`x=acos^(3)theta,y=asin^(3)theta` then find `(d^(2)y)/(dx^(2))`

Answer

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

### Step 1: Differentiate \(x\) with respect to \(\theta\)
Given:
\[ x = a \cos^3 \theta \]
Using the chain rule:
\[ \frac{dx}{d\theta} = a \cdot 3 \cos^2 \theta \cdot (-\sin \theta) = -3a \cos^2 \theta \sin \theta \]

### Step 2: Differentiate \(y\) with respect to \(\theta\)
Given:
\[ y = a \sin^3 \theta \]
Using the chain rule:
\[ \frac{dy}{d\theta} = a \cdot 3 \sin^2 \theta \cdot \cos \theta = 3a \sin^2 \theta \cos \theta \]

### Step 3: Find \(\frac{dy}{dx}\)
Using the formula for derivatives:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]
Substituting the values from Steps 1 and 2:
\[ \frac{dy}{dx} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} \]
Simplifying:
\[ \frac{dy}{dx} = \frac{3 \sin^2 \theta \cos \theta}{-3 \cos^2 \theta \sin \theta} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta \]

### Step 4: Differentiate \(\frac{dy}{dx}\) with respect to \(x\)
To find \(\frac{d^2y}{dx^2}\), we differentiate \(\frac{dy}{dx}\) with respect to \(\theta\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\tan \theta\right) = \frac{d}{d\theta}\left(-\tan \theta\right) \cdot \frac{d\theta}{dx} \]
The derivative of \(-\tan \theta\) is:
\[ -\sec^2 \theta \]
Now, 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 5: Simplify the expression
Since \(\sec^2 \theta = \frac{1}{\cos^2 \theta}\), we can rewrite:
\[ \frac{d^2y}{dx^2} = \frac{1}{3a \cos^2 \theta \sin \theta \cos^2 \theta} = \frac{1}{3a \sin \theta \cos^4 \theta} \]

### Final Result
Thus, the second derivative is:
\[ \frac{d^2y}{dx^2} = \frac{1}{3a \sin \theta \cos^4 \theta} \]

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