calculus

Show that the curves (y= √2sinx) and (y=√2cosx) intersect at right angles at a certain point with 0<x<π/2

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asked by Imad
  1. let's find their intersection points
    √2sinx = √2cosx
    sinx/cosx = 1
    tanx = 1
    x = π/4 in the given interval

    for y = √2sinx
    dy/dx = √2cosx
    for x= π/4
    dy/dx = √2sinπ/4 = √2(1/√2) = 1

    for y = √2cosx
    dy/dx = -√2sinx
    at x=π/4
    dy/dx = -√2(1/√2) = - 1

    since the slopes are negative reciprocals of each other when x = π/4, the tangents are perpendicular at that point.

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    posted by Reiny

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