Find y''_xx for x = e^tcos t, y = e^tsin t

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

Step 1: Derivatives with respect to $t$

\[ \dot x = e^{t}\bigl(\cos t-\sin t\bigr), \qquad \dot y = e^{t}\bigl(\sin t+\cos t\bigr). \]

Step 2: First derivative

The exponential cancels: \[y'_{x} = \frac{\sin t+\cos t}{\cos t-\sin t}. \tag{1}\]

Step 3: Differentiate (1) with respect to $t$

Put $u = \sin t+\cos t$ and $v = \cos t-\sin t$. These two are linked: $u' = v$ and $v' = -u$. The quotient rule therefore gives

\[ \bigl(y'_{x}\bigr)'_{t} = \frac{u'v-uv'}{v^{2}} = \frac{v^{2}+u^{2}}{v^{2}}. \]

The numerator collapses by the Pythagorean identity: \[u^{2}+v^{2} = \bigl(\sin t+\cos t\bigr)^{2}+\bigl(\cos t-\sin t\bigr)^{2} = 2\sin^{2}t+2\cos^{2}t = 2,\]

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

Step 4: Divide by $\dot x$

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

Step 5: Verification

\[t = 0:\ x = 1,\ y = 0,\ y'_{x} = 1, \qquad y''_{xx} = \frac{2}{1\cdot1} = 2 \;\;\checkmark\]

\[t = \pi:\ x = -e^{\pi},\ y = 0,\ y''_{xx} = \frac{2}{e^{\pi}\bigl(-1\bigr)^{3}} = -\frac{2}{e^{\pi}} \approx -0.0864 \;\;\checkmark\]

\[\text{Undefined at } t = \frac{\pi}{4}+k\pi, \text{ where } \dot x = 0 \text{ and the tangent is vertical} \;\;\checkmark\]