To find $\frac{dy}{dx}$ for the given parametric equations:

$x = a \cos^4(t) \quad \text{and} \quad y = a \sin^4(t)$ where $a$ is a positive constant, we use the following formula:

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Step 1: Differentiate $x$ with respect to $t$

$x = a \cos^4(t)$ Using the chain rule: $\frac{dx}{dt} = a \cdot 4 \cos^3(t) \cdot (-\sin(t)) = -4a \cos^3(t) \sin(t)$

Step 2: Differentiate $y$ with respect to $t$

$y = a \sin^4(t)$ Using the chain rule: $\frac{dy}{dt} = a \cdot 4 \sin^3(t) \cdot \cos(t) = 4a \sin^3(t) \cos(t)$

Step 3: Find $\frac{dy}{dx}$

Now, we compute the ratio of $\frac{dy}{dt}$ and $\frac{dx}{dt}$: $\frac{dy}{dx} = \frac{4a \sin^3(t) \cos(t)}{-4a \cos^3(t) \sin(t)}$

Simplifying the expression: $\frac{dy}{dx} = - \frac{\sin^2(t)}{\cos^2(t)}$ $\frac{dy}{dx} = - \tan^2(t)$

Thus, the derivative $\frac{dy}{dx}$ is: $\boxed{- \tan^2(t)}$

Would you like further details or have any questions?

Here are some related questions:

1. What is the geometric interpretation of the curve described by these parametric equations?
2. How would the result change if $a$ were a negative constant?
3. How do we find higher-order derivatives, such as $\frac{d^2y}{dx^2}$, for this curve?
4. What happens when $t = 0$ or $t = \frac{\pi}{2}$ in terms of the slope $\frac{dy}{dx}$?
5. How do these parametric equations relate to polar coordinates?

Tip: Parametric derivatives often require applying the chain rule carefully, as you need to handle multiple variables through $t$.