x/dx (du) +u

Answer 30

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

Answer 31

yes~

Answer 32

:|

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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!

Answer 43

y=x-c?

Answer 44

^ that's what I got for my final answer

Answer 45

\[\frac{ 1 }{ 2} lnx=-\frac{ 1 }{ 2}\ln(1-u^2)+\frac{ 1 }{ 2}lnC \]
picked that constant since it's completely arbitrary.

Nope, that's not how logs work!

Answer 46

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

Answer 47

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

Answer 48

you forgot the negative sign on the lnc ^^

Answer 49

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

Answer 50

hmmmm ok~

Answer 51

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

Answer 52

Or did you get a positive c?

Answer 53

@Kainui ???

Answer 54

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

Answer 55

@wio did you work this out?

Answer 56

which one?

Answer 57

the one that was first mentioned ^^

Answer 58

what is the equation?

Answer 59

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

Answer 60

but what was the original equation?

Answer 61

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

Answer 62

Oh ok.

Answer 63

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

Answer 64

y/x

Answer 65

well, undo the substitution I guess.

Answer 66

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

Answer 67

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

Answer 68

did you already integrate?

Answer 69

@ganeshie8 please come and help with this last portion~

Answer 70

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

Answer 71

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

Answer 72

You need initial conditions to solve for C.

Answer 73

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,...

Answer 74

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

Answer 75

oh! right~ the plus and minus~

Answer 76

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

Answer 77

it changes how the final equaiton wud *look* like

Answer 78

where do u wana put c ?

Answer 79

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

Answer 80

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

Answer 81

looks you're fine, till this point ?

Answer 82

yup, it doesnt matter

Answer 83

okay~ sounds good~

Answer 84

(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)

Answer 85

or,

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

Answer 86

the relation between ur constant and the earlier kainui's constant is :

C = 1/k

Answer 87

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

Answer 88

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

Answer 89

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