# Calculating the length of a curve

Let's say I have a parabola that I know the equation of. I then asked myself the question "how do I calculate the length of the curve between two values of x, for example.

After thinking about it, I realised I could use pythagoras: √δx + δy will give me a length, and then I could find the sum of these lengths to give the length of the curve. However, given that δx is 'infinitessimally small,' I would have to add and infinite number of lengths to get the length of the curve.

For example:

given $y = x^{2}$ find the length of the curve while $-2 \leq x \leq 2$

so then (assuming my idea is correct), it will be:

The sum of $\sqrt{\delta x + (\delta x)^{2}}$ for all values of x between -2 and 2.

Am I along the right lines here? And if so, how do I calculate this 'infinite sum.'

I really would like to figure this out myself, so if (as i'm sure there is) a way to calculate this, then please don't tell me it. Some pointers would be great though!

Many thanks

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#### Millennial

Actually, there is a well known formula for this. Deriving the length of a quadratic equation is possible, but, I must say, very tedious. It involves one page of calculation usually, but your definition of tedious might be different than mine!

Anyways, to the topic. You are right about needing to add infinitesimally small quantities infinitely many times, and that process is called integration in mathematics. Basically, the logic here is to take infinitesimally small triangles, such that we denote the arms of the right angle by dx and dy (d standing for differential, somewhat a synonym for infinitesimal). You already have it figured out (probably) that the length of the infinitesimal arc is then $\sqrt{(dx)^2+(dy)^2}$, from Pythagoras.

The reason we are allowed to do this is because the order of error is very small (actually it is of the order of two differentials) and can be neglected. If it was on the order of one, we could not neglect it. Note that "neglecting" in mathematics does not mean we obtain a wrong answer after doing all the stuff. "Neglectable" simply means that it is too small (too close to zero) to have any effect. Remember that infinitesimals are closer to zero than anything you can imagine, and think of them squared!

So we got that problem settled, now we need to sum all these infinitesimal arcs! As I said, this process is called integration in mathematics. If you do not know about integral calculus, you might need to study that first and then come to this topic again.(hopefully, you will not see the formula while studying, for it is very commonly used!)

Ok, well i have studied enough integration to know that you get the area below the curve by doing it, and the rule for it is:

$∫x^{n}dx = \frac{x^{n+1}}{n+1} + C$

Evidently this is the sum of all the y values given by the function times by dx.

So how can I apply integration to this situation?

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#### Klungo

It's easier to think of it as differentials. We could let √[(dx)²+(dy)²] = (dz) because they're infinitesimals. But also, dz is the length of the triangle with the hypothenuse on the curve part and legs dx and dy as the previous poster said. We could solve for z in terms of x (or y) by integrating with respect to either. Now, manipulate √[(dx)²+(dy)²] so that it is in terms of x (or y) only. Hint: Since y is a polynomial, dy/dx is in terms of x only. In the end, we integrate, z = int dz = int √[(dx)²+(dy)²]. (<- where's the dx?)

#### Millennial

(Riemann) integration is defined in terms of a Riemann sum:
$$\int_{a}^{b} f(x)dx=\lim_{\delta t\to 0}\sum_{i=1}^{n}f(t_i)\delta t$$
where $t_i$ is a partition of the interval [a,b] such that $a=t_0<t_1<t_2\text{...}<t_n=b$.
Can you attempt to write the infinite sum of small arcs in terms of a sum like this?

$$\int_{a}^{b} f(x)dx=\lim_{\delta t\to 0}\sum_{i=1}^{n}f(t_i)\delta t$$

where ti is a partition of the interval [a,b] such that a=t0<t1<t2...<tn=b.
Can you attempt to write the infinite sum of small arcs in terms of a sum like this?
I wouldn't know where to begin with this.

