The given equations are:

$y = a(\sin\theta - \theta \cos\theta) \quad \text{and} \quad x = a(\cos\theta + \theta \sin\theta)$

We are asked to find the first and second derivatives of $y$ with respect to $\theta$, i.e., $y'$ and $y''$.

Step 1: Differentiate $y$ with respect to $\theta$

$y = a(\sin\theta - \theta \cos\theta)$

We apply the product and chain rules:

[ y' = a\left(\frac{d}{d\theta}[\sin\theta] - \frac{d}{d\theta}[\theta \cos\theta]\right) ]

1. The derivative of $\sin\theta$ is $\cos\theta$.
2. To differentiate $\theta \cos\theta$, use the product rule: $\frac{d}{d\theta}[\theta \cos\theta] = \cos\theta - \theta \sin\theta$

So,

$y' = a(\cos\theta - (\cos\theta - \theta \sin\theta))$ $y' = a(\theta \sin\theta)$

Step 2: Differentiate $y'$ to find $y''$

$y' = a\theta \sin\theta$

Again, apply the product rule:

[ y'' = a\left(\frac{d}{d\theta}[\theta] \sin\theta + \theta \frac{d}{d\theta}[\sin\theta]\right) ]

1. The derivative of $\theta$ is 1.
2. The derivative of $\sin\theta$ is $\cos\theta$.

Thus,

$y'' = a(\sin\theta + \theta \cos\theta)$

Final Answer:

• $y' = a \theta \sin \theta$
• $y'' = a(\sin \theta + \theta \cos \theta)$

Let me know if you'd like further details or have any questions! Here are some related queries to expand on this topic:

1. What are the steps for differentiating trigonometric functions using the product rule?
2. How do we interpret parametric equations in the context of calculus?
3. What is the geometric significance of first and second derivatives?
4. How do chain and product rules apply to more complex functions?
5. How can these derivatives be applied in physics, such as in motion equations?

Tip: Always carefully apply the product and chain rules when dealing with functions involving both products and trigonometric terms.