Find y''_xx for x = arcsin t, y = sqrt(1-t²)

Find $y''_{xx}$ for $x = \arcsin t,\ y = \sqrt{1-t^{2}}$.

Step 1: Derivatives with respect to $t$

\[ \dot x = \frac{1}{\sqrt{1-t^{2}}}, \qquad \dot y = \frac{-2t}{2\sqrt{1-t^{2}}} = \frac{-t}{\sqrt{1-t^{2}}}. \]

Both carry the same radical, which is what makes the next step so short.

Step 2: First derivative

\[ y'_{x} = \frac{-t/\sqrt{1-t^{2}}}{1/\sqrt{1-t^{2}}} = -t. \]

Step 3: Second derivative

\[ \bigl(y'_{x}\bigr)'_{t} = -1, \qquad y''_{xx} = \frac{-1}{1/\sqrt{1-t^{2}}} = -\sqrt{1-t^{2}}. \]

Step 4: Verification by eliminating the parameter

\[x = \arcsin t \;\Longrightarrow\; t = \sin x \;\Longrightarrow\; y = \sqrt{1-\sin^{2}x} = \cos x \quad\left(|x|\le\frac{\pi}{2}\right)\]

\[y'_{x} = -\sin x = -t \;\;\checkmark, \qquad y''_{xx} = -\cos x = -\sqrt{1-t^{2}} \;\;\checkmark\]

\[t = 0:\ x = 0,\ y = 1,\ y'_{x} = 0,\ y''_{xx} = -1 \;\Longrightarrow\; \text{a maximum} \;\;\checkmark\]