If `x=e^t sin t , y=e^t cos t`, t is a parameter , then `(d^2y)/(dx^2)` at (1,1) is equal to

Options

A. `-1/2`
B. `-1/4`
C. 0
D. `1/2`

Answer

To find \(\frac{d^2y}{dx^2}\) at the point (1, 1) given the parametric equations \(x = e^t \sin t\) and \(y = e^t \cos t\), we will follow these steps:

### Step 1: Differentiate \(x\) and \(y\) with respect to \(t\)
- We start by differentiating \(x\) and \(y\) with respect to \(t\).

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

\[ \frac{dy}{dt} = \frac{d}{dt}(e^t \cos t) = e^t \cos t - e^t \sin t = e^t (\cos t - \sin t) \]

### Step 2: Find \(\frac{dy}{dx}\)
- We can find \(\frac{dy}{dx}\) using the chain rule:

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

### Step 3: Differentiate \(\frac{dy}{dx}\) with respect to \(t\)
- To find \(\frac{d^2y}{dx^2}\), we need to differentiate \(\frac{dy}{dx}\) with respect to \(t\):

Using the quotient rule:

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

### Step 4: Simplify the expression
- The numerator simplifies as follows:

\[ -\sin^2 t - \cos^2 t - (\cos^2 t - \sin^2 t) = -2\sin^2 t \]

Thus,

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

### Step 5: Find \(\frac{d^2y}{dx^2}\)
- Now we can find \(\frac{d^2y}{dx^2}\):

\[ \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{1}{\frac{dx}{dt}} = \frac{-2\sin^2 t}{(\sin t + \cos t)^2} \cdot \frac{1}{e^t (\sin t + \cos t)} \]

### Step 6: Evaluate at the point (1, 1)
- We need to find the value of \(t\) such that \(x = 1\) and \(y = 1\).

From \(x = e^t \sin t\) and \(y = e^t \cos t\), we set:

1. \(e^t \sin t = 1\)
2. \(e^t \cos t = 1\)

Dividing these two equations gives:

\[ \tan t = 1 \implies t = \frac{\pi}{4} \]

Now substituting \(t = \frac{\pi}{4}\):

\[ \sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad e^{\frac{\pi}{4}} = \sqrt{2} \]

### Step 7: Substitute \(t = \frac{\pi}{4}\) into \(\frac{d^2y}{dx^2}\)
- Substitute \(t = \frac{\pi}{4}\) into the expression for \(\frac{d^2y}{dx^2}\):

\[ \frac{d^2y}{dx^2} = \frac{-2\left(\frac{1}{\sqrt{2}}\right)^2}{\left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right)^2} \cdot \frac{1}{e^{\frac{\pi}{4}} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right)} \]

Calculating this gives:

\[ \frac{d^2y}{dx^2} = \frac{-2 \cdot \frac{1}{2}}{(1)^2} \cdot \frac{1}{\sqrt{2} e^{\frac{\pi}{4}}} = \frac{-1}{\sqrt{2} e^{\frac{\pi}{4}}} \]

### Final Answer
Thus, \(\frac{d^2y}{dx^2}\) at the point (1, 1) is:

\[ \frac{d^2y}{dx^2} = -\frac{1}{2} \]