Results 1 to 7 of 7

Math Help - Partial derivatives, word problem help

  1. #1
    Super Member
    Joined
    Sep 2008
    Posts
    613

    Partial derivatives, word problem help

    Heat is being conducted radially through a cylindrical pipe. The temperature at a radius r is T(r). In Cartesian co-ordinates,  r = \sqrt{(x^{2}+ y^{2}})

    show that  \displaystyle \frac{\partial T}{\partial x} = \frac{x}{r} \frac{dT}{dr}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2009
    From
    United States
    Posts
    676
    Thanks
    19
    T(r)=[\math]T(\sqrt{x^2+y^2))

    You have to differentiate r with restpect to x, this is just \frac{\partial r}{\partial x}=\frac{2x}{2\sqrt{x^2+y^2}}=\frac{x}{r}.

    So \frac{\partial T}{\partial x}=\frac{\partial r}{\partial x}\frac{dT}{dr}=\frac{x}{r}\frac{dT}{dr}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Sep 2008
    Posts
    613
    Quote Originally Posted by adkinsjr View Post
    <b>T(r)=[\math]T(\sqrt{x^2+y^2))</b>

    You have to differentiate r with restpect to x, this is just \frac{\partial r}{\partial x}=\frac{2x}{2\sqrt{x^2+y^2}}=\frac{x}{r}.

    So \frac{\partial T}{\partial x}=\frac{\partial r}{\partial x}\frac{dT}{dr}=\frac{x}{r}\frac{dT}{dr}
    Thank you so much, but I dont understand what the first expression is and how you got it?

    your saying  T(r) = T(\sqrt{ x^{2} + y^{2}})  ?

    How comes?


    Also for \frac{\partial r}{\partial x} , you took the partial derivative of r with respect to x, from the equation  r = \sqrt{(x^{2} + y^{2}) ?

    so you get this expression ;  \frac{2x}{2\sqrt{x^2+y^2}}
    but how does that equal the expression;  \frac{x}{r} ?

    thank you.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Sep 2008
    Posts
    613
    edit; I actually understand how you got the first expression, thanks, could please explain the rest?

    Thank you
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jun 2009
    From
    United States
    Posts
    676
    Thanks
    19
    Thank you so much, but I dont understand what the first expression is and how you got it?

    your saying  T(r) = T(\sqrt{ x^{2} + y^{2}})  ?

    How comes?
    All I'm doing is substituting the equation r=\sqrt{x^2+y^2} into the expression T(r) so you can see that T is a composite function of x and y, and therefore the partial of T with respect to x can be calculated using a chain rule.

    Also for \frac{\partial r}{\partial x} , you took the partial derivative of r with respect to x, from the equation  r = \sqrt{(x^{2} + y^{2}) ?

    so you get this expression ;  \frac{2x}{2\sqrt{x^2+y^2}}
    but how does that equal the expression;  \frac{x}{r} ?

    thank you.
    r=\sqrt{x^2+y^2}

    The derivative of a function like \sqrt{u} is \frac{u'}{2\sqrt{u}}.

    You have to apply the chain rule to find the partial of r with respect to x.

    \frac{\partial r}{\partial x}=\frac{1}{2\sqrt{x^2+y^2}}\frac{\partial}{\parti  al x}(x^2+y^2)=\frac{2x}{2\sqrt{x^2+y^2}}

    Remember that r=\sqrt{x^2+y^2} so just substitute that into the partial derivative for r with respect to x, and you get \frac{x}{r}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Sep 2008
    Posts
    613
    Thank you , I understand your method now.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Sep 2008
    Posts
    613
    sorry just one question, how does  \frac{dT}{dr} come into the equation ? It seems like its just been put there?

    Thank you
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Partial Derivatives Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 3rd 2010, 04:50 AM
  2. Partial Derivatives Problem
    Posted in the Calculus Forum
    Replies: 6
    Last Post: November 28th 2010, 03:31 AM
  3. Partial derivatives problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 21st 2010, 02:58 PM
  4. Partial Derivatives problem.
    Posted in the Calculus Forum
    Replies: 0
    Last Post: April 20th 2008, 08:01 PM
  5. Word problem involving partial derivatives
    Posted in the Calculus Forum
    Replies: 0
    Last Post: March 4th 2008, 05:04 PM

Search Tags


/mathhelpforum @mathhelpforum