If `x=a(cost+tsint)\ ` and `y=a(sint-tcost),\ 0

Answer

To solve the problem step by step, we need to find the second derivatives \( \frac{d^2x}{dt^2} \) and \( \frac{d^2y}{dt^2} \), and then find \( \frac{d^2y}{dx^2} \).

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

Given:
\[ x = a(\cos t + t \sin t) \]

Differentiating \( x \) with respect to \( t \):
\[ \frac{dx}{dt} = a \left( -\sin t + \left( t \frac{d}{dt}(\sin t) + \sin t \frac{d}{dt}(t) \right) \right) \]
Using the product rule for \( t \sin t \):
\[ \frac{dx}{dt} = a \left( -\sin t + t \cos t + \sin t \right) \]
The \( -\sin t \) and \( +\sin t \) cancel out:
\[ \frac{dx}{dt} = a t \cos t \]

### Step 2: Differentiate \( y \) with respect to \( t \)

Given:
\[ y = a(\sin t - t \cos t) \]

Differentiating \( y \) with respect to \( t \):
\[ \frac{dy}{dt} = a \left( \cos t - \left( t \frac{d}{dt}(\cos t) + \cos t \frac{d}{dt}(t) \right) \right) \]
Using the product rule for \( t \cos t \):
\[ \frac{dy}{dt} = a \left( \cos t - \left( -t \sin t + \cos t \right) \right) \]
This simplifies to:
\[ \frac{dy}{dt} = a \left( \cos t + t \sin t - \cos t \right) = a t \sin t \]

### Step 3: Find the second derivatives \( \frac{d^2x}{dt^2} \) and \( \frac{d^2y}{dt^2} \)

Now, we differentiate \( \frac{dx}{dt} \) again to find \( \frac{d^2x}{dt^2} \):
\[ \frac{d^2x}{dt^2} = \frac{d}{dt}(a t \cos t) \]
Using the product rule:
\[ \frac{d^2x}{dt^2} = a \left( \cos t + t \frac{d}{dt}(\cos t) \right) = a \left( \cos t - t \sin t \right) \]

Next, we differentiate \( \frac{dy}{dt} \) to find \( \frac{d^2y}{dt^2} \):
\[ \frac{d^2y}{dt^2} = \frac{d}{dt}(a t \sin t) \]
Using the product rule:
\[ \frac{d^2y}{dt^2} = a \left( \sin t + t \cos t \right) \]

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

Using the chain rule:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dt}\right) \cdot \frac{dt}{dx} \]
We already have \( \frac{dy}{dt} = a t \sin t \) and \( \frac{dx}{dt} = a t \cos t \).

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

Now, differentiate \( \tan t \) with respect to \( t \):
\[ \frac{d}{dt}(\tan t) = \sec^2 t \]

Now, we need \( \frac{dt}{dx} \):
\[ \frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{a t \cos t} \]

Finally, we can calculate \( \frac{d^2y}{dx^2} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dt}(\tan t) \cdot \frac{dt}{dx} = \sec^2 t \cdot \frac{1}{a t \cos t} = \frac{\sec^2 t}{a t \cos t} \]

### Final Answers:
1. \( \frac{d^2x}{dt^2} = a(\cos t - t \sin t) \)
2. \( \frac{d^2y}{dt^2} = a(\sin t + t \cos t) \)
3. \( \frac{d^2y}{dx^2} = \frac{\sec^2 t}{a t \cos t} \)