To solve the problem of finding the second derivative $\frac{d^2y}{dx^2}$ for the given parametric equations:

$x = e^t (\sin t + \cos t), \quad y = e^t (\cos t - \sin t),$

we will proceed as follows:

1. First Derivative ($\frac{dy}{dx}$):

Using the chain rule for parametric equations: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}.$

• Differentiate $x(t)$ with respect to $t$: $\frac{dx}{dt} = e^t (\cos t - \sin t) + e^t (\sin t + \cos t) = 2e^t \cos t.$

• Differentiate $y(t)$ with respect to $t$: $\frac{dy}{dt} = e^t (-\sin t - \cos t) + e^t (\cos t - \sin t) = -2e^t \sin t.$

Thus: $\frac{dy}{dx} = \frac{-2e^t \sin t}{2e^t \cos t} = -\tan t.$

2. Second Derivative ($\frac{d^2y}{dx^2}$):

To find the second derivative, use the formula: $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \div \frac{dx}{dt}.$

• Differentiate $\frac{dy}{dx} = -\tan t$ with respect to $t$: $\frac{d}{dt}(-\tan t) = -\sec^2 t.$

• Recall $\frac{dx}{dt} = 2e^t \cos t$.

Thus: $\frac{d^2y}{dx^2} = \frac{-\sec^2 t}{2e^t \cos t}.$

Simplify using $\sec^2 t = \frac{1}{\cos^2 t}$: $\frac{d^2y}{dx^2} = \frac{-1}{2e^t \cos^3 t}.$

Final Answer:

$\frac{d^2y}{dx^2} = \frac{-1}{2e^t \cos^3 t}.$

Would you like further clarification or additional steps? Here are five related questions for deeper exploration:

1. How do parametric equations relate to the curvature of a curve?
2. Can you demonstrate the physical interpretation of the second derivative in this context?
3. What happens if we analyze $\frac{d^2y}{dx^2}$ for different values of $t$?
4. Could this process apply to polar coordinates for similar parametric forms?
5. How do we use this second derivative to find inflection points of the curve?

Tip:

When working with parametric equations, always check if simplifying trigonometric expressions can reduce the complexity of derivatives.