If `x = e^(t) sin t and y = e^(t)` cos t, t is a parameter , then the value of `(d^(2) x)/( dy^(2)) + (d^(2) y)/(dx^(2))` at t = 0 , is :

Options

A. `-2`
B. `1//2`
C. 2
D. 0

Answer

To solve the given problem, we need to find the value of \(\frac{d^2 x}{dy^2} + \frac{d^2 y}{dx^2}\) at \(t = 0\), where \(x = e^t \sin t\) and \(y = e^t \cos t\).

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

1. Differentiate \(x\) with respect to \(t\):
\[ \frac{dx}{dt} = \frac{d}{dt}(e^t \sin t) = e^t \sin t + e^t \cos t = e^t (\sin t + \cos t) \]

2. Differentiate \(y\) with respect to \(t\):
\[ \frac{dy}{dt} = \frac{d}{dt}(e^t \cos t) = e^t \cos t - e^t \sin t = e^t (\cos t - \sin t) \]

### Step 2: Find the second derivatives \( \frac{d^2x}{dt^2} \) and \( \frac{d^2y}{dt^2} \)

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

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

### Step 3: Find \(\frac{dy}{dx}\)

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

### Step 4: Find \(\frac{d^2y}{dx^2}\)

Using the formula:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} \]

Where:
\[ \frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{e^t (\sin t + \cos t)} \]

Now, differentiate \(\frac{dy}{dx}\) with respect to \(t\):
\[ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{(\sin t + \cos t)(-\sin t - \cos t) - (\cos t - \sin t)(\cos t - \sin t)}{(\sin t + \cos t)^2} \]

### Step 5: Evaluate at \(t = 0\)

At \(t = 0\):
- \(x(0) = e^0 \sin(0) = 0\)
- \(y(0) = e^0 \cos(0) = 1\)
- \(\frac{dx}{dt}(0) = 1\)
- \(\frac{dy}{dt}(0) = 0\)
- \(\frac{d^2x}{dt^2}(0) = 2\)
- \(\frac{d^2y}{dt^2}(0) = 0\)

Now we can calculate:
\[ \frac{d^2y}{dx^2} = \frac{0}{1} = 0 \]
\[ \frac{d^2x}{dy^2} = \frac{2}{1} = 2 \]

Finally, we find:
\[ \frac{d^2x}{dy^2} + \frac{d^2y}{dx^2} = 2 + 0 = 2 \]

### Final Answer:
The value of \(\frac{d^2 x}{dy^2} + \frac{d^2 y}{dx^2}\) at \(t = 0\) is \(2\).