The Arc Length of a Smooth, Planar Curve and Distance Traveled
Introduced| English |
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| Arc length/ɑːk leŋθ/ |
Measuring the length of a curve
- Area and volume were integrals; so is the length of a curve itself.
- Arc length 弧长 adds up the lengths of tiny straight pieces along the curve.
- On each little piece, the curve is nearly a straight line — a hypotenuse of a small right triangle.
- Sum those hypotenuses (integrate) to get the total length.
From a small triangle to the formula
- A tiny piece has horizontal run $dx$ and vertical rise $dy=f'(x)\,dx$.
- Its length is $\sqrt{dx^2+dy^2}=\sqrt{1+\big(f'(x)\big)^2}\,dx$ (Pythagoras).
- Add them from $a$ to $b$:
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$$L=\int_a^b \sqrt{1+\big(f'(x)\big)^2}\,dx$$
A curve is longer than its chord
y = a·x^{1.5}
Arc length sums tiny hypotenuses $\sqrt{1+(f')^2}\,dx$ — always at least the horizontal distance $b-a$.
The arc length of $y=f(x)$ on $[a,b]$ is...
Square the derivative, add 1, take the root.
Using the formula
- Compute $f'(x)$, square it, add $1$, take the square root, and integrate over $[a,b]$.
- The integrand is always $\ge1$, so arc length is at least the horizontal distance $b-a$ (a curve is longer than the straight line).
- Many arc-length integrals are hard by hand; on the BC exam you often set them up and evaluate numerically.
- The key skill is building the integrand correctly from $f'$.
For $y=\tfrac23 x^{3/2}$, $f'(x)=x^{1/2}$. The arc-length integrand is...
$(f')^2=x$, so $\sqrt{1+x}$.
Evaluate $\int_0^3 \sqrt{1+x}\,dx$ (a decimal).
$\tfrac23(8-1)=\tfrac{14}{3}\approx4.667$.
Because the integrand is $\ge 1$, the arc length is always at least...
A curve is at least as long as the straight run.
Arc length as distance traveled
- If a particle moves along $y=f(x)$, the arc length is the distance it travels along the path.
- (For motion given by $x(t),y(t)$, the parametric version comes in lesson 9.3.)
- So this same integral measures both a curve's geometric length and a path's traveled distance.
- Setup is identical; only the interpretation changes.
The arc-length integrand uses the derivative $f'(x)$, not $f(x)$.
It is $\sqrt{1+(f'(x))^2}$.
Along a path $y=f(x)$, the arc length also measures the ____ traveled.
Arc length = distance traveled along the path.
The integrand is $\sqrt{1+(f'(x))^2}$ — the derivative is squared, and there's a $\boldsymbol{+1}$ inside the root, not outside. Don't write $\sqrt{1+f(x)^2}$ (using $f$ instead of $f'$) or forget the $+1$. And the whole thing is under one square root before integrating.
Set up the arc length of $y=\tfrac23 x^{3/2}$ from $x=0$ to $x=3$.
- $f'(x)=x^{1/2}$, so $\big(f'(x)\big)^2=x$.
- $L=\displaystyle\int_0^3 \sqrt{1+x}\,dx=\Big[\tfrac23(1+x)^{3/2}\Big]_0^3=\tfrac23(8-1)=\tfrac{14}{3}$.
The arc length of $y=f(x)$ on $[a,b]$ is $L=\int_a^b\sqrt{1+\big(f'(x)\big)^2}\,dx$ — summing tiny hypotenuses $\sqrt{1+(f')^2}\,dx$. Square the derivative, add $1$ inside the root, integrate. It also gives the distance traveled along the path.