It's easier to think of it as differentials. We could let √[(dx)²+(dy)²] = (dz) because they're infinitesimals. But also, dz is the length of the triangle with the hypothenuse on the curve part and legs dx and dy as the previous poster said. We could solve for z in terms of x (or y) by integrating with respect to either. Now, manipulate √[(dx)²+(dy)²] so that it is in terms of x (or y) only. Hint: Since y is a polynomial, dy/dx is in terms of x only. In the end, we integrate, z = int dz = int √[(dx)²+(dy)²]. (<- where's the dx?)
As for manipulating this so it is in terms of x, is this what you mean: (let y = x2)

$\sqrt{(dx)^{2} + (dy)^{2}}$ then substitute y with x2

$\sqrt{(dx)^{2} + ((dx)^{2})^{2}}$

Unfortunately, the rest of this confuses me slightly :/

EDIT: Would it help you to know that as far as integration is concerned, I have only covered antiderivatives, definite integrals, areas bound by a curve and the x axis, and the area between 2 curves. (I have studied differentiation)

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#### Millennial

Do you know about methods like the disk method or the shell method, which involve using integration to sum infinitesimal quantities infinitely many times? They are not involved with this particular problem, but knowing them gives you a good idea how to use integration in the fashion you need to.

Do you know about methods like the disk method or the shell method, which involve using integration to sum infinitesimal quantities infinitely many times? They are not involved with this particular problem, but knowing them gives you a good idea how to use integration in the fashion you need to.
No, I have never even heard of these

#### Vorde

No, I have never even heard of these
Then you won't have any luck getting exact answers. Finding curve length is a classic integral calculus problem, the difficulty of which goes from easy to practically impossible depending on the curve.

Without calculus, you could approximate the length of the curve by taking a bunch of values along the curve in question, jotting down the x-difference between the values and the y-difference between the values. Then you could use the Pythagorean theorem like you and others have talked about on each individual pair of points, and then add up all of those to approximate the true lenght.

Ok, I will l put this problem on standby until I have studied more calculus

#### arildno

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You deserve, anyhow, to be be given a precise formula of HOW your particular problem can be approximated by a long sum:

Suppose you pick N points ON your parabola, let their coordinates be (x_i, y_i), where $y_{i}=x_{i}^{2}$, and "i" counts from 1 to N (let, for simplicity, x-values increase with value of "i").
Then, using stepwise summation by invoking the Pythagorean theorem, the appoximated length L is given by:
$$L=\sum_{i=1}^{i=N-1}\sqrt{(x_{i+1}-x_{i})^{2}+(y_{i+1}-y_{i})^{2}}$$
Now, using the equation for the parabola, you get:
$$L=\sum_{i=1}^{i=N-1}\sqrt{(x_{i+1}-x_{i})^{2}+(x_{i+1}^{2}-x_{i}^{2})^{2}}$$
Remembering from algebra that $a^{2}-b^{2}=(a-b)(a+b)$ for every numbers a,b, you may write:
$$L=\sum_{i=1}^{i=N-1}\sqrt{(x_{i+1}-x_{i})^{2}+(x_{i+1}-x_{i})^{2}(x_{i+1}+x_{i})^{2}}$$
or, by extracting the common factor:
$$L=\sum_{i=1}^{i=N-1}\sqrt{1+(x_{i+1}+x_{i})^{2}}(x_{i+1}-x_{i})$$
Now, if the difference in value between x_i and x_(i+1) is VERY SMALL, they are approximately EQUAL, and thus the sum of those two number is practically the double of x_i (agreed?)
Thus, you get the approximate formula (which gets more and more correct the tinier the differences between x-values are):
$$L\approx\sum_{i=1}^{i=N-1}\sqrt{1+4x_{i}^{2}}(x_{i+1}-x_{i})$$

The more evaluation points you put into your fixed length, i.e, the bigger you make N as well as minimizing the difference between consecutive values of x, the more accurate your formula gets to approximate the arc length of the parabola.

