If [x = 2 - 2 ] and [y = 2 - 2 ], [ [ 0,2 ]], then [d^2y dx^2] at [ ] = [ ] is :

Question

If $$x = 2\sin \theta  - \sin 2\theta $$ and $$y = 2\cos \theta  - \cos 2\theta $$,

$$\theta  \in \left[ {0,2\pi } \right]$$, then $${{{d^2}y} \over {d{x^2}}}$$ at $$\theta $$ = $$\pi $$ is :

Accepted Answer

$$x = 2\sin \theta  - \sin 2\theta $$

$$ \Rightarrow $$ $${{dx} \over {d\theta }}$$ = $$2\cos \theta  - 2\cos 2\theta $$

$$y = 2\cos \theta  - \cos 2\theta $$

$$ \Rightarrow $$ $${{dy} \over {d\theta }}$$ = --2sin$$\theta $$ + 2sin2$$\theta $$

$${{dy} \over {dx}} = {{{{dy} \over {d\theta }}} \over {{{dx} \over {d\theta }}}}$$ = $${{\sin 2\theta  - \sin \theta } \over {\cos \theta  - \cos 2\theta }}$$

This expression is simplified by recognizing that $\sin 2 \theta = 2\sin \theta \cos \theta$ and $\cos 2 \theta = 2\cos^2 \theta -1$, leading to the expression:

$$ \frac{dy}{dx} = \frac{2 \sin \frac{\theta}{2} \cdot \cos \frac{3 \theta}{2}}{2 \sin \frac{\theta}{2} \cdot \sin \frac{3 \theta}{2}}=\cot \frac{3 \theta}{2}$$

This is the rate of change of y with respect to x, as a function of \theta .

Next, we differentiate this function with respect to \theta  to find $\frac{d^2 y}{dx^2}$, yielding :

$$ \frac{d^2 y}{dx^2}=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \frac{d \theta}{d x}=-\frac{3}{2} \operatorname{cosec}^2 \frac{3 \theta}{2} \cdot \frac{d \theta}{d x}$$

Here, $\frac{d \theta}{d x}$ is the reciprocal of $\frac{dx}{d\theta}$, so the equation becomes :

$$ \frac{d^2 y}{dx^2}=\frac{-\frac{3}{2} \operatorname{cosec}^2 \frac{3 \theta}{2}}{2\left(\cos \theta-\cos2 \theta\right)}$$

Finally, we evaluate this expression at $\theta = \pi$, yielding :

$$ \frac{d^2 y}{dx^2}(\pi)=\frac{-3}{4(-1-1)}=\frac{3}{8}$$

Therefore, the correct answer is (A) $\frac{3}{8}$.

Alternate Method :

First, let's find the derivatives of x and y with respect to \theta  :

1) $$ \frac{dx}{d\theta} = 2\cos \theta - 2\cos 2\theta $$

2) $$ \frac{dy}{d\theta} = -2\sin \theta + 2\sin 2\theta $$

We know that $$ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} $$

So, we substitute 1) and 2) into this equation :

We have

$$\frac{dy}{dx} = \frac{-2\sin \theta + 2\sin 2\theta}{2\cos \theta - 2\cos 2\theta} = \frac{\sin 2\theta - \sin \theta}{\cos \theta - \cos 2\theta}$$

For simplification, let's denote the numerator as $$N = \sin 2\theta - \sin \theta$$ and the denominator as $$D = \cos \theta - \cos 2\theta$$.

We have to compute $$\frac{d}{d\theta}(\frac{dy}{dx})$$ which is $$\frac{d}{d\theta}(\frac{N}{D})$$.

We can use the quotient rule for differentiation, which states that if we have a function of the form $$\frac{u}{v}$$, then its derivative is given by $$\frac{vu' - uv'}{v^2}$$.

So here, $$N' = 2\cos 2\theta - \cos \theta$$ and $$D' = -\sin \theta + 2\sin 2\theta$$.

Applying the quotient rule :

$$\frac{d}{d\theta}(\frac{dy}{dx}) = \frac{D N' - N D'}{D^2}$$

Substituting the expressions for $N'$, $D'$, $N$, and $D$ we get :

$$\frac{d}{d\theta}(\frac{dy}{dx}) = \frac{(\cos \theta - \cos 2\theta)(2\cos 2\theta - \cos \theta) - (\sin 2\theta - \sin \theta)(-\sin \theta + 2\sin 2\theta)}{(\cos \theta - \cos 2\theta)^2}$$

Now $$
\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \times \frac{d \theta}{d x}
$$

=  $$\frac{(\cos \theta - \cos 2\theta)(2\cos 2\theta - \cos \theta) - (\sin 2\theta - \sin \theta)(-\sin \theta + 2\sin 2\theta)}{(\cos \theta - \cos 2\theta)^2}$$ $
\times \frac{1}{(2 \cos \theta-2 \cos 2 \theta)}
$$
\begin{aligned}
\left.\therefore \frac{d^2 y}{d x^2}\right|_{\theta=\pi} & =\frac{(-1-1)(2+1)-(0-0)(-0+0)}{2(-1-1)^3} \\
& =\frac{-2 \times 3}{-2 \times 8}=\frac{3}{8}
\end{aligned}
$

Solutions · 1