Math Is Fun Forum
You are not logged in.
- Topics: Active | Unanswered
Pages: 1
#1 2006-06-13 05:25:53
- Tredici
- Member
- Registered: 2005-12-12
- Posts: 28
Parametric Differentiation
Please help me find dy/dx. I've tried everything. I thought it'd be a case of simple quotient rule application, then chain rule by dividing dy/dt by dx/dt. Maybe it is, I may just be continuously slipping somewhere. Any help at all would be really appreciated. Thanks guys.
Offline
#2 2006-06-13 07:57:27
- John E. Franklin
- Member
- Registered: 2005-08-29
- Posts: 3,588
Re: Parametric Differentiation
I think you do (t*dy)/dx, so answer is -t for slope.
I guess I said that wrong.
I mean you might do a dt to yt/x, so perhaps
slope is D(yt/x))dt.
I don't know how to say it.
This may be wrong because I made it up based on a few examples like y = t and x = t.
And the example of y = t^2 and x = t.
But what if y = t^3 and x = t^3, slope is 1, yeah, still works. Maybe it's right, maybe not...
Last edited by John E. Franklin (2006-06-13 09:03:53)
igloo myrtilles fourmis
Offline
#3 2006-06-13 14:01:44
- George,Y
- Member
- Registered: 2006-03-12
- Posts: 1,379
Re: Parametric Differentiation
X'(y-Xβ)=0
Offline
#4 2006-06-13 14:14:28
- Zhylliolom
- Real Member
- Registered: 2005-09-05
- Posts: 412
Re: Parametric Differentiation
For parametric functions,
To solve your specific problem, we find both dy/dt and dx/dt, and then simply put them in the numerator and denominator, respectively.
Then
Offline
#5 2006-06-13 15:27:11
- John E. Franklin
- Member
- Registered: 2005-08-29
- Posts: 3,588
Re: Parametric Differentiation
Silly me! He's right, I checked a web page about it.
dy/dx is (dy/dt) / (dx/dt).
Oh well. I think Zhyl is using the quotient rule, but I can't remember it, square in denominator and some combinations subtracted in numerator?
igloo myrtilles fourmis
Offline
#6 2006-06-13 15:33:32
- Zhylliolom
- Real Member
- Registered: 2005-09-05
- Posts: 412
Re: Parametric Differentiation
Silly me! He's right, I checked a web page about it.
dy/dx is (dy/dt) / (dx/dt).
Oh well. I think Zhyl is using the quotient rule, but I can't remember it, square in denominator and some combinations subtracted in numerator?
Yes, the quotient rule was used to determine dy/dt and dx/dt, but dy/dx was determined by simply putting dy/dt over dx/dt and simplifying. Just to spark your memory,
Somewhat pointless edit: Assuming v ≠ 0.
Last edited by Zhylliolom (2006-06-13 15:34:44)
Offline
#7 2006-06-14 01:25:19
- John E. Franklin
- Member
- Registered: 2005-08-29
- Posts: 3,588
Re: Parametric Differentiation
Thanks, I finally got it memorized now, again, many, many years later.
igloo myrtilles fourmis
Offline
Pages: 1