Find y''_xx for x = cos³t, y = sin³t

Find $y''_{xx}$ for $x = \cos^{3}t,\ y = \sin^{3}t$.

Step 1: Derivatives with respect to $t$

\[ \dot x = -3\cos^{2}t\sin t, \qquad \dot y = 3\sin^{2}t\cos t. \]

Step 2: First derivative

Almost everything cancels: \[y'_{x} = \frac{3\sin^{2}t\cos t}{-3\cos^{2}t\sin t} = -\frac{\sin t}{\cos t} = -\tan t.\]

Step 3: Differentiate with respect to $t$

\[ \bigl(y'_{x}\bigr)'_{t} = -\frac{1}{\cos^{2}t}. \]

Step 4: Divide by $\dot x$

The two minus signs cancel: $y''_{xx} = \frac{-1/\cos^{2}t}{-3\cos^{2}t\sin t} = \frac{1}{3\cos^{4}t\,\sin t}$.

Step 5: The same thing in $x$ and $y$

The curve is the astroid $x^{2/3}+y^{2/3} = 1$. Implicit differentiation gives

\[ \tfrac23x^{-1/3}+\tfrac23y^{-1/3}y'_{x} = 0 \;\Longrightarrow\; y'_{x} = -\frac{y^{1/3}}{x^{1/3}} = -\tan t, \]

and differentiating this quotient, then using $x^{2/3}+y^{2/3} = 1$ to collapse the numerator, gives the compact form

\[ y''_{xx} = \frac{1}{3\,x^{4/3}y^{1/3}}. \]

Since $x^{4/3} = \cos^{4}t$ and $y^{1/3} = \sin t$, this is exactly the boxed answer.

Step 6: Verification

\[t = \frac{\pi}{4}:\ x = y = \frac{1}{2\sqrt2},\ y'_{x} = -1,\ y''_{xx} = \frac{1}{3\cdot\frac14\cdot\frac{1}{\sqrt2}} = \frac{4\sqrt2}{3} \approx 1.8856\]

\[\text{From the } x,y \text{ form: } \frac{1}{3\bigl(2\sqrt2\bigr)^{-4/3}\bigl(2\sqrt2\bigr)^{-1/3}} = \frac{\bigl(2\sqrt2\bigr)^{5/3}}{3} = \frac{4\sqrt2}{3} \;\;\checkmark\]

\[y''_{xx}>0 \text{ for } 0\lt t\lt \frac{\pi}{2}: \text{ the first-quadrant arc is convex} \;\;\checkmark\]