Question

If $x = a\left( {\cos t + \log \tan \dfrac{t}{2}} \right)$ , and $y = a\sin t$ , then find $\dfrac{{{d^2}y}}{{d{x^2}}}$ at $t = \dfrac{\pi }{3}$ .

Answer 1

Hint : To find $\dfrac{{{d^2}y}}{{d{x^2}}}$ we will differentiate $x$ and $y$ with respect to $t$ . Then, we will divide expressions to get the expression for $\dfrac{{dy}}{{dx}}$ .We will again differentiate $\dfrac{{dy}}{{dx}}$ with respect to $x$ to get the expression for $\dfrac{{{d^2}y}}{{d{x^2}}}$ . Then we will substitute the value of $\dfrac{{dx}}{{dt}}$ in the expression of $\dfrac{{{d^2}y}}{{d{x^2}}}$ . After that we will substitute the value of $t$ given in question to get the final answer.
Formula used:
We are using following trigonometric formulas:
${\mathop{\rm Sin}\nolimits} 2x = 2\sin x\cos x$
${\cos ^2}x + {\sin ^2}x = 1$

Complete step-by-step answer :
We have given expressions as $x = a\left( {\cos t + \log \tan \dfrac{t}{2}} \right)$ , and $y = a\sin t$ .
We will differentiate $x$ with respect to $t$ which can be expressed as:
$\dfrac{{dx}}{{dt}} = a\left( { - \sin t + \dfrac{1}{{\tan \dfrac{t}{2}}} \cdot {{\sec }^2}\dfrac{t}{2} \cdot \dfrac{1}{2}} \right)$
We will write $\tan \dfrac{t}{2}$ as $\dfrac{{\sin \dfrac{t}{2}}}{{\cos \dfrac{t}{2}}}$ and ${\sec ^2}\dfrac{t}{2}$ as $\dfrac{1}{{{{\cos }^2}\dfrac{t}{2}}}$ in the above expression, we will get

\Rightarrow \dfrac{{dx}}{{dt}} = a\left( { - \sin t + \dfrac{{\cos \dfrac{t}{2}}}{{\sin \dfrac{t}{2}}} \cdot \dfrac{1}{{{{\cos }^2}\dfrac{t}{2}}} \cdot \dfrac{1}{2}} \right)\\\ \Rightarrow \dfrac{{dx}}{{dt}} = a\left( { - \sin t + \dfrac{1}{{2\sin \dfrac{t}{2}\cos \dfrac{t}{2}}}} \right) \end{array}$$ We know that $ {\mathop{\rm Sin}\nolimits} 2x = 2\sin x\cos x $ , hence using this expression in the above expression we get $$\begin{array}{l} \Rightarrow \dfrac{{dx}}{{dt}} = a\left( { - \sin t + \dfrac{1}{{\sin 2\left( {\dfrac{t}{2}} \right)}}} \right)\\\ \dfrac{{dx}}{{dt}} = a\left( { - \sin t + \dfrac{1}{{\sin t}}} \right)\\\ \Rightarrow \dfrac{{dx}}{{dt}} = a\left( {\dfrac{{ - {{\sin }^2}t + 1}}{{\sin t}}} \right) \end{array}$$ We know that $ {\cos ^2}x + {\sin ^2}x = 1 $ , we will use this expression in the above expression: $$\begin{array}{l} \Rightarrow \dfrac{{dx}}{{dt}} = a\left( {\dfrac{{{{\cos }^2}t}}{{\sin t}}} \right)\\\ \end{array}$$……(i) Now, we will differentiate $ y = a\sin t $ with respect to $ t $ . $ \dfrac{{dy}}{{dt}} = a \cdot \cos t $ ……(ii) We will divide equation (ii) by (i) to find $ \dfrac{{dy}}{{dx}} $ . $$\begin{array}{l} \dfrac{{{{dy} {\left/ {\vphantom {{dy} {dt}}} \right. } {dt}}}}{{{{dx} {\left/ {\vphantom {{dx} {dt}}} \right. } {dt}}}} = \dfrac{{a\cos t}}{{a\left( {\dfrac{{{{\cos }^2}t}}{{\sin t}}} \right)}}\\\ \dfrac{{dy}}{{dx}} = \dfrac{{\sin t}}{{\cos t}}\\\ \dfrac{{dy}}{{dx}} = \tan t \end{array}$$ We will differentiate the above expression with respect to $ x $ we will get $$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = {\sec ^2}t \cdot \dfrac{{dt}}{{dx}}$$ We will substitute $$a\left( {\dfrac{{{{\cos }^2}t}}{{\sin t}}} \right)$$for $ \dfrac{{dx}}{{dt}} $ in the above expression, we will get $$\begin{array}{l} \dfrac{{{d^2}y}}{{d{x^2}}} = {\sec ^2}t \cdot \dfrac{{\sin t}}{{a\left( {{{\cos }^2}t} \right)}}\\\ \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{{{\sec }^4}t\sin t}}{a} \end{array}$$ Now, we will substitute $ \dfrac{\pi }{3} $ in the above expression,

\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{{{\sec }^4}\left( {\dfrac{\pi }{3}} \right).\sin \dfrac{\pi }{3}}}{a}\\
\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{1}{a} \cdot {\left( 2 \right)^4} \cdot \dfrac{{\sqrt 3 }}{2}\\
\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{8\sqrt 3 }}{a}

Hence $ \dfrac{{{d^2}y}}{{d{x^2}}} $ at $ t = \dfrac{\pi }{3} $ is $ \dfrac{{8\sqrt 3 }}{a} $ . **Note** : The equation of $ y $ and $ x $ in the question are in the terms of $ t $ but we want $ \dfrac{{{d^2}y}}{{d{x^2}}} $ . Hence , we always need to differentiate equations separately with respect to $ t $ and then we can divide the expressions to get the required result. We are also using chain rule for the differentiation of different terms in the expression. We can get the final result by substituting the values given in question for different variables.