<s>[INST] Equation of curve $x=a(\theta- \sin\theta), y=a(1-\cos\theta)$ for varying $\theta$ Let the equation of a curve be $x=a(\theta- \sin\theta), y=a(1-\cos\theta)$. If $\theta $ changes at a constant rate $k$ then the rate of change of the slope of the tangent to the curve at $\theta = \dfrac{\pi}{3}$ is_______.
 [/INST] Notice, we have $$x=a(\theta-\sin\theta)$$ $$\implies \frac{dx}{d\theta}=a(1-\cos\theta)$$ 
$$y=a(1-\cos\theta)$$ $$\implies \frac{dy}{d\theta}=a\sin\theta$$
Hence, the slope $m$ of the tangent to the curve $$m=\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$$ $$=\frac{a\sin\theta}{a(1-\cos\theta)}$$$$m=\frac{\sin\theta}{1-\cos\theta}$$ Since, $\theta$ is changing at a rate of $\frac{d\theta}{dt}=k$ hence, by differentiating above equation w.r.t. $t$, we get $$\frac{d}{dt}(m)=\frac{d}{dt}\left(\frac{\sin\theta}{1-\cos\theta}\right)$$ 
$$\frac{dm}{dt}=\frac{(1-\cos\theta)\cos\theta\frac{d\theta}{dt}-\sin^2\theta\frac{d\theta}{dt}}{(1-\cos\theta)^2}$$ 
Now, rate of change of slope at $\theta=\frac{\pi}{3}$
$$\left(\frac{dm}{dt}\right)_{\theta=\frac{\pi}{3}}=\frac{(1-\cos\frac{\pi}{3})\cos\frac{\pi}{3}(k)-\sin^2\frac{\pi}{3}(k)}{\left(1-\cos\frac{\pi}{3}\right)^2}$$ 
$$=k\frac{(1-\frac{1}{2})\frac{1}{2}-\left(\frac{\sqrt 3}{2}\right)^2}{\left(1-\frac{1}{2}\right)^2}$$  $$=\color{blue}{-\frac{k}{2}}$$
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