if `x=a cos^(4) theta, y= a sin^(4) theta, "then" (dy)/(dx)"at" theta=(3pi)/(4)` is

Options

A. `-1`
B. 1
C. `-a^(2)`
D. `a^(2)`

Explanation

To find \(\frac{dy}{dx}\) at \(\theta = \frac{3\pi}{4}\) given \(x = a \cos^4 \theta\) and \(y = a \sin^4 \theta\), we will follow these steps:

### Step 1: Differentiate \(x\) and \(y\) with respect to \(\theta\)

1. **Differentiate \(x\)**:
\[ x = a \cos^4 \theta \]
Using the chain rule:
\[ \frac{dx}{d\theta} = a \cdot 4 \cos^3 \theta \cdot (-\sin \theta) = -4a \cos^3 \theta \sin \theta \]

2. **Differentiate \(y\)**:
\[ y = a \sin^4 \theta \]
Again using the chain rule:
\[ \frac{dy}{d\theta} = a \cdot 4 \sin^3 \theta \cdot \cos \theta = 4a \sin^3 \theta \cos \theta \]

### Step 2: Find \(\frac{dy}{dx}\)

Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{4a \sin^3 \theta \cos \theta}{-4a \cos^3 \theta \sin \theta} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{4a \sin^3 \theta \cos \theta}{-4a \cos^3 \theta \sin \theta} = -\frac{\sin^2 \theta}{\cos^2 \theta} = -\tan^2 \theta \]

### Step 3: Evaluate \(\frac{dy}{dx}\) at \(\theta = \frac{3\pi}{4}\)

Now we substitute \(\theta = \frac{3\pi}{4}\):
\[ \tan\left(\frac{3\pi}{4}\right) = -1 \quad \text{(since \(\tan\) is negative in the second quadrant)} \]
Thus:
\[ \tan^2\left(\frac{3\pi}{4}\right) = 1 \]
Therefore:
\[ \frac{dy}{dx} = -\tan^2\left(\frac{3\pi}{4}\right) = -1 \]

### Final Answer
\[ \frac{dy}{dx} \text{ at } \theta = \frac{3\pi}{4} \text{ is } -1. \]