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? asked in Science & MathematicsMathematics · 2 weeks ago

Finding the equation of the normal to the curve?

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5 Answers

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  • 2 weeks ago

    y=4cos^-1(x/2)

    =>

    cos(y/4)=x/2

    =>

    -sin(y/4)(y'/4)=1/2

    =>

    y'=-2csc(y/4)

    At x=sqr(3),

    y(sqr(3))=4cos^-1(sqr(3)/2)=4(pi/6)=2pi/3

    y'(sqr(3))=-2csc(pi/6)=-4

    =>

    the slope of the normal line=1/4. Thus

    the equation of the normal line is

    y-2pi/3=(x-sqr(3))/4

    =>

    3x-12y+(8pi-3sqr(3))=0

  • ?
    Lv 7
    2 weeks ago

    y = 4arcos(x/2)

    x/2 = cos(y/4)

    dx/dy = -(1/2)sin(y/4)

    sin(y/4) = √[1 – cos^2(y/4)]

    dx/dy = -(1/2)√[1 – x^2/4] = -(1/4)√[4 – x^2]

    dy/dx = -4/√(4 – x^2)

    When x = √(3), y = 4arcos[√(3)/2] = 4*π/6

    Also, dy/dx = -4 = m the slope of the tangent there

    Slope of the normal, -1/m = 1/4

    If y = mx + c is the normal

    2π/3 = (1/4)√(3) + c

    y = x/4 + 2π/3 - √(3)/4

    Bryce found this result first

  • Bryce
    Lv 7
    2 weeks ago

    y'= -4/√(4 - x²)

    At x= √3, y'= -4

    -1/m= 1/4

    y=(1/4)(x - √3) + 2π/3

  • Anonymous
    2 weeks ago

    No, thank you - not today, sweetie. I do not help people appear smarter than they actually are.

  • ?
    Lv 7
    2 weeks ago

    First, find the slope of the curve at that point. Can you do that part?

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