Question

Given y=sin^2(x). Find the equation for the tangent an normal lines to the graph where x=pi/6

Answer 1

first find the slope
which is = dy/dx
dy/dx=2*sinx*cosx
at x=pi/6 dy/dx=2 *sin (pi/6)*cos(pi/6)=2*(1/2)*(sqrt3)/2=(sqrt3)/2
now at x= pi/6 y= sin^2(pi/6)=(1/2)^2=1/4
so normal line pass through (pi/6,1/4) and slope =(sqrt3)/2

Answer 2

equation of a line
y=mx+c
m=slope and c=constant

Answer 3

I see my mistake, I took the derivative wrong

Answer 4

okkkkk :)

Answer 5

Ok so i got y=sqrt(3)/2(x-pi/6)+1/4, and for the normal lines its just perpendicular to the tangent line right