Arc Length | Math Coach Amy

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Key Ideas Easy Medium Hard

\(L=\int_a^b \sqrt{1+(y')^2}\,dx\)

What to Remember

Arc length of a graph

\(L=\int_a^b \sqrt{1+\left(y'\right)^2}\,dx\)

This is the length of the curve, not the straight-line change in \(y\).

A line segment still uses the formula

If \(y'=m\) is constant, the integral is the distance formula.

Look for a perfect square under the radical

Simplify \(1+(y')^2\) before you integrate. Some derivatives are built so that expression is a square.

Leave a setup when the antiderivative is not basic

An integral such as \(\int_0^1 \sqrt{1+4x^2}\,dx\) is a complete answer when the problem asks only for the length expression.

Level 1

Easy

The derivative is constant, so the radical is constant.

Example

Find the length of \(y=2x\) from \(x=0\) to \(x=3\).

\(y'=2\)

\(L=\int_0^3 \sqrt{1+4}\,dx\)

\(\int_0^3 \sqrt{5}\,dx\)

\(3\sqrt{5}\)

Answer: \(3\sqrt{5}\)

Find the length of \(y=x\) from \((0,0)\) to \((3,3)\).

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\(y'=1\)

\(\int_0^3 \sqrt{2}\,dx\)

\(3\sqrt{2}\)

Answer: \(3\sqrt{2}\)

Find the length of the horizontal segment \(y=4\) from \(x=1\) to \(x=6\).

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\(y'=0\)

\(\int_1^6 1\,dx\)

\(5\)

Answer: \(5\)

A curve has \(y'=0\) on \([2,9]\). What is its length?

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\(7\)

Answer: \(7\)

Which integral is the length of \(y=f(x)\) from \(a\) to \(b\)?

A. \(\int_a^b \sqrt{1+(f'(x))^2}\,dx\)

B. \(\int_a^b \sqrt{1+f(x)}\,dx\)

C. \(f(b)-f(a)\)

D. \(\int_a^b (f'(x))^2\,dx\)

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The formula uses 1 plus the square of the derivative, all under a square root.

Answer: A. \(\int_a^b\sqrt{1+(f')^2}\,dx\)

The segment from \((1,2)\) to \((4,6)\) is part of a line. Find its length with the distance formula, which matches the arc-length integral.

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\(\sqrt{(4-1)^2+(6-2)^2}\)

\(\sqrt{9+16}\)

\(5\)

Answer: \(5\)

Level 2

Medium

Simplify 1 + (y')² before integrating.

Example

Let \(y=\dfrac{1}{3}(x^2+2)^{3/2}\). Then \(y'=x\sqrt{x^2+2}\). Find the length from \(x=0\) to \(x=2\).

\(1+(y')^2=1+x^2(x^2+2)\)

\(1+x^4+2x^2\)

\((x^2+1)^2\)

\(\sqrt{1+(y')^2}=x^2+1\)

\(\int_0^2 (x^2+1)\,dx\)

\(\left[\dfrac{x^3}{3}+x\right]_0^2\)

\(\dfrac{8}{3}+2\)

\(\dfrac{14}{3}\)

Answer: \(\dfrac{14}{3}\)

Find the length of \(y=\dfrac{2}{3}x^{3/2}\) from \(x=0\) to \(x=3\). Here \(y'=\sqrt{x}\).

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\(1+(y')^2=1+x\)

\(L=\int_0^3 \sqrt{1+x}\,dx\)

Let \(u=1+x\). Then \(u\) runs from 1 to 4.

\(\int_1^4 u^{1/2}\,du\)

\(\left[\dfrac{2}{3}u^{3/2}\right]_1^4\)

\(\dfrac{2}{3}(8-1)\)

\(\dfrac{14}{3}\)

Answer: \(\dfrac{14}{3}\)

Find the length of \(y=\dfrac{1}{2}(e^{x}+e^{-x})\) from \(x=0\) to \(x=\ln 2\). Use \(y'=\dfrac{1}{2}(e^{x}-e^{-x})\).

