Injenueengado1zy8

2023-03-24

How to find the length of a curve in calculus?

marallocajcyb

Beginner2023-03-25Added 6 answers

In Cartesian coordinates for y = f(x) defined on interval [a,b] the length of the curve is

$\Rightarrow L={\int}_{a}^{b}\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}}dx$

In general, we could just write:

$\Rightarrow L={\int}_{a}^{b}ds$Solution:

Let's use Cartesian coordinates for this explanation.

If we consider an arbitrary curve defined as $y=f\left(x\right)$ and are interested in the interval $x\in [a,b]$, we can approximate the length of the curve using very tiny line segments.

Consider a point on the curve $P}_{i$. We can compute the distance of a line segment by finding the difference between two consecutive points on the line $|{P}_{i}-{P}_{i-1}|$ for $i\in [1,n]$ where n is the number of points we've defined on the curve.

This means that the approximate total length of curve is simply a sum of all of these line segments:

$L\approx \sum _{i=1}^{n}|{P}_{i}-{P}_{i-1}|$

If we want the exact length of the curve, then we can make the assumption that all of the points are infinitesimally separated. We now take the limit of our sum as $n\to \infty$.

$L=\underset{n\to \infty}{lim}\sum _{i=1}^{n}|{P}_{i}-{P}_{i-1}|$

Since we are working in the xy-plane, we can redefine our distance between points to take on the typical definition of Euclidean distance.

$|{P}_{i}-{P}_{i-1}|=\sqrt{{({y}_{i}-{y}_{i-1})}^{2}+{({x}_{i}-{x}_{i-1})}^{2}}=\sqrt{\delta {y}^{2}+\delta {x}^{2}}$

We can now apply the Mean Value Theorem, which states there exists a point $x}_{i}^{\prime$ lying in the interval $[{x}_{i-1},{x}_{i}]$ such that

$\Rightarrow f\left({x}_{i}\right)-f\left({x}_{i-1}\right)=f\prime \left({x}_{i}^{\prime}\right)({x}_{i}-{x}_{i-1})$

which we could also write (in the same notation) as

$\Rightarrow \delta y=f\prime \left({x}_{i}^{\prime}\right)\delta x$

This means that we now have

$|{P}_{i}-{P}_{i-1}|=\sqrt{{[f\prime \left({x}_{i}^{\prime}\right)\delta x]}^{2}+\delta {x}^{2}}$

Simplifying this expression a bit gives us

$|{P}_{i}-{P}_{i-1}|=\sqrt{{[f\prime \left({x}_{i}^{\prime}\right)]}^{2}\delta {x}^{2}+\delta {x}^{2}}$

$|{P}_{i}-{P}_{i-1}|=\sqrt{({[f\prime \left({x}_{i}^{\prime}\right)]}^{2}+1)\delta {x}^{2}}$

$|{P}_{i}-{P}_{i-1}|=\sqrt{(1+{[f\prime \left({x}_{i}^{\prime}\right)]}^{2})}\delta x$

We can now use this new distance definition in our summation for our points.

$L=\underset{n\to \infty}{lim}\sum _{i=1}^{n}\sqrt{(1+{[f\prime \left({x}_{i}^{\prime}\right)]}^{2})}\delta x$

Sums are nice, but integrals are better in continuous situations! Because integrals and sums are both "summation" tools, it's simple to write this as a definite integral. We can also remove our sum index from the integral.

$L={\int}_{a}^{b}\sqrt{(1+{[f\prime \left(x\right)]}^{2})}\delta x$

Writing this a little bit more typically yields

$L={\int}_{a}^{b}\sqrt{(1+{\left(\frac{dy}{dx}\right)}^{2})}dx$

We have arrived at our result! In general, the length is usually defined for a differential of arclength ds

$L={\int}_{a}^{b}ds$

where ds is defined accordingly for whatever type of coordinate system you are working in. However, I wanted the explanation to be clearer, so I just chose Cartesian ones for simplicity. You could also use polar or spherical coordinates by simply making the necessary substitutions.

