To find the radius of curvature for the parametric equations $x = a \cos^3 \theta$ and $y = a \sin^3 \theta$ at $\theta = \frac{\pi}{4}$, we follow these steps:

1. Definitions and Formula

The radius of curvature $R$ for parametric curves is given by:

$R = \frac{\left( \dot{x}^2 + \dot{y}^2 \right)^{3/2}}{\lvert \dot{x} \ddot{y} - \dot{y} \ddot{x} \rvert}$

Where:

• $\dot{x} = \frac{dx}{d\theta}$, $\dot{y} = \frac{dy}{d\theta}$
• $\ddot{x} = \frac{d^2x}{d\theta^2}$, $\ddot{y} = \frac{d^2y}{d\theta^2}$

2. Derivatives of $x$ and $y$

Given $x = a \cos^3 \theta$ and $y = a \sin^3 \theta$:

• $\dot{x} = \frac{dx}{d\theta} = 3a \cos^2 \theta (-\sin \theta) = -3a \cos^2 \theta \sin \theta$
• $\dot{y} = \frac{dy}{d\theta} = 3a \sin^2 \theta \cos \theta$

Now calculate the second derivatives:

• $\ddot{x} = \frac{d}{d\theta} (-3a \cos^2 \theta \sin \theta) = -3a \left( -2\cos \theta \sin^2 \theta - \cos^3 \theta \right) = 3a \cos \theta (2\sin^2 \theta + \cos^2 \theta)$
• $\ddot{y} = \frac{d}{d\theta} (3a \sin^2 \theta \cos \theta) = 3a \left( 2\sin \theta \cos^2 \theta - \sin^3 \theta \right)$

3. Simplifications at $\theta = \frac{\pi}{4}$

At $\theta = \frac{\pi}{4}$:

• $\cos \theta = \sin \theta = \frac{\sqrt{2}}{2}$
• $\cos^2 \theta = \sin^2 \theta = \frac{1}{2}$

Substitute these values:

• $\dot{x} = -3a \left( \frac{1}{2} \right)\left( \frac{\sqrt{2}}{2} \right) = -\frac{3a\sqrt{2}}{4}$

• $\dot{y} = 3a \left( \frac{1}{2} \right)\left( \frac{\sqrt{2}}{2} \right) = \frac{3a\sqrt{2}}{4}$

• $\ddot{x} = 3a \left( \frac{\sqrt{2}}{2} \right)\left( 2\left( \frac{1}{2} \right) + \frac{1}{2} \right) = 3a \left( \frac{\sqrt{2}}{2} \right)\left( \frac{3}{2} \right) = \frac{9a\sqrt{2}}{4}$

• $\ddot{y} = 3a \left( 2\left( \frac{\sqrt{2}}{2} \right)\left( \frac{1}{2} \right) - \left( \frac{\sqrt{2}}{2} \right)^3 \right) = 3a \left( \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{8} \right) = \frac{9a\sqrt{2}}{8}$

4. Radius of Curvature

Substitute into the formula for $R$:

• $\dot{x}^2 + \dot{y}^2 = \left( -\frac{3a\sqrt{2}}{4} \right)^2 + \left( \frac{3a\sqrt{2}}{4} \right)^2 = \frac{9a^2}{8} + \frac{9a^2}{8} = \frac{18a^2}{8} = \frac{9a^2}{4}$
• $\dot{x} \ddot{y} - \dot{y} \ddot{x} = \left( -\frac{3a\sqrt{2}}{4} \right)\left( \frac{9a\sqrt{2}}{8} \right) - \left( \frac{3a\sqrt{2}}{4} \right)\left( \frac{9a\sqrt{2}}{4} \right) = -\frac{27a^2}{16} - \frac{27a^2}{16} = -\frac{54a^2}{16} = -\frac{27a^2}{8}$

Now compute $R$: $R = \frac{\left( \frac{9a^2}{4} \right)^{3/2}}{\left| -\frac{27a^2}{8} \right|} = \frac{\left( \frac{9a^2}{4} \right)\sqrt{\frac{9a^2}{4}}}{\frac{27a^2}{8}} = \frac{\frac{9a^2}{4} \cdot \frac{3a}{2}}{\frac{27a^2}{8}} = \frac{\frac{27a^3}{8}}{\frac{27a^2}{8}} = a$

Final Answer:

The radius of curvature at $\theta = \frac{\pi}{4}$ is:

$\boxed{a}$

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Would you like further clarifications or steps explained in detail? 😊
Here are 5 related questions for you to explore:

1. What is the geometric significance of the radius of curvature?
2. How would the calculation differ for Cartesian (non-parametric) equations?
3. Can we calculate the radius of curvature for polar coordinates?
4. What happens to the radius of curvature at an inflection point?
5. How is the radius of curvature related to the osculating circle?

Tip: Always double-check units and dimensions in curvature problems to ensure consistency!