Prove that the arc length of the curve y_1 = f(x) equals

Prove that the arc length of the curve $y_1 = f(x)$ equals the arc length of the curve $y_2 = f(x) + c$ between points with abscissas $x = a$ and $x = b$. Give a geometric interpretation.

We compare the arc-length integrands of the two curves directly.

Step 1: Arc length formula

\[L = \int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\]

Step 2: Derivative of $y_1=f(x)$

\[\frac{dy_1}{dx} = f'(x) \implies L_1 = \int_a^b\sqrt{1+[f'(x)]^2}\,dx\]

Step 3: Derivative of $y_2=f(x)+c$

Since $c$ is a constant, $\dfrac{d}{dx}[f(x)+c] = f'(x)$ — identical to $y_1$'s derivative. So:

\[L_2 = \int_a^b\sqrt{1+[f'(x)]^2}\,dx\]

Step 4: Conclude

Since the two integrands are identical:

\[\boxed{L_1 = L_2}\]

Step 5: Geometric interpretation

$y_2=f(x)+c$ is $y_1=f(x)$ shifted vertically by $c$ units — a rigid motion (translation) that preserves shape, distances, and angles exactly. The arc-length element $ds=\sqrt{1+(y')^2}\,dx$ depends only on the curve's slope $y'$ at each $x$, never on its absolute height; sliding a curve up or down doesn't stretch or compress it, so its length cannot change. This is the same intuition as lifting a piece of string vertically without pulling on it — its length stays the same.

Step 6: Illustration

For $y_1=x^2$ and $y_2=x^2+5$ on $[0,1]$: both have $y'=2x$, so both give $L=\int_0^1\sqrt{1+4x^2}\,dx$ — the same number, despite $y_2$'s graph sitting five units higher.

Summary

\[\boxed{L[f(x)] = L[f(x)+c] = \int_a^b\sqrt{1+[f'(x)]^2}\,dx \text{ for any constant } c}\]

Key insight: Arc length is a purely intrinsic, "how much the curve bends and rises" quantity — it is blind to a curve's absolute vertical position, exactly because a vertical shift is a distance-preserving rigid motion of the plane.