math

how do i do this.


find the equation for the lines that are tangent and normal to the curve y= (square root of 2)(cosx)at the point (pi/4, 1)

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  1. y = √2 cosx
    I usually check if the given point is actually on the line
    LS = 1
    RS = √2 cos(pi/4) = √2(1/√2) = 1 = LS

    dy/dx = -√2(sinx)

    at (pi/4,1)
    dy/dx = -√2(1/√2) = -1

    So the slope of the tangent is -1
    and the slope of the normal is +1

    Take it from there,
    every student taking Calculus surely has to know how to find the equation of a line given the slope and a point.

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    posted by Reiny
  2. use product rule, then substitute value of x in dy/dx. the obtain then find equation of tangent y=mx+c.

    For normal,

    gradient of curve*gradient of normal=-1

    find gradient of normal from equation and once again obtain normal equation in y=mx+c form

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    posted by Salman

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