If `x=acos^3theta,y=bsin^3theta,fin d(d^3y)/(dx^3)` at `theta=0.`

Explanation

To solve the problem of finding the third derivative of \( y \) with respect to \( x \) at \( \theta = 0 \), we start with the given equations:

1. \( x = a \cos^3 \theta \)
2. \( y = b \sin^3 \theta \)

We need to find \( \frac{d^3y}{dx^3} \) at \( \theta = 0 \).

### Step 1: Find the first derivative \( \frac{dy}{dx} \)

Using the chain rule, we can express \( \frac{dy}{dx} \) as:

\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]

**Calculating \( \frac{dy}{d\theta} \):**

\[ y = b \sin^3 \theta \implies \frac{dy}{d\theta} = 3b \sin^2 \theta \cos \theta \]

**Calculating \( \frac{dx}{d\theta} \):**

\[ x = a \cos^3 \theta \implies \frac{dx}{d\theta} = -3a \cos^2 \theta \sin \theta \]

Now substituting these into the derivative:

\[ \frac{dy}{dx} = \frac{3b \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} \]

This simplifies to:

\[ \frac{dy}{dx} = -\frac{b}{a} \cdot \frac{\sin^2 \theta}{\cos^2 \theta} = -\frac{b}{a} \tan^2 \theta \]

### Step 2: Evaluate \( \frac{dy}{dx} \) at \( \theta = 0 \)

At \( \theta = 0 \):

\[ \sin(0) = 0 \quad \text{and} \quad \cos(0) = 1 \]

Thus,

\[ \frac{dy}{dx} = -\frac{b}{a} \tan^2(0) = -\frac{b}{a} \cdot 0 = 0 \]

### Step 3: Find the second derivative \( \frac{d^2y}{dx^2} \)

Using the formula for the second derivative:

\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{d\theta} \left( \frac{dy}{dx} \right) \cdot \frac{d\theta}{dx} \]

First, we need \( \frac{d\theta}{dx} \):

\[ \frac{d\theta}{dx} = \frac{1}{\frac{dx}{d\theta}} = \frac{-1}{3a \cos^2 \theta \sin \theta} \]

Now, we differentiate \( \frac{dy}{dx} \):

\[ \frac{d}{d\theta} \left( -\frac{b}{a} \tan^2 \theta \right) = -\frac{b}{a} \cdot 2 \tan \theta \sec^2 \theta \]

Thus,

\[ \frac{d^2y}{dx^2} = -\frac{b}{a} \cdot 2 \tan \theta \sec^2 \theta \cdot \frac{-1}{3a \cos^2 \theta \sin \theta} \]

### Step 4: Evaluate \( \frac{d^2y}{dx^2} \) at \( \theta = 0 \)

At \( \theta = 0 \):

\[ \tan(0) = 0 \quad \text{and} \quad \sec(0) = 1 \]

Thus,

\[ \frac{d^2y}{dx^2} = -\frac{b}{a} \cdot 2 \cdot 0 \cdot 1 \cdot \frac{-1}{3a \cdot 1 \cdot 0} \text{ (undefined)} \]

### Step 5: Conclusion for the third derivative \( \frac{d^3y}{dx^3} \)

Since \( \frac{d^2y}{dx^2} \) is undefined at \( \theta = 0 \), it follows that \( \frac{d^3y}{dx^3} \) also does not exist at \( \theta = 0 \).

Thus, we conclude:

\[ \frac{d^3y}{dx^3} \text{ does not exist at } \theta = 0. \]