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4 November, 05:17

Suppose a curve is defined by the equation (x + y) 2 = 1. what is the equation of the line tangent to the curve at (3, - 2) ?

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Answers (2)
  1. 4 November, 07:07
    0
    (x + y) ^2 = 1

    Take the derivative of both sides wrt x.

    d/dx (x + y) ^2 = 0

    d/dx (x^2 + 2xy + y^2) = 0

    2x + 2y + 2x*dy/dx + 2y*dy/dx = 0

    dy/dx (2x + 2y) = - 2x - 2y

    dy/dx = (-x - y) / (x + y)

    dy/dx = - (x + y) / (x + y) = - 1

    So anywhere on the curve (x + y) ^2 will have a tangent of - 1

    Eqn of tangent: y + 2 = - (x - 3)

    y = 3 - x - 2

    y = 1 - x
  2. 4 November, 08:44
    0
    Don't you mean (x + y) ^2 = 1?

    Take the derivative with respect to x of both sides of this equation:

    2 (x + y (dy/dx) = 0. Then y (dy/dx) = - x, and (dy/dx) = - x/y.

    Thus, the slope of the tangent line to this curve at (3, - 2) is m = - 3 / (-2), or 3/2.

    The eqn of the T. L. is then y - (-2) = (3/2) (x-3) (answer)
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