Taking the second derivative of a parametric curve

Question

I understand that for the parametric equations

$$\begin{align*}x&=f(t)\\ y&=g(t)\end{align*}$$

If $F(x)$ is the function with parameter removed then $\displaystyle F'(x) = \frac{\text{d}y}{\text{d}t}\big/\frac{\text{d}x}{\text{d}t}$

But the procedure for taking the second derivative is just described as " replace $y$ with dy/dx " to get

$$\frac{\text{d}^2y}{\text{d}x^2}=\frac{\text{d}}{\text{d}x}\left(\frac{\text{d}y}{\text{d}x}\right)=\frac{\left[\frac{\text{d}}{\text{d}t}\left(\frac{\text{d}y}{\text{d}t}\right)\right]}{\left(\frac{\text{d}x}{\text{d}t}\right)}$$

I don't understand the justification for this step. Not at all.

But that's all my book says on the matter then it launches in to plugging things in to this formula, and it seems to work well enough, but I don't know why.

I often find answers about question on differentials are beyond my level, I'd really like to get this, it'd mean a lot to me if someone could break it down.

Answer 1

Consider

$$ \begin{align*} \frac{\text{d}^2y}{\text{d}x^2}&=\frac{\text{d}}{\text{d}x}\left(\frac{\text{d}y}{\text{d}x}\right)\\ &=\frac{\text{d}}{\text{d}t}\left(\frac{\text{d}y}{\text{d}x}\right).\frac{\text{d}t}{\text{d}x}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot \frac{1}{\frac{dx}{dt}}\\ \end{align*}$$
where the last equality is as a result of applying the chain rule.

Answer 2

Their justification is that you can use the same process for $\frac{dy}{dx}$ as for $Y$ since you can now consider $Y_2 = g_2(t) = \frac{dy}{dx}(t)$, that is, you once again have a parametric equation in terms of the parameter $t$, and the parametric equation for $x$ stays the same.

Answer 3

I've come up with a 'simple enough' proof for the parametric derivative. Begin by consider a curve parameterized in the way :$<x(t), y(t) >$, then I will rewrite derivative operator with $x$ :

$$ \frac{dy(\theta)}{dx} = \frac{d \theta}{dx} \frac{d y}{d \theta}$$ The above equality holds by the chain rule, derivative of inner function times derivative of outer function, now I will remove the $y$:

$$ \frac{d}{dx} = \frac{d \theta}{dx} \frac{d}{ d \theta} \tag{1}$$

Now, the above can be thought of as an operator which eats a function and maps to another function, for example if we multiply both side on right by $y$ , then it gives the $ \frac{dy}{dx}$ with the parametric derivative of $y$.

And, I will introduce one more identity:

$$ \theta( x( \theta) ) = \theta \tag{2}$$

The above can be thought of in the following way: The function $x(\theta)$ maps from $ \theta \to x$ , the function $ \theta(x)$ maps from $ x \to \theta$, simply the inverse mapping and hence above equality holds true for at least a small interval of where we can say that $ \theta $ and $x$ are bijective functions of each other. Taking the derivate of $(2)$ with $ \theta$,:

$$ \frac{ d \theta}{dx} \frac{dx}{d \theta} = 1$$

Or,

$$ \frac{d\theta}{dx} = \frac{1}{ \frac{ d x}{d \theta}} \tag{2}$$

Hence, we can rewrite (1) as:

$$ \frac{d}{dx} = \frac{ \frac{d }{d \theta} }{\frac{dx}{d \theta}} \tag{3}$$

Now, multiply $y$ on right side of (3): $$ \frac{dy}{dx} = \frac{ \frac{dy }{d \theta} }{\frac{dx}{d \theta}} \tag{4}$$

This is the first parameteric derivative. Now, use (3) on equation (4):

$$ \frac{d^2 y}{dx^2} = \frac{1}{ \frac{dx}{d \theta}} \left[ \frac{ \frac{d^2 y}{d \theta^2} \frac{dx}{d \theta} - \frac{dy}{d \theta} \frac{dx}{ d \theta}}{ ( \frac{dx}{d \theta})^2} \right]= \left[ \frac{ \frac{d^2 y}{d \theta^2} \frac{dx}{d \theta} - \frac{dy}{d \theta} \frac{dx}{ d \theta}}{ ( \frac{dx}{d \theta})^3} \right]$$

Which is what we sought to prove.

Answer 4

I'm not strong enough to know when you can treat dy/dx as a "fraction" and when you can't. But the way I reasoned it was as follows...

dy/dx = dy/dt * dt/dx (very chain rule-esque)

d2y/dx2 = d2y/dxdt * dt/dx

Which seems entirely consistent with Nana's answer!