Playing with intgerals: The Length of the Function

How can we find a formula for the length of the function?

As the topic suggests we'll need integrals. First of all lets recall a well-known Pythagorean Theorem. It states that in a right triangle we have c2=a2+b2c^2=a^2+b^2, with cc denoting length of the hypotenuse and a,ba,b - lengths of catheti. Now we'll consider a very tiny bit of function, so tiny than we can aproximate the function with a line. Then according to Pythagorean Identity we'll have s2=x2+y2s^2=x^2+y^2, where ss is the length of the function and x,yx,y - Δx,Δf(x)\Delta x,\Delta f(x) respectively. Now lets use our knowledge of derivatives to transform our identity into ds2=dx2+dy2ds^2=dx^2+dy^2, which may be rearanged to get ds=1+(dydx)2dx=1+[f(x)]2dxds=\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dx=\sqrt{1+[f^\prime(x)]^2}\,dx. After integrating both sides we arrive to the desired result

s=ab1+[f(x)]2dx.s=\displaystyle\int^b_a{\sqrt{1+[f^\prime(x)]^2}}\,dx.

Note by Nicolae Sapoval
5 years, 10 months ago

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My dear infinitesimal, how I hate thee!

This is best done using metric tensors, in my opinion. In fact, we define arclength using metric tensors in most contexts.

Jacob Erickson - 5 years, 10 months ago

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Very nice! However, would you be able to derive it without using infinitesimals? I feel slightly uneasy, even though it is correct.

Isak Falk - 5 years, 10 months ago

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Yep, we can do this without infinitesmals and Pythagorean Theorem. However, this approach would be a bit harder. You will need to consider the line integral, so I think this approach is a bit easier to comprehed for non-college students. If there will be more requests on altrnative solutions I will release a note discussing them (or maybe someone else will).

Nicolae Sapoval - 5 years, 10 months ago

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How can we find a formula for the length of the function?

As the topic suggests we'll need integrals. First of all lets recall a well-known Pythagorean Theorem. It states that in a right triangle we have c2=a2+b2, with c denoting length of the hypotenuse and a,b - lengths of catheti. Now we'll consider a very tiny bit of function, so tiny than we can aproximate the function with a line. Then according to Pythagorean Identity we'll have s2=x2+y2, where s is the length of the function and x,y - Δx,Δf(x) respectively. Now lets use our knowledge of derivatives to transform our identity into ds2=dx2+dy2, which may be rearanged to get ds=1+(dydx)2−−−−−−−−−√dx=1+[f′(x)]2−−−−−−−−−√dx. After integrating both sides we arrive to the desired result

s=∫ba1+[f′(x)]2−−−−−−−−−√dx.

Mekael Alhammod - 5 years, 9 months ago

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