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

Multivariable functions - chain rule

  • Thread starter hex.halo
  • Start date
  • #1
13
0

Homework Statement



Since both my questions are on the same topic, i'll throw them both in here

1. Find dz/dt for z=(x^2)(t^2), x^2+3xt+2t^2=1

2. Show that if u=xy, v=xy and z=f(u,v) then:
x.dz/dx-y.dz/dy=(x-y)dz/dv

Homework Equations





The Attempt at a Solution



1. I only know how to do this when I have something like:

z=x+y, x=t+..., y=t+...

2. I realise that after I get the partial derivatives I just have to substitute back to prove this, but i'm not sure how to get the derivatives. I think it's the z=f(u,v) that has me lost
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
For 1. find dz/dt using the product rule. Then you realize you need dx/dt. Find this by implicitly differentiating the second expression. For the second one review the chain rule for partial derivatives. But it looks to me like the left hand side of the expression you want to prove evaluates to zero. Are you sure you transcribed that right?
 
  • #3
378
2
The second question seems wrong!
Is this from Glyn james and that#2 has two parts.. and is like at the bottom of the page on left side column?
One of them should be x+y or x-y
 
  • #4
13
0
You're right. It should be u=x+y

This still doesn't help me though. I may be wrong, but i've got:

dz/dx=dz/du.du/dx+dz/dv.dv/dx

dz/dy=dz/du.du/dy+dz/dv.dv/dy

as my partial fractions. But how do I get my dz/du and dz/dv when I have z=f(u,v) and not, for example, z=f(u,v)=2u+3v^2 - i'm lost without this little equation on the end like in previous problems i've done.

Also, I don't think i've understood your explanation of my first question properly. I tried:

dz/dt=dz/dx.dx/dt

dz/dx=2xt^2
dx/dt=(-3x-4t)/(2x+3t)

And therefore:

dz/dt=(-6(x^2)(t^2)-8xt^3)/(2x+3t)

but my textbook's telling me it should be:

4xt(x^2-2t^2)/(2x+3t)
 
  • #5
Defennder
Homework Helper
2,591
5
[tex]\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} \ + \ \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}[/tex]

Let [tex]y=t^2.[/tex]

Hence [tex]\frac{dz}{dt} = 2xt^2 \left( \frac{-3x-4t}{2x+3t} \right) + x^2(2t)[/tex]

Just simplify the above and you're done.

For the 2nd question, is it supposed to be u=xy, v=x+y? I can get the answer if so, but not if the values of u and v are switched as you have put it.
 
  • #6
13
0
My sheet's telling me it should be u=x+y, v=xy, but let's just say it's u=xy and v=x+y for the hell of it. How would you go about solving this problem now?
 
  • #7
Defennder
Homework Helper
2,591
5
I don't know if you have copied the question wrongly or if there is something wrong with my working, but here it is:

[tex]\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial x} \ + \ \frac{\partial z}{\partial v} \frac{\partial v}{\partial x} = \frac{\partial z}{\partial u} + y\frac{\partial z}{\partial v}[/tex]

[tex]\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial y} \ + \ \frac{\partial z}{\partial v} \frac{\partial v}{\partial y} \ = \ \frac{\partial z}{\partial u} \ + \ x \frac{\partial z}{\partial v}[/tex]

[tex]\frac{\partial u}{\partial x} = 1 \\ \frac{\partial u}{\partial y} = 1[/tex]
[tex]\frac{\partial v}{\partial x} = y \\ \frac{\partial v}{\partial y} = x[/tex]

Hence:

[tex]x\frac{\partial z}{\partial x} - y\frac{\partial z}{\partial y} = (x-y)\frac{\partial z}{\partial u}[/tex]
 

Related Threads for: Multivariable functions - chain rule

  • Last Post
Replies
1
Views
828
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
3
Views
887
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
1
Views
3K
Top