Question

Find the value of derivatives $\dfrac{{{d}^{2}}x}{d{{t}^{2}}},\dfrac{{{d}^{2}}y}{d{{t}^{2}}},\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ from the given the parametric equations $x=a\left( \cos t+t\sin t \right)$ and $y=a\left( \sin t-t\cos t \right)$ where the domain of $t$ is given as $0< t <\dfrac{\pi }{2}$.

Answer 1

The given pair equations are in parameterized form where both dependent and independent variables are expressed in terms of a parameter $t$. Determine the parametric derivative of $x$ with respect to $t$ and derivative of $y$ with respect to $t$. Differentiate second time to determine the values of $\dfrac{{{d}^{2}}x}{d{{t}^{2}}}\text{ and }\dfrac{{{d}^{2}}y}{d{{t}^{2}}}$. Use the values of $\dfrac{dx}{dt}\text{ and }\dfrac{dy}{dt}$ obtained during calculation to determine the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$.

Complete step by step answer:
The first given parametric equation is $x=a\left( \cos t+t\sin t \right)$. If $f\left( x \right)\text{ and }g\left( x \right)$ are two differentiable functions then by product rule $$\dfrac{d}{dx}\left( f\left( x \right)g\left( x \right) \right)=g\left( x \right)\dfrac{d}{dx}f\left( x \right)+f\left( x \right)\dfrac{d}{dx}g\left( x \right)$$ Differentiating both side with respect to $t$ using product rule if needed, $\begin{aligned} & x=a\left( \cos t+t\sin t \right) \\\ & \Rightarrow \dfrac{dx}{dt}=a\dfrac{d}{dt}\left( \cos t+t\sin t \right) \\\ & \Rightarrow \dfrac{dx}{dt}=a\left( -\sin t+t\cos t+\sin t \right)=at\cos t....(1) \\\ \end{aligned}$
Differentiating again with respect to $t$ both side
$\dfrac{{{d}^{2}}x}{d{{t}^{2}}}=\dfrac{d}{dt}\left( at\cos t \right)=a\left( -t\sin t+\cos t \right)$
The second given parametric equation is$y=a\left( \sin t-t\cos t \right)$.$$$$
Differentiating both side with respect to $t$ using product rule if needed,

& y=a\left( \sin t-t\cos t \right) \\\ & \Rightarrow \dfrac{dy}{dt}=a\dfrac{d}{dt}\left( \sin t-t\cos t \right) \\\ & \Rightarrow \dfrac{dy}{dt}=a\left( \cos t-\cos t+t\sin t \right) \\\ & \Rightarrow \dfrac{dy}{dt}=at\sin t....(2) \\\ \end{aligned}$$ Differentiating again with respect to $t$ both side $$\dfrac{{{d}^{2}}y}{d{{t}^{2}}}=\dfrac{d}{dt}\left( at\sin t \right)=a\left( \sin t+t\cos t \right)$$ We know that in case of parameterized equations the domain of the parameter is same as the domain of the dependent variable. Here the domains of $x\text{ and }y$ are same as the domain of $t$, that means $0< x,y,t <\dfrac{\pi }{2}$. Hence we can express $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ in terms of $\dfrac{dx}{dt}\text{ and }\dfrac{dy}{dt}$. $$\dfrac{dy}{dx}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\text{ }$$ Now we shall substitute the values $\dfrac{dx}{dt}\text{ and }\dfrac{dy}{dt}$ obtained from (1) and (2). $$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{at\sin t}{at\cos t}=\tan t$$ Differentiating again with respect to $t$, $$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \tan t \right)=\dfrac{d}{dt}\left( \tan t \right)\cdot \dfrac{dt}{dx}=\dfrac{{{\sec }^{2}}t}{\dfrac{dx}{dt}}$$ Again substituting $\dfrac{dx}{dt}\text{ }$above, $$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{{{\sec }^{2}}t}{\dfrac{dx}{dt}}=\dfrac{{{\sec }^{2}}t}{at\cos t}=\dfrac{{{\sec }^{2}}t}{at\dfrac{1}{\sec t}}=\dfrac{{{\sec }^{3}}t}{at}$$ **Therefore the obtained values are $$\dfrac{{{d}^{2}}x}{d{{t}^{2}}}=a\left( -t\sin t+\cos t \right),\dfrac{{{d}^{2}}y}{d{{t}^{2}}}=a\left( \sin t+t\cos t \right)$$ and $$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{{{\sec }^{3}}t}{at}$$** **Note:** The question tests your concept of parametric equations. Parametric equations are used when an equation cannot be expressed either in implicit or explicit form. The domains of $x,y$and $t$ are the same. That is exactly why we can express $\dfrac{dy}{dx}$ in terms of $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$.