Calculate the length of parameterized curve. (Lacking intuitive understanding of subject)

Question

Problem

Calculate the length of parameterized curve which is:

$$ r(t)=(\frac{\sqrt{7}t^3}{3},2t^2)$$

in which $1 \le t \le 5$

Attempt to solve

We can express our parameterized curve in vector form.

$$ r(t)=\begin{bmatrix} \frac{\sqrt{7}t^3}{3} \\ 2t^2 \end{bmatrix} $$ where $r(t) \in \text{R}^2$

We should be able to figure out curve length by evaluating integral of length of tangent vector in $1 \le t \le 5$. Tangent vector would be simply defined as:

$$ r'(t)=\begin{bmatrix} \frac{d}{dt}(\frac{\sqrt{7}t^3}{3}) \\ \frac{d}{dt}(2t^2) \end{bmatrix} $$

$$ r'(t)=\begin{bmatrix}\sqrt{7}t^2 \\ 4t \end{bmatrix}$$

Length of tangent vector would be defined as:

$$ ||r'(t)||=\sqrt{(\sqrt{7}t^2)^2+(4t)^2} $$ $$ ||r'(t)||=\sqrt{7t^4+16t^2} $$

We have formula for curve length defined as:

$$ L=\int_{a}^{b}||r'(t)||dt$$ $$ [a,b]\rightarrow [1,5] $$ $$ L=\int_{1}^{5}\sqrt{7t^4+16t^2}dt $$ $$ L=\big|_{1}^{5} \frac{(t^2(7t^2+16))^{3/2}}{21t^3}+C $$ $$ L=\frac{(5^2(7\cdot 5^2+16))^{3/2}}{21\cdot 5^3}-\frac{(1^2(7\cdot1^2+16))^{3/2}}{21\cdot 1^3} $$ $$L=\frac{191\cdot\sqrt{191}}{21}-\frac{23\cdot\sqrt{23}}{21} $$ $$L\approx 120.4461139$$

---

I don't good intuitive understanding of how integrating the length of tangent vector of this curve would give us length of the curve in between this range ? Essentially we are calculating area in between x-axis and our tangent length function $||r'(t)||$ but how this area relates to the curve length that i dont know.

Now the problem could be that i don't understand the concept of integration as well as derivation. My understanding of integration is that we are summing up areas of infinitely small rectangles under given curve.

I understand derivation intuitively as when our $\frac{\Delta Y}{\Delta X}$ of given function or curve approaches zero. $\Delta$ as symbol of change. This can be expressed in terms of limits:

$$=\frac{\Delta f(a)}{\Delta (a)}=\frac{f(a+h)-f(a)}{(a+h)-(a)}=\frac{f(a+h)-f(a)}{h}$$ $$ f'(a)=\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}$$

If someone knows a good source that could give more of a intuitive insight on how this works or alternatively if someone wants to try explaining this that would be highly appreciated.

Answer 1

Consider the path $f(t) = (x(t), y(t))$. The length of the segment with endpoints $f(t)$ and $f(t+\Delta t)$ is

\begin{align} |f(t+\Delta t) - f(t)| &= |(x(t+\Delta t), y(t+\Delta t)) - (x(t), y(t))| \\ &= \sqrt{(x(t+\Delta t) - x(t))^2 + (y(t+\Delta t) - y(t))^2} \\ &= \Delta t \sqrt{ \left(\dfrac{x(t+\Delta t) - x(t)}{\Delta t}\right)^2 + \left(\dfrac{y(t+\Delta t) - y(t)}{\Delta t}\right)^2} \end{align}

So arc length becomes a (Reimann) sum of the form

\begin{align} \lim_{N \to \infty} \frac 1N\sum_{i=1}^N |f(t_i+\Delta t_i) - f(t_i)| &= \lim_{\substack N \to \infty \\ |\Delta t| \to 0} \frac 1N \sum_{i=1}^N \Delta t_i \sqrt{ \left(\frac{x(t_i+\Delta t_i) - x(t_i)}{\Delta t_i}\right)^2 + \left(\frac{y(t_i+\Delta t_i) - y(t_i)}{\Delta t_i}\right)^2} \\ &= \int_{t_i}^{t_f} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \ dt \end{align}

Another way to look at it

There is an unwritten rule for real-valued, differentiable functions $f:D \to \mathbb R$ where $D$ is an open subset of $\mathbb R$.

$$\text{If you look closely enough, $y=f(x)$ is a straight line.}$$

This is almost always a completely false statement. But it is also almost true most of the time and it is a very good source of "intuition". Physicists have been using this rule for a long time.

In this case, suppose we have a curve $\alpha(t) = (x(t), y(t))$, and we wish to compute $\dfrac{ds}{dt}$ where $s(t)$ is the arc length function. We consider the infinitesimal piece of arc from $\alpha = \alpha(t)$ to $\alpha + d\alpha = \alpha(t+dt)$

It follows that $ds = \sqrt{dx^2 + dy^2}$ and so $\dfrac{ds}{dt} = dt\sqrt{ \left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2}$

Answer 2

The tangent vector you are talking about is actually velocity in a sense. We have that $\mathbf{v} = \mathbf{v_x} + \mathbf{v_y}$ and also that $\mathbf{v_x} = \frac{dx}{dt} \mathbf{\hat{e_x}}$ and similarly $\mathbf{v_y}$.

$$\mathbf{v}= \frac{dx}{dt} \mathbf{\hat{e_x}}+\frac{dy}{dt} \mathbf{\hat{e_y}}$$

Also we know from pythagoras theorem that path length $dl = \sqrt{(dx)^2 + (dy)^2}$

$$|\mathbf{v}| dt = \sqrt{(dx)^2 + (dy)^2} = dl$$

So it follows from here that path length is given as $\int_{t_1}^{t_2} |\mathbf{v}| dt $.

Answer 3

When you increase $t$ by $dt$, $x$ and $y$ correspondingly increase by $dx$ and $dy$, and by Pythagoras the length travelled is the hypothenuses, $\sqrt{dx^2+dy^2}=\sqrt{\dot x^2+\dot y^2}dt$.

You replace the curve by a polyline of many segments and integrate.