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Differential of triangles and anlges

  1. Oct 25, 2008 #1
    Given cosine rule: L=[tex]\sqrt{(r_{1})^{2}+(r_{2})^{2}-2r_{1}r_{2}cosx}[/tex]

    Consider a triangle with side lengths measured at [tex]r_{1}=3, r_{2}=4[/tex], and included angle x=[tex]\pi/2[/tex], each measured accurate to within 1%. Write down the differential dL in terms of [tex]dr_{1}, dr_{2}[/tex] and [tex]dx[/tex], and use this to estimate the maximum possible percentage error in L.

    Any help? :S
  2. jcsd
  3. Oct 26, 2008 #2


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    Re: differentials

    What kind of help do you want? You haven't done anything at all. You are asked to write the differential dL. What is the derivative of L with respect to each of the variables? You might find it easier to use [itex]L^2= r_1^2+ r_2^2- 2r_1r_2 cos(x)[/itex] and find the differential from that.
  4. Oct 26, 2008 #3
    Re: differentials

    I got dL= 1/L (0.03(r1-r2cosx) + 0.04(r2-r1cosx) + 0.005pi(r1r2sinx)).. wasn't sure if that was right.
    Last edited: Oct 26, 2008
  5. Oct 26, 2008 #4
    Re: differentials

    My bad..
    I got [tex]|dL|\leq\ 1/L[0.03(r_{1}+r_{2}cosx)+0.04(r_{2}+r_{1}cosx)-0.005\pi(r_{1}r_{2}sinx)][/tex]

    If that's right, any hints on what I do next?
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