$\Rightarrow L={\int}_{a}^{b}\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}}dx$

In general, we could just write:

$\Rightarrow L={\int}_{a}^{b}ds$Solution:

Let's use Cartesian coordinates for this explanation.

If we consider an arbitrary curve defined as $y=f\left(x\right)$ and are interested in the interval $x\in [a,b]$, we can approximate the length of the curve using very tiny line segments.

Consider a point on the curve $P}_{i$. We can compute the distance of a line segment by finding the difference between two consecutive points on the line $|{P}_{i}-{P}_{i-1}|$ for $i\in [1,n]$ where n is the number of points we've defined on the curve.

This means that the approximate total length of curve is simply a sum of all of these line segments:

$L\approx \sum _{i=1}^{n}|{P}_{i}-{P}_{i-1}|$

If we want the exact length of the curve, then we can make the assumption that all of the points are infinitesimally separated. We now take the limit of our sum as $n\to \infty$.

$L=\underset{n\to \infty}{lim}\sum _{i=1}^{n}|{P}_{i}-{P}_{i-1}|$

Since we are working in the xy-plane, we can redefine our distance between points to take on the typical definition of Euclidean distance.

$|{P}_{i}-{P}_{i-1}|=\sqrt{{({y}_{i}-{y}_{i-1})}^{2}+{({x}_{i}-{x}_{i-1})}^{2}}=\sqrt{\delta {y}^{2}+\delta {x}^{2}}$

We can now apply the Mean Value Theorem, which states there exists a point $x}_{i}^{\prime$ lying in the interval $[{x}_{i-1},{x}_{i}]$ such that

$\Rightarrow f\left({x}_{i}\right)-f\left({x}_{i-1}\right)=f\prime \left({x}_{i}^{\prime}\right)({x}_{i}-{x}_{i-1})$

which we could also write (in the same notation) as

$\Rightarrow \delta y=f\prime \left({x}_{i}^{\prime}\right)\delta x$

This means that we now have

$|{P}_{i}-{P}_{i-1}|=\sqrt{{[f\prime \left({x}_{i}^{\prime}\right)\delta x]}^{2}+\delta {x}^{2}}$

Simplifying this expression a bit gives us

$|{P}_{i}-{P}_{i-1}|=\sqrt{{[f\prime \left({x}_{i}^{\prime}\right)]}^{2}\delta {x}^{2}+\delta {x}^{2}}$

$|{P}_{i}-{P}_{i-1}|=\sqrt{({[f\prime \left({x}_{i}^{\prime}\right)]}^{2}+1)\delta {x}^{2}}$

$|{P}_{i}-{P}_{i-1}|=\sqrt{(1+{[f\prime \left({x}_{i}^{\prime}\right)]}^{2})}\delta x$

We can now use this new distance definition in our summation for our points.

$L=\underset{n\to \infty}{lim}\sum _{i=1}^{n}\sqrt{(1+{[f\prime \left({x}_{i}^{\prime}\right)]}^{2})}\delta x$

Sums are nice, but integrals are better in continuous situations! Because integrals and sums are both "summation" tools, it's simple to write this as a definite integral. We can also remove our sum index from the integral.

$L={\int}_{a}^{b}\sqrt{(1+{[f\prime \left(x\right)]}^{2})}\delta x$

Writing this a little bit more typically yields

$L={\int}_{a}^{b}\sqrt{(1+{\left(\frac{dy}{dx}\right)}^{2})}dx$

We have arrived at our result! In general, the length is usually defined for a differential of arclength ds

$L={\int}_{a}^{b}ds$

where ds is defined accordingly for whatever type of coordinate system you are working in. However, I wanted the explanation to be clearer, so I just chose Cartesian ones for simplicity. You could also use polar or spherical coordinates by simply making the necessary substitutions.

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