A curve is represented paramtrically by the equations `x=e^(t)cost` and `y=e^(t)sint` where `t` is a parameter. Then
The value of `(d^(2)y)/(dx^(2))` at the point where `t=0` is

Options

A. 1
B. 2
C. -2
D. 3

Answer

To find the value of \(\frac{d^2y}{dx^2}\) at the point where \(t=0\) for the given parametric equations \(x = e^t \cos t\) and \(y = e^t \sin t\), we will follow these steps:

### Step 1: Differentiate \(x\) and \(y\) with respect to \(t\)

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

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

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

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

### Step 3: Differentiate \(\frac{dy}{dx}\) to find \(\frac{d^2y}{dx^2}\)

Using the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{1}{\frac{dx}{dt}} \]

Let \(u = \sin t + \cos t\) and \(v = \cos t - \sin t\):
\[ \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]

Calculating \(\frac{du}{dt}\) and \(\frac{dv}{dt}\):
\[ \frac{du}{dt} = \cos t - \sin t \]
\[ \frac{dv}{dt} = -\sin t - \cos t \]

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

### Step 4: Substitute \(t = 0\)

At \(t = 0\):
- \(x = e^0 \cos(0) = 1\)
- \(y = e^0 \sin(0) = 0\)

Now substituting \(t = 0\) into the derivatives:
\[ \frac{du}{dt} = \cos(0) - \sin(0) = 1 \]
\[ \frac{dv}{dt} = -\sin(0) - \cos(0) = -1 \]

Now substituting back into the expression for \(\frac{d^2y}{dx^2}\):
\[ \frac{d^2y}{dx^2} = \frac{(\cos(0) - \sin(0))^2 - (\sin(0) + \cos(0))(-\sin(0) - \cos(0))}{(\cos(0) - \sin(0))^2} \cdot \frac{1}{\frac{dx}{dt}} \]

Calculating:
\[ \frac{d^2y}{dx^2} = \frac{(1)^2 - (0)(-1)}{(1)^2} \cdot \frac{1}{1} = \frac{1}{1} = 2 \]

### Final Answer:
Thus, the value of \(\frac{d^2y}{dx^2}\) at \(t = 0\) is \(2\).

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