Explanation behind Second Derivative of a Parametric Equation Formula

Question

I am looking to find out how the second derivative formula works. I can blindly apply it but I don't have a grasp of what is going on or why.

The first derivitive has the formula $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. I told this was because if you divide them you cancel out the dts and get dy/dx.

The second derivative has the formula $\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{ \frac{d}{dt}\left(\frac{dy}{dx}\right) }{\frac{dx}{dt}}$. Is this something that can be explained with only a knowledge of calculus or is this something I would have to take on faith until I learn more. Any help would be greatly appreciated.

Answer 1

Note $$\frac{d}{dx}\left(\frac{dy}{dt}\right)=\frac{dt}{dx}\frac{d}{dt}\left(\frac{dy}{dt}\right)=\frac{dt}{dx}\frac{d^2y}{dt^2}$$ and $$\frac{d}{dx}\left(\frac{dx}{dt}\right)=\frac{dt}{dx}\frac{d^2x}{dt^2}$$

Use the quotient rule $$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy/dt}{dx/dt}\right)=\frac{\frac{d}{dx}(\frac{dy}{dt})\frac{dx}{dt}-\frac{d}{dx}(\frac{dx}{dt})\frac{dy}{dt}}{(\frac{dx}{dt})^2}=\frac{\frac{dt}{dx}\frac{d^2y}{dt^2}\frac{dx}{dt}-\frac{dt}{dx}\frac{d^2x}{dt^2}\frac{dy}{dt}}{(\frac{dx}{dt})^2}$$ $$=\frac{\frac{d^2y}{dt^2}-\frac{d^2x}{dt^2}\frac{dy}{dx}}{(\frac{dx}{dt})^2}$$ Now multiply top and bottom of the fraction by $\frac{dx}{dt}$: $$=\frac{\frac{dx}{dt}\frac{d^2y}{dt^2}-\frac{dy}{dt}\frac{d^2x}{dt^2}}{(\frac{dx}{dt})^3}$$ This is the correct formula for the second derivative of a parametric equation.

Answer 2

$f’(x)=\frac{dy}{dx}$ is a function of $x$, applying the chain rule gives:

$$\frac{df’}{dx} \frac{dx}{dt}=\frac{df’}{dt}=\frac{d}{dt} f’ $$

Now solve for $f’’(x)= \frac{df’}{dx} $ assuming $\frac{dx}{dt} \neq 0$.

Answer 3

So we have two functions $x = x(t)$ and $y = y(t)$ defined for $t \in [a,b]$.

For every $t$ we can combine the values of those functions into a point with coordinates $(x(t),y(t))$. Then, if we consider all the points for every $t\in[a,b]$, we'll get some curve $L$, which is called a parametric curve with parameter $t$.

Choose some $t_0$ from $[a,b]$, which would give you a point on a curve $(x(t_0),y(t_0)) = (x_0,y_0) \in L$. The whole curve can be really wild - intersecting itself, going vertically up or down, etc. - but in some small enough neighbourhood of $t_0$ it will (usually) look like a graph of some function $Y(x)$. Knowing that both $x$ and $y$ are actually functions of $t$ we can write that $x = x(t)$ and $y=y(t)=Y(x(t))$ (near $(x_0,y_0)$).

Now, we can analyse $Y(x)$ using plain old calculus and, in doing that, we can analyse the parametric curve near $(x_0,y_0)$.

Let's find the first derivative $Y'(x)$: $$ \begin{array}{l} y(t) = Y(x(t)) \\ y'(t) = Y'(x(t)) \cdot x'(t) \\ Y'(x(t)) = \frac{x'(t)}{y'(t)} \end{array} $$

or $\frac{dY}{dx}$, in dx-dy notation:

$$ \begin{array}{l} \frac{dy}{dt} = \frac{dY}{dx} \frac{dx}{dt} \\ \frac{dY}{dx} = \frac{dx}{dt} / \frac{dy}{dt} \end{array} $$

So now we can plug $t$'s in the formula we got and that would give us the rise-over-run values for $(x(t),y(t))$, so we can judge whether the curve's going up or down, draw our tangent lines, etc.

Now, for the second derivative $Y''(x)$ ! $$ \begin{array}{l} (Y'(x(t)))^{'}_t = (Y'(x(t)))^{'}_x \cdot x'(t) \\ (Y'(x(t)))^{'}_t = Y''(x(t)) \cdot x'(t) \\ Y''(x(t)) = \frac{(Y'(x(t)))^{'}_t}{x'(t)} \end{array} $$

or, using dx-xy notation, we want to find $\frac{d^2 Y}{dx^2}$:

$$ \begin{array}{l} \frac{d}{dt}\left[\frac{dY}{dx}\right] = \frac{d^2 Y}{dx^2} \frac{dx}{dy} \\ \frac{d^2 Y}{dx^2} = \frac{d}{dt}\left[\frac{dY}{dx}\right] / \frac{dx}{dy} \end{array} $$

So now you can tell whether your $Y(x)$ is convex or concave, which helps to better understand how $L$ looks.

We never actually fixed $t_0$ to some number, so these derivatives would be valid for every $t_0$, for which they actually exist (no division by zero and stuff like that). The only problem is there might exist another value $t_1$, such that $x(t_1)=x(t_0)=x_0$ and $y(t_1)=y(t_0)=y_0$, i.e. the curve might intersect itself at $(x_0,y_0)$. If so, there would be multiple values of $Y'(x(t))$ and $Y''(x(t))$ at $(x_0,y_0)$ - one for every $t_i$, such that $(x(t_i),y(t_i)) = (x_0,y_0)$. Which makes sense: if a curve intersects itself at a point, you can draw multiple tangent lines to it there.