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

Explanation

To find \(\frac{d^3y}{dx^3}\) at \(\theta = 0\) given \(x = a \cos^3 \theta\) and \(y = b \sin^3 \theta\), we will follow these steps:

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

Given:
\[ x = a \cos^3 \theta \]
Using the chain rule, we differentiate:
\[ \frac{dx}{d\theta} = a \cdot 3 \cos^2 \theta \cdot (-\sin \theta) = -3a \cos^2 \theta \sin \theta \]

### Step 2: Differentiate \(y\) with respect to \(\theta\)

Given:
\[ y = b \sin^3 \theta \]
Using the chain rule, we differentiate:
\[ \frac{dy}{d\theta} = b \cdot 3 \sin^2 \theta \cdot \cos \theta = 3b \sin^2 \theta \cos \theta \]

### Step 3: Find \(\frac{dy}{dx}\)

Now we can find \(\frac{dy}{dx}\) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3b \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} \]
Simplifying this gives:
\[ \frac{dy}{dx} = -\frac{b}{a} \cdot \frac{\sin \theta}{\cos \theta} = -\frac{b}{a} \tan \theta \]

### Step 4: Evaluate \(\frac{dy}{dx}\) at \(\theta = 0\)

Now, substituting \(\theta = 0\):
\[ \frac{dy}{dx} = -\frac{b}{a} \tan(0) = -\frac{b}{a} \cdot 0 = 0 \]

### Step 5: Find \(\frac{d^2y}{dx^2}\)

To find \(\frac{d^2y}{dx^2}\), we differentiate \(\frac{dy}{dx}\) with respect to \(\theta\) and then divide by \(\frac{dx}{d\theta}\):
\[ \frac{d^2y}{dx^2} = \frac{d}{d\theta}\left(-\frac{b}{a} \tan \theta\right) \cdot \frac{1}{\frac{dx}{d\theta}} \]
Calculating the derivative:
\[ \frac{d}{d\theta}\left(-\frac{b}{a} \tan \theta\right) = -\frac{b}{a} \sec^2 \theta \]
Thus,
\[ \frac{d^2y}{dx^2} = \frac{-\frac{b}{a} \sec^2 \theta}{-3a \cos^2 \theta \sin \theta} = \frac{\frac{b}{a} \sec^2 \theta}{3a \cos^2 \theta \sin \theta} \]
At \(\theta = 0\):
\[ \sec^2(0) = 1 \quad \text{and} \quad \sin(0) = 0 \]
This means \(\frac{d^2y}{dx^2}\) is undefined at \(\theta = 0\).

### Step 6: Find \(\frac{d^3y}{dx^3}\)

Since \(\frac{d^2y}{dx^2}\) is undefined at \(\theta = 0\), \(\frac{d^3y}{dx^3}\) will also be undefined at this point.

### Conclusion

Thus, the final result is:
\[ \frac{d^3y}{dx^3} \text{ is undefined at } \theta = 0. \]