Consider the following.x = cos(2t),y = cos(t),0 < t < π

Consider the following.
x = cos(2t)
y = cos(t)
0 < t < π

x = cos(2t)
y = cos(t)
0 < t < π
(a) Find the following.
dy/dx = 1/4 sect
d^2y/dx^2= ? I get 1/16 sec^3(t)(wrong answer)
(b) For which values of t is the curve concave upward???

dy/dx = (dy/dt) / (dx/dt) = -sin(t) / (-2*sin(2t)) =

(1/2)*sin(t) / (2*sin(t)*cos(t)) = (1/4)*sec(t). So far, so good.

Next, d^2y/dx^2 = d/dx(dy/dx) = d/dt(dy/dx) / (dx/dt) =

d/dt((1/4)*sec(t)) / (-2*sin(2t)) =

(1/4)*(sec(t)*tan(t)) / (-4*sin(t)*cos(t)) = (-1/16)*sec^3(t),

so I think you are just missing a minus sign.

The curve will be concave upward when (-1/16)*sec^3(t) > 0, i.e., when

sec^3(t) < 0 ----> sec(t) < 0 ----> cos(t) < 0, which on (0, pi) is just

the interval (pi/2, pi).