If ` x =3 cos t-2cos ^(3) t,y =3 sin t- 2 sin ^(3) t,then (dy)/(dx) =`

Options

A. ` cot t`
B. ` -cott `
C. ` tant`
D. ` -tant `

Answer

To find \(\frac{dy}{dx}\) given the parametric equations \(x = 3 \cos t - 2 \cos^3 t\) and \(y = 3 \sin t - 2 \sin^3 t\), we will use the chain rule of differentiation.

### Step-by-step Solution:

1. **Differentiate \(x\) with respect to \(t\)**:
\[ x = 3 \cos t - 2 \cos^3 t \]
Using the product and chain rule, we differentiate:
\[ \frac{dx}{dt} = -3 \sin t - 2 \cdot 3 \cos^2 t \cdot (-\sin t) = -3 \sin t + 6 \cos^2 t \sin t \]
Factoring out \(\sin t\):
\[ \frac{dx}{dt} = \sin t (6 \cos^2 t - 3) = 3 \sin t (2 \cos^2 t - 1) \]

2. **Differentiate \(y\) with respect to \(t\)**:
\[ y = 3 \sin t - 2 \sin^3 t \]
Again, using the product and chain rule:
\[ \frac{dy}{dt} = 3 \cos t - 2 \cdot 3 \sin^2 t \cdot \cos t = 3 \cos t - 6 \sin^2 t \cos t \]
Factoring out \(\cos t\):
\[ \frac{dy}{dt} = \cos t (3 - 6 \sin^2 t) = 3 \cos t (1 - 2 \sin^2 t) \]

3. **Find \(\frac{dy}{dx}\)**:
Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3 \cos t (1 - 2 \sin^2 t)}{3 \sin t (2 \cos^2 t - 1)} \]
The \(3\) cancels out:
\[ \frac{dy}{dx} = \frac{\cos t (1 - 2 \sin^2 t)}{\sin t (2 \cos^2 t - 1)} \]
Recognizing that \(1 - 2 \sin^2 t = \cos 2t\) and \(2 \cos^2 t - 1 = \cos 2t\):
\[ \frac{dy}{dx} = \frac{\cos t \cos 2t}{\sin t \cos 2t} \]
The \(\cos 2t\) cancels out (assuming \(\cos 2t \neq 0\)):
\[ \frac{dy}{dx} = \frac{\cos t}{\sin t} = \cot t \]

### Final Answer:
\[ \frac{dy}{dx} = \cot t \]