Let us know come to face the problem of computing the second derivative for functions defined through a pair of parametric equations:

Utilizing the results of the previous theorem on the first derivative, we can proceed -after having obtained the first derivative $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ as a function of the parameter $t$- to the computation of the second derivative either as

\begin{equation} \label{secder1st}

\frac{d^{2}y}{dx^{2}} = \frac{dy'}{dx} = \frac{dy'/dt}{dx/dt}

\end{equation}

where $y'=\frac{dy}{dx}$, or as

\begin{equation} \label{secder2nd}

\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\Big[ \frac{dy}{dx} \Big] = \frac{d}{dt}\Big[ \frac{dy}{dx} \Big] \cdot \frac{dt}{dx}

\end{equation}

Let us now procceed in a couple of clarifying examples:

\begin{array}{l}

x = sint \\

y = cos2t

\end{array}

\right. \ \ $, for $\ t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

It is clear that $x=sint$ is invertible in the domain $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and thus we can directly apply \eqref{secder2nd}:

Since $\frac{dx}{dt}=cost$ and $\frac{dy}{dt}=-2sin2t$, we have:

$$

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2sin2t}{cost} = \frac{-4 sint cost}{cost} = -4 sint

$$

therefore:

$$

\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\Big[ \frac{dy}{dx} \Big] = \frac{d}{dx}(-4 sint) = \frac{d}{dt}(-4 sint) \cdot \frac{dt}{dx} = (-4 cost) \cdot (\frac{1}{cost}) = -4

$$

where we have made use of the fact that $x=sint \Leftrightarrow t=arcsinx$ for $t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ and thus $\frac{dt}{dx} = (arcsinx)' = \frac{1}{cost} = \frac{1}{\sqrt{1-x^{2}}}$ for $t \in (-\frac{\pi}{2}, \frac{\pi}{2})$. (see previous post on the derivatives of the inverse trigonometric functions).

Notice that $\frac{dx}{dt} \Big|_{\pm \frac{\pi}{2}} = \frac{dy}{dt} \Big|_{\pm \frac{\pi}{2}}=0$ so these points are singular points and the above result does not apply at these points. Can you figure out what is happening at these points ? (plotting a graph of the parametric equations will probably help you understand the behavior at these singular points).

\begin{array}{l}

x = t^{2} \\

y = t^{3}

\end{array}

\right. \ \ $, for $\ t \in [-\infty, \infty]$ is called a

This curve (and the functions defined by it) can be equivalently described in implicit form by the equation $y^{2}=x^{3}$ (square $y(t)$ and cube $x(t)$ to see this!).

The graph of these parametric equations consists of two branches: the upper branch corresponding to the function $y=x^{3/2}$ while the lower branch is the graph of the function $y=-x^{3/2}$. These two branches meet at the origin, which corresponds to the value $t=0$.

Since $\frac{dx}{dt}=2t$ and $\frac{dy}{dt}=3t^{2}$ we can clearly see that the origin corresponds to a singular point of the graph.

We can readily apply the theorem to compute the first derivative (for either branch):

$$

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{d}{dt}(t^{3})}{\frac{d}{dt}(t^{2})} = \frac{3t^{2}}{2t} = \frac{3}{2}t

$$

while we will follow \eqref{secder1st} to compute the second derivative:

$$

\frac{d^{2}y}{dx^{2}} = \frac{dy'}{dx} = \frac{dy'/dt}{dx/dt} = \frac{\frac{d}{dt}(3t/2)}{\frac{d}{dt}(t^{2})} = \frac{3}{4t}

$$

Can you apply \eqref{secder2nd}, after suitably dividing the domain $(-\infty, \infty)$, to obtain the same results?

Can you figure out what is happening at the singular point $O(0,0)$ ?

