Results 1 to 5 of 5

Math Help - temperature problem

  1. #1
    Junior Member pirateboy's Avatar
    Joined
    Jul 2010
    From
    Portland, OR
    Posts
    34

    temperature problem

    Problem Statement

    The temperature, T, of a cooling fin of a finite length, L, satisfies
    \frac{d^2T}{dx^2} = -k^2\left( T_L-T \right),\quad 0\leq x\leqL,\quad T(0) = 0,\quad T(L) = T_L
    where k, T_0, and T_L are constants


    a) Find the complimentary function T_c(x), stating your answer in terms of unknown constants, c_1 and c_2


    b)Complete the following identity
    \sinh(kl-kx)=


    c) write down your starting point for the particular solution, T_p(x)


    d) find T_p(x) and the coefficients  c_1, and c_1 and hence show that hte solution of the above problem is
    T(x) = T_L + (T_0 - T_L)\frac{\sinh(k(L-x))}{\sinh(kL)}

    My attempt

    a)
    The first thing I do here (which may be where I'm going wrong) is that I rearrange the equation. I take

    T''(x) = -k^2(T_L-T)

    T''(x) -k^2T(x) = -k^2T_L

    characteristic equation:
    r^2 - k^2 = 0

    \therefore r = \pm k

    So

    T_c(x) = c_1 \cosh(kx) + c_2\sinh(kx) + c_3 k^2T_L

    With this, I jump straight to

    T_c''(x) = k^2c_1\cosh(kx) +k^2c_2\sinh(kx)

    I can then verify that T''(x) -k^2T(x) + c_3k^2T_L = -k^2T_L is a solution, with c_3=-1. But this helps me very little when I try to solve for my initial values and helps me less when I try to complete part (d).

    Grinding through a bit further I can solve for c_1 with initial values with

    T(0) = T_0 = c_1\cosh(k\cdot 0) + c_2\sinh(k\cdot 0) -k^2T_L

     c_1 - k^2T_L = T_0

    \therefore c_1 = T_0 + k^2T_L

    So my new T(x)

    T(x) = (T_0 + k^2T_L)\cosh(kx)+c_2\sinh(kx)-k^2T_L

    But then attempting to solve for c_2 I get stuck. I end up with

    T(L) = T_L = (T_0+k^2T_L)\cosh(kL)+c_2\sinh(kL)-k^2T_L


    And I think I'm stuck.
    Am I missing some simple algebra trick here, or have I gone down the wrong road?



    b) \sinh(kL-kx) = \sinh(kL)\cosh(kx)-\cosh(kL)\sinh(kx)
    I'm pretty sure that's right, at least.

    c)???

    d)???


    So, I'm thinking I'm starting this wrong, but I can't think of another way to attempt this. Any help would be greatly appreciated.
    Last edited by pirateboy; August 3rd 2011 at 04:21 PM. Reason: oops forgot something.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Feb 2011
    Posts
    4

    Re: temperature problem

    Everything in your equation for c_2 is constant, so you should just be able to rearrange it to get c_2 on its own.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member pirateboy's Avatar
    Joined
    Jul 2010
    From
    Portland, OR
    Posts
    34

    Re: temperature problem

    Quote Originally Posted by TheBigH View Post
    Everything in your equation for c_2 is constant, so you should just be able to rearrange it to get c_2 on its own.
    Ah, right, but solving that I get a monster. How do I get that to look like (d), is the question.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member pirateboy's Avatar
    Joined
    Jul 2010
    From
    Portland, OR
    Posts
    34

    Re: temperature problem

    ok, got it. I was starting off wrong. If anyone is interested, I'll post the solution when I get a chance.

    hint:

    T(x) = c_1\cosh(kx) + c_2\sinh(kx) +c_3 T_L

    c_3 = 1

    The rest is algebra.
    Last edited by Ackbeet; August 3rd 2011 at 06:02 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,234
    Thanks
    1795

    Re: temperature problem

    Quote Originally Posted by pirateboy View Post
    Problem Statement

    The temperature, T, of a cooling fin of a finite length, L, satisfies
    \frac{d^2T}{dx^2} = -k^2\left( T_L-T \right),\quad 0\leq x\leqL,\quad T(0) = 0,\quad T(L) = T_L
    where k, T_0, and T_L are constants


    a) Find the complimentary function T_c(x), stating your answer in terms of unknown constants, c_1 and c_2


    b)Complete the following identity
    \sinh(kl-kx)=


    c) write down your starting point for the particular solution, T_p(x)


    d) find T_p(x) and the coefficients  c_1, and c_1 and hence show that hte solution of the above problem is
    T(x) = T_L + (T_0 - T_L)\frac{\sinh(k(L-x))}{\sinh(kL)}

    My attempt

    a)
    The first thing I do here (which may be where I'm going wrong) is that I rearrange the equation. I take

    T''(x) = -k^2(T_L-T)

    T''(x) -k^2T(x) = -k^2T_L

    characteristic equation:
    r^2 - k^2 = 0

    \therefore r = \pm k

    So

    T_c(x) = c_1 \cosh(kx) + c_2\sinh(kx) + c_3 k^2T_L

    With this, I jump straight to

    T_c''(x) = k^2c_1\cosh(kx) +k^2c_2\sinh(kx)

    I can then verify that T''(x) -k^2T(x) + c_3k^2T_L = -k^2T_L is a solution, with c_3=-1. But this helps me very little when I try to solve for my initial values and helps me less when I try to complete part (d).

    Grinding through a bit further I can solve for c_1 with initial values with

    T(0) = T_0 = c_1\cosh(k\cdot 0) + c_2\sinh(k\cdot 0) -k^2T_L

     c_1 - k^2T_L = T_0

    \therefore c_1 = T_0 + k^2T_L

    So my new T(x)

    T(x) = (T_0 + k^2T_L)\cosh(kx)+c_2\sinh(kx)-k^2T_L

    But then attempting to solve for c_2 I get stuck. I end up with

    T(L) = T_L = (T_0+k^2T_L)\cosh(kL)+c_2\sinh(kL)-k^2T_L
    That's a simple linear equation:
    c_2\sinh(kL)= T_L+ k^2T- (T_0+ k^2T_L)\cosh(kL)
    c_2= \frac{T_L+ k^2T- (T_0+ k^2T_L)\cosh(kL)}{\sinh(kL)}


    And I think I'm stuck.
    Am I missing some simple algebra trick here, or have I gone down the wrong road?



    b) \sinh(kL-kx) = \sinh(kL)\cosh(kx)-\cosh(kL)\sinh(kx)
    I'm pretty sure that's right, at least.

    c)???

    d)???


    So, I'm thinking I'm starting this wrong, but I can't think of another way to attempt this. Any help would be greatly appreciated.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. temperature problem
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: September 5th 2011, 03:29 AM
  2. Temperature Problem
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: January 30th 2011, 04:07 PM
  3. Corpse Temperature Problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 5th 2010, 03:21 PM
  4. Temperature problem
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: January 29th 2009, 09:43 PM
  5. Cooling/temperature Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 27th 2007, 02:53 PM

Search Tags


/mathhelpforum @mathhelpforum