To calculate the derivative of the parametric equation given by:

• $x = \theta^2$
• $y = \sin(2\theta)$

we need to find $\frac{dy}{dx}$. This requires applying the chain rule and calculating:

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

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

$\frac{dx}{d\theta} = \frac{d}{d\theta}(\theta^2) = 2\theta$

Step 2: Differentiate $y$ with respect to $\theta$

$\frac{dy}{d\theta} = \frac{d}{d\theta}(\sin(2\theta)) = 2\cos(2\theta)$

Step 3: Compute the derivative $\frac{dy}{dx}$

$\frac{dy}{dx} = \frac{2\cos(2\theta)}{2\theta} = \frac{\cos(2\theta)}{\theta}$

Step 4: Evaluate the derivative at $\theta = \frac{\pi}{2}$

Substitute $\theta = \frac{\pi}{2}$ into $\frac{dy}{dx}$:

$\frac{dy}{dx}\Big|_{\theta = \frac{\pi}{2}} = \frac{\cos\left(2 \times \frac{\pi}{2}\right)}{\frac{\pi}{2}} = \frac{\cos(\pi)}{\frac{\pi}{2}} = \frac{-1}{\frac{\pi}{2}} = \frac{-2}{\pi}$

Step 5: Evaluate the derivative at $\theta = -\frac{\pi}{4}$

Substitute $\theta = -\frac{\pi}{4}$ into $\frac{dy}{dx}$:

$\frac{dy}{dx}\Big|_{\theta = -\frac{\pi}{4}} = \frac{\cos\left(2 \times -\frac{\pi}{4}\right)}{-\frac{\pi}{4}} = \frac{\cos(-\frac{\pi}{2})}{-\frac{\pi}{4}} = \frac{0}{-\frac{\pi}{4}} = 0$

Final Results:

• The derivative at $\theta = \frac{\pi}{2}$ is $\frac{-2}{\pi}$.
• The derivative at $\theta = -\frac{\pi}{4}$ is $0$.

Would you like more details on any step? Here are some related questions to explore further:

1. How is the chain rule applied in parametric derivatives?
2. Can you explain why the derivative at $\theta = -\frac{\pi}{4}$ is 0?
3. How does parametric differentiation differ from regular differentiation?
4. What happens if we change the parametric equation to $x = \sin(\theta)$ and $y = \cos(\theta)$?
5. Can you graph the parametric curve for $x = \theta^2$ and $y = \sin(2\theta)$?

Tip: When differentiating parametric equations, always remember to use the chain rule by finding both $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$.