Arc Length In A Rectilinear Metric?

The distance for infinitesimal neighborhood points $(x,y)$ and $(x+dx,y+dy)$ using this metric is given by $ds = |dx| + |dy| = \left(1+\left|\frac{dy}{dx}\right|\right) |dx|$

Then the arc length of curve $y=f(x)$ from $x=a$ to $x=b$ with $a<b$ respect to this metric is given by $S=\int_{a}^{b}\:\left(1+\left|\frac{dy}{dx}\right|\right) dx$ since $a<b$ , we have $dx >0$ .

For $y=x^2$ , we have $\frac{dy}{dx} = 2x$ which is positive for $x>0$ and negative for $x<0$ . Then, $S=\int_{-1}^{1}\:\left(1+\left|2x\right|\right) dx = \int_{-1}^{0}\:\left(1-2x\right) dx + \int_{0}^{1}\:\left(1+2x\right) dx = 4$

We will prove the general result for any function $f(x)$ for its arc length between $a$ and $b$ . Note that this proof is quite similar to proving the result for the Euclidean plane (Read this if you want since it should be known to understand the reasoning behind this method).

We assume an interval of $h$ between $a$ and $a+h$ . We approximate the arc length by using line segments. We have the line segment have endpoints (f(a)) and $f(a+h)$ . This length would be $h+|f(a+h)-f(a)|=h \left ( \left | \dfrac{f(a+h)-f(a)}{h} \right | + 1 \right ) = h\left ( \left |\frac{\mathrm{d}f}{\mathrm{d}x} \right | + 1 \right ).$

The last expression is attained as we wish $h$ to tend to 0, so we get a more accurate result as possible. We get that as the number of intervals tend to infinity, the sum of this expression would add up to become the area under a curve, but we only wish to ignore the height (the expression is equivalent to a rectangle with height $h$ and length $\left |\frac{\mathrm{d}f}{\mathrm{d}x} \right | + 1$ . Thus, the arc length is

$\displaystyle \int_a^b \left |\frac{\mathrm{d}f}{\mathrm{d}x} \right | + 1 \quad \mathrm{d}x$

We substitute the values from the question to get the answer as 4.