Module 4 \u2013 CIN Problem.docx - Module 4 Discussion Application CIN problem CIN a=2 x 2 y 2 =2 x 2 y First Differentiate 2 x 2 y dy dy =2 2 dx dx 2 x

Module 4 u2013 CIN Problem.docx - Module 4 Discussion...

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Module 4 – Discussion: Application CIN problem CIN #: a=2 x 2 + y 2 = 2 x + 2 y First, Differentiate 2 x + 2 y dy dx = 2 + 2 dy dx 2 x 2 = 2 dy dx 2 y dy dx 2 x 2 =( 2 2 ) dy dx 2 x 2 = dy dx Second, Set dy/dx =0, tangent are parallel to the x-axis when the derivative is 0 2 x 2 = 0 2 x = 2 x = 1 Three, Substitute x in the original formula. ( 1 ) 2 + y 2 = 2 ( 1 )+ 2 y 1 + y 2 = 2 + 2 y 1 + y 2 ( 2 + 2 y ) = 2 + 2 y −( 2 + 2 y ) y 2 2 y 1 = 0 Four, Solve for y using the quadratic equation. y = −(− 2 ) ± (− 2 ) 2 4 ( 1 )(− 1 ) 2 ( 1 )
¿ 2 ± 4 + 4 2 y = 2 ± 2 2 y = 1 + 2 y = 1 2 y ≈ 2.4242 y≈ 0.4142 The horizontal tangent points are at points, (1, 2.41) and (1,-0.4141).

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