x=2sin(t), y=1+cos(2t)
a) By chain rule, dy/dx = (dy/dt)/(dx/dt)
dy/dt = -2sin(2t), dx/dt= 2cos(t)
dy/dx= -sin(2t)/cos(t)
dy/dx=-2sin(t)cos(t)/cos(t)
dy/dx=-2sin(t)
when t = pi/3,
dy/dx= -sqrt(3)
b) dy/dx = -2sin(t)= -x
y= -x^2/2
k=2 as x=2sin(t) has max and min values at 2, -2
c) Draw a sketch
Sketch shows symmetrical quadratic with min value 0, max values of 2.
0<=f(x)<=2