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# 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 $$c^2=a^2+b^2$$, 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 $$s^2=x^2+y^2$$, where $$s$$ is the length of the function and $$x,y$$ - $$\Delta x,\Delta f(x)$$ respectively. Now lets use our knowledge of derivatives to transform our identity into $$ds^2=dx^2+dy^2$$, which may be rearanged to get $$ds=\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dx=\sqrt{1+[f^\prime(x)]^2}\,dx$$. After integrating both sides we arrive to the desired result

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

Note by Nicolae Sapoval
3 years, 1 month 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. · 3 years, 1 month ago

Very nice! However, would you be able to derive it without using infinitesimals? I feel slightly uneasy, even though it is correct. · 3 years, 1 month ago

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). · 3 years, 1 month ago

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. · 3 years ago