If `x = e ^(t ) sin t, y = e ^(t) cos t,` then `(d ^(2) y )/(dx ^(2))` at `t = pi` is

Options

A. `2 e ^(pi)`
B. `1/2 e ^(pi)`
C. `(1)/(2e ^(pi ))`
D. `(2)/( e ^(pi))`

Answer

To find the value of \(\frac{d^2y}{dx^2}\) at \(t = \pi\) for the given parametric equations \(x = e^t \sin t\) and \(y = e^t \cos t\), we will follow these steps:

### Step 1: Find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\)

1. **Differentiate \(y\) with respect to \(t\)**:
\[ y = e^t \cos t \]
Using the product rule:
\[ \frac{dy}{dt} = \frac{d}{dt}(e^t) \cos t + e^t \frac{d}{dt}(\cos t) = e^t \cos t - e^t \sin t = e^t (\cos t - \sin t) \]

2. **Differentiate \(x\) with respect to \(t\)**:
\[ x = e^t \sin t \]
Using the product rule:
\[ \frac{dx}{dt} = \frac{d}{dt}(e^t) \sin t + e^t \frac{d}{dt}(\sin t) = e^t \sin t + e^t \cos t = e^t (\sin t + \cos t) \]

### Step 2: Find \(\frac{dy}{dx}\)

Using the chain rule for parametric equations:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{e^t (\cos t - \sin t)}{e^t (\sin t + \cos t)} = \frac{\cos t - \sin t}{\sin t + \cos t} \]

### Step 3: Find \(\frac{d^2y}{dx^2}\)

To find \(\frac{d^2y}{dx^2}\), we need to differentiate \(\frac{dy}{dx}\) with respect to \(x\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} \]

1. **Differentiate \(\frac{dy}{dx}\) with respect to \(t\)**:
\[ \frac{dy}{dx} = \frac{\cos t - \sin t}{\sin t + \cos t} \]
Using the quotient rule:
\[ \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]
where \(u = \cos t - \sin t\) and \(v = \sin t + \cos t\).

- \( \frac{du}{dt} = -\sin t - \cos t \)
- \( \frac{dv}{dt} = \cos t - \sin t \)

Therefore,
\[ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{(\sin t + \cos t)(-\sin t - \cos t) - (\cos t - \sin t)(\cos t - \sin t)}{(\sin t + \cos t)^2} \]

2. **Find \(\frac{dt}{dx}\)**:
\[ \frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{e^t (\sin t + \cos t)} \]

### Step 4: Evaluate at \(t = \pi\)

1. **Calculate \(\frac{dy}{dx}\) at \(t = \pi\)**:
\[ \frac{dy}{dx} = \frac{\cos(\pi) - \sin(\pi)}{\sin(\pi) + \cos(\pi)} = \frac{-1 - 0}{0 - 1} = 1 \]

2. **Calculate \(\frac{dx}{dt}\) at \(t = \pi\)**:
\[ \frac{dx}{dt} = e^\pi (0 - 1) = -e^\pi \]
Thus, \(\frac{dt}{dx} = -\frac{1}{e^\pi}\).

3. **Calculate \(\frac{d^2y}{dx^2}\) at \(t = \pi\)**:
Substitute into the expression for \(\frac{d^2y}{dx^2}\):
\[ \frac{d^2y}{dx^2} = \left(\text{Evaluate the derivative at } t = \pi\right) \cdot \left(-\frac{1}{e^\pi}\right) \]

After performing the calculations, we find:
\[ \frac{d^2y}{dx^2} = \frac{2}{e^\pi} \]

### Final Answer:
\[ \frac{d^2y}{dx^2} \text{ at } t = \pi \text{ is } \frac{2}{e^\pi} \]