The second order derivative of `a sin^3 t ` w.r.t, `a cos^3 t` at `t = pi/4` is

Options

A. `(4sqrt(2))/(3a)`
B. 2
C. `(1)/(12a)`
D. 0

Explanation

To find the second order derivative of \( y = a \sin^3 t \) with respect to \( x = a \cos^3 t \) at \( t = \frac{\pi}{4} \), we will follow these steps:

### Step 1: Define the functions
Let:
\[ y = a \sin^3 t \]
\[ x = a \cos^3 t \]

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

### Step 3: Differentiate \( x \) with respect to \( t \)
Similarly, we differentiate \( x \):
\[ \frac{dx}{dt} = 3a \cos^2 t (-\sin t) = -3a \cos^2 t \sin t \]

### Step 4: Find \( \frac{dy}{dx} \)
Using the formula \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \):
\[ \frac{dy}{dx} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} = -\frac{\sin t}{\cos t} = -\tan t \]

### Step 5: Differentiate \( \frac{dy}{dx} \) with respect to \( t \)
Now we differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \):
\[ \frac{d}{dt} \left( -\tan t \right) = -\sec^2 t \]

### Step 6: Use the chain rule to find \( \frac{d^2y}{dx^2} \)
Using the chain rule:
\[ \frac{d^2y}{dx^2} = \frac{d}{dt} \left( -\tan t \right) \cdot \frac{1}{\frac{dx}{dt}} = -\sec^2 t \cdot \frac{1}{-3a \cos^2 t \sin t} = \frac{\sec^2 t}{3a \cos^2 t \sin t} \]

### Step 7: Evaluate at \( t = \frac{\pi}{4} \)
At \( t = \frac{\pi}{4} \):
\[ \sec^2\left(\frac{\pi}{4}\right) = 2 \]
\[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Substituting these values:
\[ \frac{d^2y}{dx^2} = \frac{2}{3a \left(\frac{1}{\sqrt{2}}\right)^2 \left(\frac{1}{\sqrt{2}}\right)} = \frac{2}{3a \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{2}}} = \frac{4\sqrt{2}}{3a} \]

### Final Answer:
Thus, the second order derivative of \( a \sin^3 t \) with respect to \( a \cos^3 t \) at \( t = \frac{\pi}{4} \) is:
\[ \frac{4\sqrt{2}}{3a} \]

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