Question

If x = a $(cos\theta +logtan\frac{\theta }{2})andy=asin\theta$, then find the value of $\frac{d^{2}y}{d{x}^2}at\theta =\frac{\pi }{4}.$

Answer 1

$x=a[cos\theta +logtan\frac{\theta }{2}]$
$\frac{dx}{d\theta }=a[-sin\theta +\frac{1}{tan\frac{\theta }{2}}se{c}^2\frac{\theta }{2}\frac{1}{2}]$
$=-asin\theta +\frac{a}{2sin\frac{\theta }{2}cos\frac{\theta }{2}}$
$=\frac{a}{sin\theta }-asin\theta =\frac{a(1-si{n}^2\theta )}{sin\theta }$
$=\frac{aco{s}^2\theta }{sin\theta }$
$\frac{d^{2}x}{d{\theta }^2}=a\left[\frac{sin\theta \left(2cos\theta \right(-sin\theta \left)\right)-co{s}^2\theta \left(cos\theta \right)}{si{n}^2\theta }\right]$
$=a\left[\frac{-2si{n}^2\theta cos\theta -co{s}^3\theta }{si{n}^2\theta }\right]$
$=a\left[\frac{-2cos\theta (1-co{s}^2\theta )-co{s}^3\theta }{si{n}^2\theta }\right]$
$=a\left[\frac{-2cos\theta +2co{s}^3\theta -co{s}^3\theta }{si{n}^2\theta }\right]$
$=a\left[\frac{co{s}^3\theta -2cos\theta }{si{n}^2\theta }\right]$
$y=asin\theta$
$\frac{dy}{d\theta }=acos\theta$
$\frac{d^{2}y}{d{\theta }^2}=-asin\theta$
$\frac{d^{2}y}{d{\theta }^2}=\frac{-asin\theta }{a\left[\frac{co{s}^3\theta -2cos\theta }{si{n}^2\theta }\right]}$
$=\frac{-si{n}^3\theta }{co{s}^3\theta -2cos\theta }$
$\frac{d^{2}y}{d{x}^2}{|}_{\theta =\frac{\pi }{4}}=\frac{-si{n}^3\frac{\pi }{4}}{co{s}^3\frac{\pi }{4}-2cos\frac{\pi }{4}}$
$=\frac{-{\left(\frac{1}{\sqrt{2}}\right)}^3}{{\left(\frac{1}{\sqrt{2}}\right)}^3-2\left(\frac{1}{\sqrt{2}}\right)}=\frac{-\frac{1}{2\sqrt{2}}}{\frac{1}{2\sqrt{2}}-\sqrt{2}}$
$=\frac{-\frac{1}{2\sqrt{2}}}{\frac{1-2}{2\sqrt{2}}}=\frac{-1}{-1}=1$.