2. For the curve $c(t)=\left(\cos ^{3} t, \sin ^{3} t\right)$.
(a) Find the equation of the tangent line to the curve when $t=\frac{\pi}{4}$
(b) Find the speed at $t=\frac{\pi}{4}$.
(c) Find the length of the curve for $0 \leq t \leq \frac{\pi}{2}$.

Step 1 ：Let's denote the curve as \(c(t) = (\cos^3 t, \sin^3 t)\).

Step 2 ：We need to find the derivative of \(c(t)\) to get the slope of the tangent line. The derivative of \(c(t)\) is \(c'(t) = (-3\sin(t)\cos^2(t), 3\sin^2(t)\cos(t))\).

Step 3 ：Let's evaluate \(c(t)\) and \(c'(t)\) at \(t = \frac{\pi}{4}\). We get \(c(\frac{\pi}{4}) = (\frac{\sqrt{2}}{4}, \frac{\sqrt{2}}{4})\) and \(c'(\frac{\pi}{4}) = (-\frac{3\sqrt{2}}{4}, \frac{3\sqrt{2}}{4})\).

Step 4 ：The slope of the tangent line is the ratio of the y-coordinate to the x-coordinate of \(c'(\frac{\pi}{4})\), which is -1.

Step 5 ：The equation of the tangent line is given by \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the point on the curve at \(t = \frac{\pi}{4}\) and \(m\) is the slope of the tangent line. Substituting the values we get, \(y = -x + \frac{\sqrt{2}}{2}\).

Step 6 ：The speed at a given time \(t\) is the magnitude of the velocity vector at that time. The velocity vector is \(c'(t)\), so the speed at \(t = \frac{\pi}{4}\) is \(\frac{3}{2}\).

Step 7 ：The length of the curve from \(t = 0\) to \(t = \frac{\pi}{2}\) is given by the integral \(\int_{0}^{\frac{\pi}{2}} ||c'(t)|| dt\), where \(||c'(t)||\) is the magnitude of \(c'(t)\). The length of the curve is \(\frac{3}{2}\).

Step 8 ：Final Answer: (a) The equation of the tangent line to the curve when \(t = \frac{\pi}{4}\) is \(\boxed{x + y = \frac{\sqrt{2}}{2}}\). (b) The speed at \(t = \frac{\pi}{4}\) is \(\boxed{\frac{3}{2}}\). (c) The length of the curve for \(0 \leq t \leq \frac{\pi}{2}\) is \(\boxed{\frac{3}{2}}\).