If `x=acostheta,y=bsintheta," then"(d^(3)y)/(dx^(3))` is equal to

Options

A. `-(3b)/(a^(3))"cosec"^(4)thetacot^(4)theta`
B. `(3b)/(a^(3))"cosec"^(4)thetacottheta`
C. `-(3b)/(a^(3))"cosec"^(4)thetacottheta`
D. none of these

Answer

To solve the problem, we need to find the third derivative of \( y \) with respect to \( x \), given the parametric equations \( x = a \cos \theta \) and \( y = b \sin \theta \).

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

We start by differentiating both equations with respect to \( \theta \):

\[ \frac{dx}{d\theta} = \frac{d}{d\theta}(a \cos \theta) = -a \sin \theta \]

\[ \frac{dy}{d\theta} = \frac{d}{d\theta}(b \sin \theta) = b \cos \theta \]

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

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

\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{b \cos \theta}{-a \sin \theta} = -\frac{b}{a} \cot \theta \]

### Step 3: Differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \)

Now, we differentiate \( \frac{dy}{dx} \) with respect to \( x \):

Using the chain rule again:

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{b}{a} \cot \theta\right) = -\frac{b}{a} \frac{d}{d\theta}(\cot \theta) \cdot \frac{d\theta}{dx} \]

We know that:

\[ \frac{d}{d\theta}(\cot \theta) = -\csc^2 \theta \]

And from our earlier calculation:

\[ \frac{d\theta}{dx} = \frac{1}{\frac{dx}{d\theta}} = -\frac{1}{a \sin \theta} \]

Thus,

\[ \frac{d^2y}{dx^2} = -\frac{b}{a} (-\csc^2 \theta) \left(-\frac{1}{a \sin \theta}\right) = \frac{b \csc^2 \theta}{a^2 \sin \theta} \]

This simplifies to:

\[ \frac{d^2y}{dx^2} = -\frac{b}{a^2} \cos^2 \theta \]

### Step 4: Differentiate \( \frac{d^2y}{dx^2} \) to find \( \frac{d^3y}{dx^3} \)

Now, we differentiate \( \frac{d^2y}{dx^2} \):

\[ \frac{d^3y}{dx^3} = \frac{d}{dx}\left(-\frac{b}{a^2} \cos^2 \theta\right) = -\frac{b}{a^2} \cdot 2 \cos \theta \cdot \frac{d\cos \theta}{d\theta} \cdot \frac{d\theta}{dx} \]

Using \( \frac{d\cos \theta}{d\theta} = -\sin \theta \):

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

This simplifies to:

\[ \frac{d^3y}{dx^3} = \frac{2b \cos^2 \theta}{a^3} \]

### Final Result

Thus, the third derivative \( \frac{d^3y}{dx^3} \) is:

\[ \frac{d^3y}{dx^3} = -\frac{3b \cos^4 \theta \cot \theta}{a^3} \]