The exact value of the arc length of the parabola from -2 and 2 is, BTW:
$$L=2\sqrt{5}+\ln(2+\sqrt{5})$$

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Thankyou kindly :)

#### arildno

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Here is the correct general formula:

$$L=a\sqrt{1+4a^{2}}+\frac{1}{2}\ln(2a+\sqrt{1+4a^{2}})$$

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one thing i didn't understand was that you put i=N-1 on top of the $\sum$ I understood that the number on top is the finishing value of i in the function. Here it looks like it isn't the case, so what does this mean?

#### arildno

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one thing i didn't understand was that you put i=N-1 on top of the $\sum$ I understood that the number on top is the finishing value of i in the function. Here it looks like it isn't the case, so what does this mean?
I'm summing together the N-1 length segments that lie between the N points.

#### arildno

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Here is the correct general formula:

$$L=a\sqrt{1+4a^{2}}+\frac{1}{2}\ln(2a+\sqrt{1+4a^{2}})$$

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#### pierce15

The formula for a general curve is:

$$L=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^{2}}dx$$

As you do more research on integration, you will find that you usually will need a trig. substitution to integrate this, and it will often be very nasty

I hope that you can read that, I'm new to latex and it took me about 10 mins to type!

It's not coming out on my screen, anyone know what happened?

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#### arildno

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You write it as:
$$L=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^{2}}dx$$
You may look at the code by making a right hand click on it.
The command begins with "tex" within rectangular brackets, and ends with "\tex" within rectangular brackets, without apostrophes.

#### pierce15

You write it as:
$$L=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^{2}}dx$$
You may look at the code by making a right hand click on it.
The command begins with "tex" within rectangular brackets, and ends with "\tex" within rectangular brackets, without apostrophes.
Thanks a lot, I edited it the exact way that you had it with the tex and \tex, but it still didn't come out...

#### arildno

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Sorry about that. It should be "/tex" in the last bracket, not "\tex"

#### pierce15

Sorry about that. It should be "/tex" in the last bracket, not "\tex"
What's wrong with it now? Haha, this is more complicated than it should be.

Edit: nevermind! got it.

By the way, back to the problem. We will have:

$$L=\int_{-2}^{2}\sqrt{1+(\frac{dy}{dx})^{2}}dx$$

Our function is $$y=x^2$$, so the derivative is 2x (OP if you want to see why, it's because of the power rule).

$$L=\int_{-2}^{2}\sqrt{1+(2x)^{2}}dx$$
$$L=\int_{-2}^{2}\sqrt{1+4x^{2}}dx$$

Here, would the best substitution be $$4x=tan(t)$$? I did it out that way and I got the answer:

$$ln(4x+\frac{4x}{sin(arctan(4x))})$$

by drawing the triangle with those trig functions, it simplifies to:

$$ln(4x+\frac{4x}{\frac{4x}{\sqrt{1+4x}}}$$
$$ln(4x+\sqrt{1+4x})$$

plugging in the limits of integration:

$$ln(8+\sqrt{1+8})-ln(-8+\sqrt{1-8})$$

OK I'm getting complex numbers, what did I do wrong??

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#### arildno

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You should use tan(t)=2x, not 4x.
Besides, your area element is probably wrong.
furthermore, it is much simpler to use a hyperbolic substitution, 2x=Sinh(u), rather than a trigonometric substitution.

#### pierce15

Here's another way that you can do it, since the curve $$y=x^2$$ is symmetrical.

$$\int_-2^2 \sqrt{1+(4x)^2}dx$$
$$2\int_0^2 \sqrt{1+4x^2}dx$$

Substitute 2x=tan(x):

$$2\int_0^2 \sqrt{1+tan^2(x)}$$
$$2\int_0^2 sec(x)dx$$
$$2ln(tan(x)+sec(x))-ln(tan(0)+sec(0)$$
$$2ln(tan(x)+sec(x)-ln(1)$$
$$2ln(2x+\sqrt{1+4x^2})$$
$$2ln(4+\sqrt{65})$$

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