Find `(d^(2)y)/(dx^(2))` in the following
`x=a(cost+tsint),y=a(sint-tcost)`

Answer

To find \(\frac{d^2y}{dx^2}\) for the given parametric equations \(x = a(\cos t + t \sin t)\) and \(y = a(\sin t - t \cos t)\), we will follow these steps:

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

First, we differentiate \(x\) and \(y\) with respect to \(t\).

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

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

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

Using the chain rule, we can find \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a(t \sin t)}{a(t \cos t)} = \frac{t \sin t}{t \cos t} = \tan t \]

### Step 3: Differentiate \(\frac{dy}{dx}\) with respect to \(t\)

Next, we differentiate \(\frac{dy}{dx}\) with respect to \(t\):

\[ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}(\tan t) = \sec^2 t \]

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

Now, we can find \(\frac{d^2y}{dx^2}\) using the formula:

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

Substituting the values we found:

\[ \frac{d^2y}{dx^2} = \sec^2 t \cdot \frac{1}{a(t \cos t)} = \frac{\sec^2 t}{a(t \cos t)} \]

### Final Result

Thus, the second derivative \(\frac{d^2y}{dx^2}\) is:

\[ \frac{d^2y}{dx^2} = \frac{\sec^2 t}{a(t \cos t)} \]

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