Calculus

posted by Caleb

Use implicit differentiation to find an equation
of the tangent line to the curve
y^2 = x^3 (26 − x)
at the point (1, 5).

  1. Damon

    2 y dy = x^3(-dx)+3x^2dx(26-x)

    2 y dy = -x^3 dx + 78 x^2 dx - 3 x^3 dx

    dy/dx = (-4 x^3 + 78 x^2)/ (2 y)

    dy/dx = m = (39 x^2 - 2 x^3)/y
    at (1,5)
    m = (39-2)/5 = 37/5

    so y = (37/5) x + b
    5 = 37/5 + b
    b = 25/5 - 37/5 = -12/5
    so
    y = (37/5) x - 12/5
    5 y = 37 x - 12

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