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Partial derivatives

  • Thread starter Lyn
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  • #1
Lyn
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I am stuck on the question,

'If f is a twice differentiable function of a single variable, find f = z(sqrt(x^2+y^2)) that satisfies d^2z/dx^2 +d^2z/dy^2 = x^2 +y^2

(ALL d's ARE MEANT TO BE PARTIAL DERIVATIVES)

i know dz/dx=(dz/du).(du/dx)
i can find du/dx but i don't know how to find dz/du
 

Answers and Replies

  • #2
Hootenanny
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Welcome to Physics Forums Lyn.

HINT: Let u = (x2 + y2)1/2

HINT (2): Using you definition of dz/dx above, can you write down an expression for d2z/dx2?
 
  • #3
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Isn't this suppose to be [itex]z=f(\sqrt{x^2+y^2})[/itex] In which f is a twice differentiable function?
 
  • #4
Lyn
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Welcome to Physics Forums Lyn.

HINT: Let u = (x2 + y2)1/2

HINT (2): Using you definition of dz/dx above, can you write down an expression for d2z/dx2?
Thank you for replying.

Yes I have set u=sqrt(x^2+y^2)

Would d^2z/dx^2=d/dx.(dz/dx)?

So far i have dz/dx=(x/u).dz/du but i am unsure of how to find dz/du so i can't carry on my calculation
 
  • #5
Hootenanny
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Thank you for replying.

Yes I have set u=sqrt(x^2+y^2)

Would d^2z/dx^2=d/dx.(dz/dx)?

So far i have dz/dx=(x/u).dz/du but i am unsure of how to find dz/du so i can't carry on my calculation
You don't need to find dz/du. Let's just start working though it. We have

[tex]\frac{dz}{dx} = \frac{dz}{du}\frac{du}{dx}[/tex]

Hence,

[tex]\begin{aligned}
\frac{d^2 z}{dx^2} & = \frac{d}{dx}\left(\frac{dz}{dx}\right) \\
& =\frac{d}{dx}\left(\frac{dz}{du}\frac{du}{dx}\right)
\end{aligned}[/tex]

Can you take the next couple of steps?
 
  • #6
Lyn
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You don't need to find dz/du. Let's just start working though it. We have

[tex]\frac{dz}{dx} = \frac{dz}{du}\frac{du}{dx}[/tex]

Hence,

[tex]\begin{aligned}
\frac{d^2 z}{dx^2} & = \frac{d}{dx}\left(\frac{dz}{dx}\right) \\
& =\frac{d}{dx}\left(\frac{dz}{du}\frac{du}{dx}\right)
\end{aligned}[/tex]

Can you take the next couple of steps?
i'm slightly confused. how do i find dz/dx because i don't have an equation with z equal to an equation with x variables?
 
  • #7
Hootenanny
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i'm slightly confused. how do i find dz/dx because i don't have an equation with z equal to an equation with x variables?
It doesn't matter, if you work the problem through you'll find that you don't actually need to know the explicit form of z=z(x,y).

Simply expand the expression in the final line of my previous post.
 
  • #8
Lyn
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ok i tried to expand your expression, not really sure but i get

d^2z/dx^2=d/dx(dz/du.du/dx)
=d/dx((dz/dx.dx/du+dz/dy.dy/du).du/dx)

then i could change dz/dx again but then i get a dz/du again and then i keep going round in circles
 
  • #9
Hootenanny
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ok i tried to expand your expression, not really sure but i get

d^2z/dx^2=d/dx(dz/du.du/dx)
=d/dx((dz/dx.dx/du+dz/dy.dy/du).du/dx)

then i could change dz/dx again but then i get a dz/du again and then i keep going round in circles
Not entirely sure what you're doing there. Let me do it for dz/dx,

[tex]\begin{aligned}
\frac{d^2 z}{dx^2} & = \frac{d}{dx}\left(\frac{dz}{dx}\right) \\
& =\frac{d}{dx}\left(\frac{dz}{du}\frac{du}{dx}\right) \\
& = \frac{du}{dx}\frac{d}{dx}\frac{dz}{du} + \frac{dz}{du}\frac{d}{dx}\frac{du}{dx} \\
& = \frac{du}{dx}\left(\frac{d}{du}\frac{dz}{du}\right)\frac{du}{dx} + \frac{dz}{du}\frac{d^2u}{dx^2} \\
& = \left(\frac{du}{dx}\right)^2\frac{d^2 z}{du^2} + \frac{dz}{du}\frac{d^2u}{dx^2}
\end{aligned}[/tex]

Do you follow?

