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Differential problem

  1. Nov 29, 2008 #1
    1. The problem statement, all variables and given/known data

    A side of an equilateral triangle is measured to be 10 cm. Estimate the change in the area of the triangle when the side shrinks to 9.8 cm.

    2. Relevant equations

    3. The attempt at a solution

    [tex] A = 1/2*bh [/tex]

    [tex] x = 10, dx = 0.2[/tex]

    [tex]b = x/2, h = \sqrt{x^{2} - x^{2}/4}[/tex]

    [tex]1/2(x/2)(\sqrt{3x^{2}/4} = \sqrt{3x^{2}}b / 8 [/tex]

    [tex]dy = f'(x)dx = f'(10)(-0.2)[/tex]

    [tex]f'(x) = 8[(3x^{1/2})^{1/2} + 1/2(3x^{2})^{-1/2})(6b)(b)] / 64[/tex]

    [tex] = 8(\sqrt{3x^{2}} + 3x^{2} / \sqrt{3x^{2}) / 64[/tex]

    [tex] = 8(6x^{2} / \sqrt{3x^{2}}) / 64 = 48x^{2}/\sqrt{3x^{2}} / 64 = 48x^{2} / 64\sqrt{3x^{2}}[/tex]

    [tex]dy = (48(10)^{2} / 64\sqrt{3(10)^{2}})(-0.2) = -0.866[/tex]

    [tex]Delta A = f(x + Delta x) - f(x)[/tex]

    [tex]= f(10-0.2) - f(10)[/tex]

    [tex]= f(9.8) - f(10)[/tex]

    [tex]= 20.793 - 21.651[/tex]

    [tex]= -.857[/tex]

    Answer Key: -1.732
  2. jcsd
  3. Nov 29, 2008 #2


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    Homework Helper

    You seem to be using 'b' to mean both the base of the triangle (which is actually x), and the base of right triangle you get by splitting the equilateral triangle in two. You should have gotten that the area is A=(1/2)*x*sqrt(3x^2/4). For future reference, you can make life a lot easier if you simplify expressions before proceeding onwards. Like you could simplify the area to A=sqrt(3)*x^2/4. Doesn't that make it easier?
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