Find `(d^(2)y)/(dx^(2))` in the following
If `x=acos^(3)thetaandy=asin^(3)theta`, then find the value of `(d^(2)y)/(dx^(2))" at "theta=pi/6`.

Answer

To find the second derivative \(\frac{d^2y}{dx^2}\) given the parametric equations \(x = a \cos^3 \theta\) and \(y = a \sin^3 \theta\), we will follow these steps:

### Step 1: Find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\)

1. Differentiate \(x = a \cos^3 \theta\) with respect to \(\theta\):
\[ \frac{dx}{d\theta} = a \cdot 3 \cos^2 \theta \cdot (-\sin \theta) = -3a \cos^2 \theta \sin \theta \]

2. Differentiate \(y = a \sin^3 \theta\) with respect to \(\theta\):
\[ \frac{dy}{d\theta} = a \cdot 3 \sin^2 \theta \cdot \cos \theta = 3a \sin^2 \theta \cos \theta \]

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

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} \]

This simplifies to:
\[ \frac{dy}{dx} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta \]

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

To find the second derivative, we differentiate \(\frac{dy}{dx}\) with respect to \(x\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\tan \theta\right) \]

Using the chain rule again:
\[ \frac{d^2y}{dx^2} = \frac{d}{d\theta}(-\tan \theta) \cdot \frac{d\theta}{dx} \]

Now, differentiate \(-\tan \theta\):
\[ \frac{d}{d\theta}(-\tan \theta) = -\sec^2 \theta \]

Next, we need \(\frac{d\theta}{dx}\):
\[ \frac{d\theta}{dx} = \frac{1}{\frac{dx}{d\theta}} = \frac{1}{-3a \cos^2 \theta \sin \theta} \]

Thus,
\[ \frac{d^2y}{dx^2} = -\sec^2 \theta \cdot \left(\frac{1}{-3a \cos^2 \theta \sin \theta}\right) = \frac{\sec^2 \theta}{3a \cos^2 \theta \sin \theta} \]

### Step 4: Evaluate at \(\theta = \frac{\pi}{6}\)

1. Calculate \(\sec^2\left(\frac{\pi}{6}\right)\):
\[ \sec\left(\frac{\pi}{6}\right) = \frac{1}{\cos\left(\frac{\pi}{6}\right)} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} \quad \Rightarrow \quad \sec^2\left(\frac{\pi}{6}\right) = \frac{4}{3} \]

2. Calculate \(\cos^2\left(\frac{\pi}{6}\right)\) and \(\sin\left(\frac{\pi}{6}\right)\):
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \quad \Rightarrow \quad \cos^2\left(\frac{\pi}{6}\right) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]
\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]

3. Substitute these values into \(\frac{d^2y}{dx^2}\):
\[ \frac{d^2y}{dx^2} = \frac{\frac{4}{3}}{3a \cdot \frac{3}{4} \cdot \frac{1}{2}} = \frac{\frac{4}{3}}{\frac{9a}{4}} = \frac{4 \cdot 4}{3 \cdot 9a} = \frac{16}{27a} \]

### Final Answer
Thus, the value of \(\frac{d^2y}{dx^2}\) at \(\theta = \frac{\pi}{6}\) is:
\[ \frac{d^2y}{dx^2} = \frac{16}{27a} \]