Geometric intuition for length of path in 3D

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $h<\delta$.