 2023-03-24

How to find the length of a curve in calculus? marallocajcyb

In Cartesian coordinates for y = f(x) defined on interval [a,b] the length of the curve is
$⇒L={\int }_{a}^{b}\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}}dx$
In general, we could just write:
$⇒L={\int }_{a}^{b}ds$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 $x\in \left[a,b\right]$, 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 $i\in \left[1,n\right]$ 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.
$|{P}_{i}-{P}_{i-1}|=\sqrt{{\left({y}_{i}-{y}_{i-1}\right)}^{2}+{\left({x}_{i}-{x}_{i-1}\right)}^{2}}=\sqrt{\delta {y}^{2}+\delta {x}^{2}}$
We can now apply the Mean Value Theorem, which states there exists a point ${x}_{i}^{\prime }$ lying in the interval $\left[{x}_{i-1},{x}_{i}\right]$ such that
$⇒f\left({x}_{i}\right)-f\left({x}_{i-1}\right)=f\prime \left({x}_{i}^{\prime }\right)\left({x}_{i}-{x}_{i-1}\right)$
which we could also write (in the same notation) as
$⇒\delta y=f\prime \left({x}_{i}^{\prime }\right)\delta x$
This means that we now have
$|{P}_{i}-{P}_{i-1}|=\sqrt{{\left[f\prime \left({x}_{i}^{\prime }\right)\delta x\right]}^{2}+\delta {x}^{2}}$
Simplifying this expression a bit gives us
$|{P}_{i}-{P}_{i-1}|=\sqrt{{\left[f\prime \left({x}_{i}^{\prime }\right)\right]}^{2}\delta {x}^{2}+\delta {x}^{2}}$
$|{P}_{i}-{P}_{i-1}|=\sqrt{\left({\left[f\prime \left({x}_{i}^{\prime }\right)\right]}^{2}+1\right)\delta {x}^{2}}$
$|{P}_{i}-{P}_{i-1}|=\sqrt{\left(1+{\left[f\prime \left({x}_{i}^{\prime }\right)\right]}^{2}\right)}\delta x$
We can now use this new distance definition in our summation for our points.
$L=\underset{n\to \infty }{lim}\sum _{i=1}^{n}\sqrt{\left(1+{\left[f\prime \left({x}_{i}^{\prime }\right)\right]}^{2}\right)}\delta x$
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.
$L={\int }_{a}^{b}\sqrt{\left(1+{\left[f\prime \left(x\right)\right]}^{2}\right)}\delta x$
Writing this a little bit more typically yields
$L={\int }_{a}^{b}\sqrt{\left(1+{\left(\frac{dy}{dx}\right)}^{2}\right)}dx$
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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