Find dy/dx and d2y/dx2. x = cos(2t), y = cos(t), 0 < t < π dy dx = Correct: Your answer is correct. d2y dx2 = Correct: Your answer is correct. For which values of t is the curve concave upward? (Enter your answer using interval notation.) Incorrect: Your answer is incorrect.

By the chain rule, the first derivative is[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}[/tex]We have[tex]x=\cos2t\implies\dfrac{\mathrm dx}{\mathrm dt}=-2\sin2t[/tex][tex]y=\cos t\implies\dfrac{\mathrm dy}{\mathrm dt}=-2\sin t[/tex]so that[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{-2\sin t}{-2\sin2t}=\dfrac{\sin t}{2\sin t\cos t}=\boxed{\dfrac{\sec t}2}[/tex]The second derivative is[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{\mathrm dy}{\mathrm dx}\right][/tex]where [tex]\dfrac{\mathrm dy}{\mathrm dx}[/tex] is a function of [tex]t[/tex]; denote it by [tex]f(t)[/tex]. Then by the chain rule,[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\mathrm df}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}[/tex]Since[tex]f=\dfrac{\sec t}2\implies\dfrac{\mathrm df}{\mathrm dt}=\dfrac{\sec t\tan t}2[/tex]we end up with[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac{\sec t\tan t}2}{-2\sin2t}=\boxed{-\dfrac{\sec^3t}8}[/tex]The curve is concave upward whenever the second derivative is positive. We have this for[tex]-\dfrac{\sec^3t}8>0\implies\sec^3t=\dfrac1{\cos^3t}<0\implies\cos^3t<0[/tex]We have [tex]\cos t=0[/tex] for [tex]t=\dfrac{n\pi}2[/tex] where [tex]n[/tex] is any integer; for the given interval, this happens at [tex]t=\dfrac\pi2[/tex]. For [tex]0<t<\dfrac\pi2[/tex], we have [tex]\cos t>0[/tex], while for [tex]\boxed{\dfrac\pi2<t<\pi}[/tex], we have [tex]\cos t<0[/tex]; it's the latter that we care about, since over this interval we have [tex]\sec^3t<0[/tex].