there is a parametric equation with x = theta ^ 2 and y = sin (2theta), calculate the derivative at pi/2 and -pi/4

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 \(\frac{dy}{d\theta}\).

There is a parametric equation with x = theta^2 and y = sin(2theta), calculate the derivative at pi/2 and -pi/4.

Calculate the derivative dy/dx for the parametric equations x = θ^2 and y = sin(2θ) at θ = π/2 and -π/4. Learn step-by-step how to use the chain rule, differentiate trigonometric functions, and compute derivatives at specific values of θ. Suitable for undergraduate calculus students.

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