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Consider the curve given by x^ 2 +sin(xy)+3y^ 2 =C, where Cis a constant. The point (1, 1) lies on this curve. Use the tangent line approximation to approximate the y-coordinate when x = 1.01 .

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Sarah

  1. dy/dx = - (y*cos(xy)+2x)/(x*cos(xy)+6y)
    At (1,1), then y' = -(cos(1)+2)/(cos(1)+6) ≈ -0.388
    Now, using the tangent line as an approximation, we have
    y(1+∆x) = y(1) + y' ∆x = 1 + (-0.388)(.01) = 0.996

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    oobleck

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