Parametrization and arc length

Parametrization: case study. Consider our favorite example $y = x^{2}$. We may consider the curve from $(0, 0)$ to $(1, 1)$. Naturally, you think about a particle moving along the curve. We may consider the particle to be $(x(t), y(t)) = (t, t^{2})$ because then it will satisfy $y(t) = x(t)^{2}$. Notice that $$\frac{y'(t)}{x'(t)} = 2x,$$ which is equal to $y'(x) = 2x.$

In general, when $y$ is a differentialble function in $x$ and $x$ is a differentiable function in $t$ so that

• $y = y(x)$;
• $x = x(t)$.

Then we have $y(t) = y(x(t))$, so $y'(t) = y'(x(t))x'(t)$ by Chain Rule. Hence, we get $$y'(x) = y'(x(t)) = \frac{y'(t)}{x'(t)},$$ as long as $x'(t) \neq 0.$ In the above example, we had $x(t) = t$ and $y(x) = x^{2}.$ This formula can be remembered as: $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.$$ This gives a nice way to take derivatives.

Example. Consider the implicit function $y(x)$ that is given by the unit circle $x^{2} + y^{2} = 1.$ We can locally compute $y'(x)$ as follows: first, consider the parametrization given by

• $x(t) = \cos(t)$ and
• $y(t) = \sin(t)$.

Then $y'(\cos(t)) = y'(x(t)) = x'(t)/y'(t) = -\sin(t)/\cos(t) = -y(t)/x(t),$ so you can see that $y'(x) = -y/x$ when $x \neq 0$.

Arc length: case study. You know the length from $(0,0)$ to $(1,1)$ along the line $y = x$. Can you compute the length from $(0,0)$ to $(1,1)$ along the curve $y = x^{2}$?

Write $f(x) = x^{2}$. To compute the length, let's break the interval $[0, 1]$ into pieces. Each subinterval would have length $\Delta x$ and you can estimate the length you want by summing up lengths of line segments $\sqrt{(\Delta x)^{2} +(f(x + \Delta x) - f(x))^{2}}.$ Note that $$\sqrt{(\Delta x)^{2} +(f(x + \Delta x) - f(x))^{2}} = \sqrt{1 + \left(\dfrac{f(x + \Delta x) - f(x)}{\Delta x}\right)^{2}}\Delta x,$$ so when you sum them up and take $\Delta x \rightarrow 0$, you should get $$\textbf{Arc length} = \int_{0}^{1}\sqrt{1 + f'(x)^{2}}dx = \int_{0}^{1}\sqrt{1 + 4x^{2}} dx.$$

In general, if you have a smooth function $y = f(x)$, then $$\textbf{Arc length from } x = a \textbf{ to }  x = b \text{ is } \int_{a}^{b}\sqrt{1 + f'(x)^{2}}dx.$$

If you have a parametrization $(x(t), y(t)),$ again, thinking $y(t) = y(x(t))$, you may apply the change of variable in the above formula. That is, you consider $x = x(t)$ so that $dx = x'(t)dt$. Find $t = c$ such that $x(c) = a$ and $t = d$ such that $x(d) = b$, you get

\[\int_{c}^{d}\sqrt{1 + (y'(t)/x'(t))^{2}}x'(t)dt = \int_{c}^{d}\sqrt{x'(t)^{2} + y'(t)^{2}}dt.\]

You can consider the above as the arc length of the particle $(x(t), y(t))$ from time $t = c$ to $t = d$.

Memorization trick. Given a particle $(x(t), y(t))$, its velocity is $(x'(t), y'(t))$ and its speed is the size of the velocity, which is $\sqrt{x'(t)^{2} + y'(t)^{2}}$. You notice that:

Arc length is the integral of speed!

Exercise. Using arc length of the parametrization $(2\cos(t), 2\sin(t))$ of the circle of radius $2$ for $0 \leq t \leq 2\pi$, derive the usual fact that its circumference is equal to $4\pi$.

Exercise. Solve #4b of Exam 1, Winter 2017.

Exercise. Solve #6c of Exam 1, Fall 2016.

Exercise. Solve #3 of Exam 1, Fall 2015.

Exercise. Solve #8 of Exam 1, Fall 2015.