Finding the equation of the normal to the curve?

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

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

y'= -4/√(4 - x²)

At x= √3, y'= -4

-1/m= 1/4

y=(1/4)(x - √3) + 2π/3

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

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