Can you now do the same for dz/dy?
 
  • #10
Lyn
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Not entirely sure what you're doing there. Let me do it for dz/dx,

[tex]\begin{aligned}
\frac{d^2 z}{dx^2} & = \frac{d}{dx}\left(\frac{dz}{dx}\right) \\
& =\frac{d}{dx}\left(\frac{dz}{du}\frac{du}{dx}\right) \\
& = \frac{du}{dx}\frac{d}{dx}\frac{dz}{du} + \frac{dz}{du}\frac{d}{dx}\frac{du}{dx} \\
& = \frac{du}{dx}\left(\frac{d}{du}\frac{dz}{du}\right)\frac{du}{dx} + \frac{dz}{du}\frac{d^2u}{dx^2} \\
& = \left(\frac{du}{dx}\right)^2\frac{d^2 z}{du^2} + \frac{dz}{du}\frac{d^2u}{dx^2}
\end{aligned}[/tex]

Do you follow?

Can you now do the same for dz/dy?
I haven't actually seen expanding like that before, like i don't really get your third line of working and where the addition sign came from but i can follow the rest.

I have done the same for dz/dy

d^2z/dy^2=d/dy(dz/dy)=d/dy(dz/du.du/dy)=d/dy.dz/du.du/dy+dz/du.d/dy.du/dy
=d^2z/du^2(du/dy)^2+dz/du.d^2u/dy^2

but there is still the problem that i can't work out dz/du. i have a feeling i'm misunderstanding or missing a really important point but i don't know what?
 
  • #11
Hootenanny
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I haven't actually seen expanding like that before, like i don't really get your third line of working and where the addition sign came from but i can follow the rest.
In going from the second line to the third, I have simply used the product rule since both dz/du and du/dx are functions of x. Does that make sense?
I have done the same for dz/dy

d^2z/dy^2=d/dy(dz/dy)=d/dy(dz/du.du/dy)=d/dy.dz/du.du/dy+dz/du.d/dy.du/dy
=d^2z/du^2(du/dy)^2+dz/du.d^2u/dy^2

but there is still the problem that i can't work out dz/du. i have a feeling i'm misunderstanding or missing a really important point but i don't know what?
Good. Don't worry about dz/du, we'll come to that later. So now we have,

[tex]\frac{d^2 z}{dx^2} + \frac{d^2 z}{dx^2} = \left[\left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2\right]\frac{d^2 z}{d u^2} + \left[\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\right]\frac{d z}{du}[/tex]

Do you agree?

Note that I have change the ordinary differentials du/dx and du/dx to partial differentials since u is a function of two variables, u=u(x,y). Technically I should have used partial differentials from the start.
 
  • #12
Lyn
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In going from the second line to the third, I have simply used the product rule since both dz/du and du/dx are functions of x. Does that make sense?

Good. Don't worry about dz/du, we'll come to that later. So now we have,

[tex]\frac{d^2 z}{dx^2} + \frac{d^2 z}{dx^2} = \left[\left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2\right]\frac{d^2 z}{d u^2} + \left[\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\right]\frac{d z}{du}[/tex]

Do you agree?

Note that I have change the ordinary differentials du/dx and du/dx to partial differentials since u is a function of two variables, u=u(x,y). Technically I should have used partial differentials from the start.

Yes thank you, up to here i understand. now i just don't see how i can carry on the calculation to get it equal to x^2+y^2 i.e. u^2
 
  • #13
Hootenanny
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Yes thank you, up to here i understand. now i just don't see how i can carry on the calculation to get it equal to x^2+y^2 i.e. u^2
You are told in the question that this quantity is equal to u2. So we now have,

[tex]\left[\left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2\right]\frac{d^2 z}{d u^2} + \left[\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\right]\frac{d z}{du} = u^2[/tex]

Now it's time for you to do a little work. All that's left for you to do is evaluate the partial derivatives and solve the resulting ODE.
 
  • #14
Lyn
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ok i think i'm getting there. thenk you so much for all your help, i'll try finish it off now
 

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