If $x=a \cos ^3 \theta \quad y=a \sin ^3 \theta$ Then $\sqrt{1+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2}=$

Given the parametric equations $x = a \cos^3 \theta$, $y = a \sin^3 \theta$, determine $\sqrt{1 + \left( \frac{\mathrm{d} y}{\mathrm{d} x} \right)^2}$.

Differentiate both with respect to $\theta$: $\frac{\mathrm{d} x}{\mathrm{d} \theta} = -3a \cos^2 \theta \sin \theta$, $\frac{\mathrm{d} y}{\mathrm{d} \theta} = 3a \sin^2 \theta \cos \theta$.

Using the parametric derivative formula, $\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta$.

Substitute into the expression: $\sqrt{1 + (-\tan \theta)^2} = \sqrt{1 + \tan^2 \theta} = \sqrt{\sec^2 \theta}$.

Assuming $\sec \theta \geq 0$, $\sqrt{\sec^2 \theta} = \sec \theta$.

Final answer: $\sec \theta$ (choice C).