Parametric equations of a curve are: x=acos^4(t) and y=asin^4(t) where a is a positive constant. Find dy/dx

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 \(t\).

Parametric equations of a curve are: x = a cos^4(t) and y = a sin^4(t), where a is a positive constant. Find dy/dx.

Learn how to differentiate parametric equations where x = a cos^4(t) and y = a sin^4(t). This problem applies key calculus concepts such as the chain rule and trigonometric identities, with step-by-step guidance to find dy/dx as -tan^2(t). Suitable for Grades 11-12 and college-level students.

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