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

Hint: 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,
$\begin{align}
  & x=a\left( \cos t+t\sin t \right) \\
 & \Rightarrow \dfrac{dx}{dt}=a\dfrac{d}{dt}\left( \cos t+t\sin t \right) \\
 & \Rightarrow \dfrac{dx}{dt}=a\left( -\sin t+t\cos t+\sin t \right)=at\cos t....(1) \\
\end{align}$ \[\]
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,
\[\begin{align}
& y=a\left( \sin t-t\cos t \right) \\
& \Rightarrow \dfrac{dy}{dt}=a\dfrac{d}{dt}\left( \sin t-t\cos t \right) \\
 & \Rightarrow \dfrac{dy}{dt}=a\left( \cos t-\cos t+t\sin t \right) \\
 & \Rightarrow \dfrac{dy}{dt}=at\sin t....(2) \\
\end{align}\]

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 $\dfrac{dx}{dt}$.