Utilizing the results of the previous theorem on the first derivative, we can proceed -after having obtained the first derivative $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ as a function of the parameter $t$- to the computation of the second derivative either as

\begin{equation} \label{secder1st}

\frac{d^{2}y}{dx^{2}} = \frac{dy'}{dx} = \frac{dy'/dt}{dx/dt}

\end{equation}

where $y'=\frac{dy}{dx}$, or as

\begin{equation} \label{secder2nd}

\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\Big[ \frac{dy}{dx} \Big] = \frac{d}{dt}\Big[ \frac{dy}{dx} \Big] \cdot \frac{dt}{dx}

\end{equation}

**Remarks:****1.**Notice that -according to a previous remark in the computation of the first derivative- if we wish to apply \eqref{secder2nd}, a suitable partition of the domain $E \subseteq \mathbb{R}$ of $x=f(t)$, must be considered in order for $x=f(t)$ to be bijective ("1-1"). Thus, we will have $t=f_{i}^{-1}(x)$ (in the corresponding interval in $E$'s partition) and the correct formula for $\frac{dt}{dx}$ must be replaced.**2.**Notice that in general $$\frac{d^{2}y}{dx^{2}} \neq \frac{d^{2}y/dt^{2}}{d^{2}x/dt^{2}}$$Let us now procceed in a couple of clarifying examples:

__Example1:__Let us consider the pair of parametric equations $\left\{%\begin{array}{l}

x = sint \\

y = cos2t

\end{array}

\right. \ \ $, for $\ t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

It is clear that $x=sint$ is invertible in the domain $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and thus we can directly apply \eqref{secder2nd}:

Since $\frac{dx}{dt}=cost$ and $\frac{dy}{dt}=-2sin2t$, we have:

$$

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2sin2t}{cost} = \frac{-4 sint cost}{cost} = -4 sint

$$

therefore:

$$

\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\Big[ \frac{dy}{dx} \Big] = \frac{d}{dx}(-4 sint) = \frac{d}{dt}(-4 sint) \cdot \frac{dt}{dx} = (-4 cost) \cdot (\frac{1}{cost}) = -4

$$

where we have made use of the fact that $x=sint \Leftrightarrow t=arcsinx$ for $t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ and thus $\frac{dt}{dx} = (arcsinx)' = \frac{1}{cost} = \frac{1}{\sqrt{1-x^{2}}}$ for $t \in (-\frac{\pi}{2}, \frac{\pi}{2})$. (see previous post on the derivatives of the inverse trigonometric functions).

Notice that $\frac{dx}{dt} \Big|_{\pm \frac{\pi}{2}} = \frac{dy}{dt} \Big|_{\pm \frac{\pi}{2}}=0$ so these points are singular points and the above result does not apply at these points. Can you figure out what is happening at these points ? (plotting a graph of the parametric equations will probably help you understand the behavior at these singular points).

__Example2:__The curve given by the parametric equations $\left\{%\begin{array}{l}

x = t^{2} \\

y = t^{3}

\end{array}

\right. \ \ $, for $\ t \in [-\infty, \infty]$ is called a

*semicubic parabola*.This curve (and the functions defined by it) can be equivalently described in implicit form by the equation $y^{2}=x^{3}$ (square $y(t)$ and cube $x(t)$ to see this!).

The graph of these parametric equations consists of two branches: the upper branch corresponding to the function $y=x^{3/2}$ while the lower branch is the graph of the function $y=-x^{3/2}$. These two branches meet at the origin, which corresponds to the value $t=0$.

Since $\frac{dx}{dt}=2t$ and $\frac{dy}{dt}=3t^{2}$ we can clearly see that the origin corresponds to a singular point of the graph.

We can readily apply the theorem to compute the first derivative (for either branch):

$$

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{d}{dt}(t^{3})}{\frac{d}{dt}(t^{2})} = \frac{3t^{2}}{2t} = \frac{3}{2}t

$$

while we will follow \eqref{secder1st} to compute the second derivative:

$$

\frac{d^{2}y}{dx^{2}} = \frac{dy'}{dx} = \frac{dy'/dt}{dx/dt} = \frac{\frac{d}{dt}(3t/2)}{\frac{d}{dt}(t^{2})} = \frac{3}{4t}

$$

Can you apply \eqref{secder2nd}, after suitably dividing the domain $(-\infty, \infty)$, to obtain the same results?

Can you figure out what is happening at the singular point $O(0,0)$ ?

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