# Geometric intuition for length of path in 3D

I am told that:

$\int_a^b \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2} dt$ is the length of a path. However, I can't find online or in my textbook anywhere the proof of this or any geometric intuition for this problem.

I can't just believe the formula, can someone explain how they derived this formula.

• Given the Pythagorean theorem, it is not hard to believe that $\sqrt{dx^2+dy^2+dz^2}$ is an infinitesimal element of length. If $x,y,z$ are continuously differentiable functions of the $t$-variable, say $\gamma(t)=(x(t),y(t),z(t))$, then the length of $\gamma:[a,b]\to \mathbb{R}^3$ is given by the mentioned integral since $dx = x'(t)\,dt$ and so on. – Jack D'Aurizio Feb 25 '18 at 22:12

Picture two nearby points $(x(t),y(t),z(t))$ and $(x(t+\Delta t),y(t + \Delta t),z(t + \Delta t))$ on the curve. The displacement vector from the first point to the second is $$(x(t + \Delta t) - x(t),y(t + \Delta t) - y(t),z(t + \Delta t) - z(t)) \approx (x'(t) \Delta t, y'(t) \Delta t, z'(t) \Delta t)$$ and the length of this vector is $$\sqrt{(x'(t) \Delta t)^2 + (y'(t) \Delta t)^2 + (z'(t) \Delta t)^2} = \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2} \Delta t.$$ Chop up the curve into tiny pieces and sum up the lengths of all the tiny displacement vectors to get (approximately) the length of the curve.

Note that the vector tangent to the curve at any point in time $t$ is given by $$(x'(t),y'(t),z'(t))$$ and so $$\sqrt{x'(t)^2+y'(t)^2+z'(t)^2}=||(x'(t),y'(t),z'(t))||$$ and the integral is "summing" up all of these lengths. So I like to think about it as approximating the curve at each point by something straight, i.e. a vector which is easy to measure, and "adding" up all these contributions in a continuous way, meaning through an integral.

Its the mean value theorem. Tale two points. $x(t),y(t),z(t)$ and $x(t+h),y(t+h),z(t+h)$ on the curve. The length of the secant is

$$\sqrt{(x(t+h)-x(t))^2+(y(t+h)-y(t))^2+(z(t+h)-z(t))^2}$$

And by the mean value,

$$x(t+h)-x(t)=x^{\prime}(a)h$$ $$y(t+h)-y(t)=y^{\prime}(b)h$$ $$z(t+h)-z(t)=z^{\prime}(a)h$$

So the secant length is

$$\sqrt{x^{\prime}(a)^2+y^{\prime}(b)^2+z^{\prime}(c)^2}\ h$$

And if we sum all these secant lengths and take the limit, which is the definition of the arc length we get

$$\lim\limits_{h\to 0}\sum \sqrt{x^{\prime}(a_i)^2+y^{\prime}(b_i)^2+z^{\prime}(c_i)^2}\ h$$

It is easy to see that this is equal to the integral using the continuity of $x^{\prime}(t),y^{\prime}(t),z^{\prime}(t)$.

As

$$\sqrt{x^{\prime}(a_i)^2+y^{\prime}(b_i)^2+z^{\prime}(c_i)^2} -\sqrt{x^{\prime}(t_i)^2+y^{\prime}(t_i)^2+z^{\prime}(t_i)^2}<\epsilon$$ provided $h<\delta$.