• Support PF! Buy your school textbooks, materials and every day products Here!

Partial derivative using differentials

  • Thread starter eprparadox
  • Start date
  • #1
138
2

Homework Statement


If [itex] xs^2 + yt^2 = 1 [/itex] and [itex] x^2s + y^2t = xy - 4 [/itex], find [itex] \frac{\partial x}{\partial s}, \frac{\partial x}{\partial t}, \frac{\partial y}{\partial s}, \frac{\partial y}{\partial t} [/itex], at (x, y, s, t) = (1, -3, 2, -1)


Homework Equations






The Attempt at a Solution


I took the differential of both equations and substituted in the values and I got:

[tex] 4ds + 4dx + 6dt + dy = 0 [/tex] and
[tex] ds + 7dx + 9dt + 5dy = 0 [/tex]

but now I don't know how do find the individual partial derivatives that are being asked for. For example, for [itex] \frac{\partial x}{\partial s} [/itex], I assumed t and y are constant and made dt and dy = 0 and tried to solve for ds and dx but that doesn't work.

Any ideas on how to proceed in finding these partials?
Thanks!
 

Answers and Replies

  • #2
95
13
##s## and ##t## don't happen to be independent, so that ## \frac{\partial s}{\partial t} = \frac{\partial t}{\partial s} = 0##? That would make things much easier.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,770
911
In order that this problem make any sense, s and t have to be independent so Quesadilla's comment is crucial. You should NOT assume y is constant since it is one of the dependent variables. But, taking the derivatives with respect to s you should assume t is constant (I would have said "treat t as a constant").

So from [itex]4ds+ 4dx+ 6dt+ dy= 0[/itex] and [itex]ds+ 7dx+ 9dt+ 5dy= 0[/itex] you can get
[tex]4\frac{\partial s}{\partial s}+ 4\frac{\partial x}{\partial s}+ 6\frac{\partial t}{\partial s}+ \frac{\partial y}{\partial s}= 4+ 4\frac{\partial x}{\partial s}+ \frac{\partial y}{\partial s}= 0[/tex]
and
[tex]\frac{\partial s}{\partial s}+ 7\frac{\partial x}{\partial s}+ 9\frac{\partial t}{\partial s}+ 5\frac{\partial y}{\partial s}= 1+ 7\frac{\partial x}{\partial s}+ 5\frac{\partial y}{\partial s}= 0[/tex]
Giving you two equations to solve for [itex]\partial x/\partial s[/itex] and [itex]\partial y/\partial[/itex].

Do the same thing to find the derivatives with respect to t.

Personally, I wouldn't have use "differentials". Just differentiating [itex]xs^2+ yt^2= 1[/itex] with respect to s (and treating t as a constant) gives
[tex]s^2\frac{\partial x}{\partial s}+ 2xs+ t^2\frac{\partial y}{\partial s}= 0[/tex]
and differentiating [itex]x^2s+ y^2t= xy- 4[/itex] with respect to s gives
[tex]2sx\frac{\partial x}{\partial s}+ x^2+ 2ty\frac{\partial y}{\partial s}= x\frac{\partial y}{\partial s}+ y\frac{\partial x}{\partial s}[/tex]
again giving two equations to solve for [itex]\partial x/\partial s[/itex] and [itex]\partial y/\partial s[/itex].
 
Last edited by a moderator:
  • #4
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,705
1,722
##s## and ##t## don't happen to be independent, so that ## \frac{\partial s}{\partial t} = \frac{\partial t}{\partial s} = 0##? That would make things much easier.
Why do you claim are s and t not independent? If we arbitrarily specify values of s and t we can then use the equations to find x and y. In fact, I would say we can take s and t as independent, but with x and y being functions of them: x = x(s,t) and y = y(s,t).
 
Last edited:
  • #5
95
13
Why do you claim are s and t not independent? If we arbitrarily specify values of s and t we can then use the equations to find x and y. In fact, I would say we can take s and t as independent, but with x and y being functions of them: x = x(s,t) and y = y(s,t).
If you mean that we should look to apply the implicit function theorem that sounds like an excellent idea!
 
  • #6
138
2
Great, thanks everyone!
 

Related Threads for: Partial derivative using differentials

Replies
4
Views
848
  • Last Post
Replies
4
Views
630
Replies
2
Views
960
Replies
3
Views
1K
Replies
0
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
1
Views
2K
Top