Find y''_xx for x = a cos t, y = b sin t

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

Step 1: Derivatives with respect to $t$

\[ \dot x = -a\sin t, \qquad \dot y = b\cos t. \]

Step 2: First derivative

\[ y'_{x} = \frac{b\cos t}{-a\sin t} = -\frac{b}{a}\cot t. \]

Step 3: Differentiate with respect to $t$

Since $\bigl(\cot t\bigr)' = -\dfrac{1}{\sin^{2}t}$, the two minus signs cancel: $\bigl(y'_{x}\bigr)'_{t} = \frac{b}{a\sin^{2}t}$.

Step 4: Divide by $\dot x$

\[ y''_{xx} = \frac{b/\bigl(a\sin^{2}t\bigr)}{-a\sin t} = -\frac{b}{a^{2}\sin^{3}t}. \]

Step 5: Verification from the implicit equation

The curve is $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}} = 1$. Implicit differentiation gives $y'_{x} = -\dfrac{b^{2}x}{a^{2}y}$, and differentiating again, \[y''_{xx} = -\frac{b^{2}}{a^{2}}\cdot\frac{y-xy'_{x}}{y^{2}} = -\frac{b^{2}}{a^{2}}\cdot\frac{a^{2}y^{2}+b^{2}x^{2}}{a^{2}y^{3}} = -\frac{b^{2}}{a^{2}}\cdot\frac{a^{2}b^{2}}{a^{2}y^{3}} = -\frac{b^{4}}{a^{2}y^{3}}\]

\[\text{With } y = b\sin t:\quad -\frac{b^{4}}{a^{2}b^{3}\sin^{3}t} = -\frac{b}{a^{2}\sin^{3}t} \;\;\checkmark\]

\[t = \frac{\pi}{2} \text{ (top): } y''_{xx} = -\frac{b}{a^{2}}, \text{ negative — the arc is concave down} \;\;\checkmark\]