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\(1+(y')^2=\left(\dfrac{e^{x}+e^{-x}}{2}\right)^2\)

So \(\sqrt{1+(y')^2}=\dfrac{e^{x}+e^{-x}}{2}\) for every real \(x\).

\(\int_0^{\ln 2}\dfrac{e^{x}+e^{-x}}{2}\,dx\)

\(\left[\dfrac{e^{x}-e^{-x}}{2}\right]_0^{\ln 2}\)

At \(\ln 2\): \(\dfrac{2-1/2}{2}=\dfrac{3}{4}\).

At 0: \(0\).

\(\dfrac{3}{4}\)

Answer: \(\dfrac{3}{4}\)

Set up the length of \(y=x^2\) from \(x=0\) to \(x=1\). Do not evaluate it.

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\(y'=2x\)

\(\int_0^1 \sqrt{1+4x^2}\,dx\)

Answer: \(\int_0^1 \sqrt{1+4x^2}\,dx\)

A particle moves along \(y=2x\) from \(x=0\) to \(x=3\). How far does it travel along the curve?

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Travel along the curve is the arc length.

\(3\sqrt{5}\)

Answer: \(3\sqrt{5}\)

Why is \(f(3)-f(0)\) not the length of \(y=2x\) from 0 to 3?

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\(f(3)-f(0)=6\) is the vertical change.

The curve also moves 3 units horizontally.

The length is \(3\sqrt{5}\), which is longer than 6.

Answer: vertical change is not arc length

Level 3

Hard

Build the square yourself, or compare length with a straight chord.

Example

For \(y=\dfrac{x^3}{3}+\dfrac{1}{4x}\) with \(x>0\), a classmate says \(\sqrt{1+(y')^2}=\dfrac{x^2}{2}+\dfrac{1}{2x^2}\). Check that claim, then find the length from \(x=1\) to \(x=2\).

\(y'=x^2-\dfrac{1}{4x^2}\)

\((y')^2=x^4-\dfrac{1}{2}+\dfrac{1}{16x^4}\)

\(1+(y')^2=x^4+\dfrac{1}{2}+\dfrac{1}{16x^4}\)

\(\left(x^2+\dfrac{1}{4x^2}\right)^2\)

That is not \(x^2/2+1/(2x^2)\). The square root is \(x^2+\dfrac{1}{4x^2}\).

\(L=\int_1^2 \left(x^2+\dfrac{1}{4x^2}\right)\,dx\)

\(\left[\dfrac{x^3}{3}-\dfrac{1}{4x}\right]_1^2\)

At 2: \(\dfrac{8}{3}-\dfrac{1}{8}=\dfrac{64-3}{24}=\dfrac{61}{24}\).

At 1: \(\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{1}{12}\).

\(\dfrac{61}{24}-\dfrac{2}{24}\)

\(\dfrac{59}{24}\)

Answer: \(\dfrac{59}{24}\)

Find the length of \(y=\dfrac{2}{3}(x-1)^{3/2}\) from \(x=1\) to \(x=4\).

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\(y'=\sqrt{x-1}\)

\(1+(y')^2=x\)

\(\int_1^4 \sqrt{x}\,dx\)

\(\left[\dfrac{2}{3}x^{3/2}\right]_1^4\)

\(\dfrac{2}{3}(8-1)\)

\(\dfrac{14}{3}\)

Answer: \(\dfrac{14}{3}\)

The straight chord from \((0,0)\) to \((3,6)\) on \(y=2x\) has length \(3\sqrt{5}\). Explain why a curve from the same two points, other than the line, would be longer.

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The line segment is the shortest path between two points.

Any other smooth graph connecting them has a larger arc-length integral.

Answer: the segment is the shortest path

Set up the distance a particle travels along \(y=\ln x\) from \(x=1\) to \(x=e\).

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\(y'=1/x\)

\(\int_1^e \sqrt{1+\dfrac{1}{x^2}}\,dx\)

\(\int_1^e \dfrac{\sqrt{x^2+1}}{x}\,dx\)

Answer: \(\int_1^e \dfrac{\sqrt{x^2+1}}{x}\,dx\)

A curve has \(dy/dx=\dfrac{3}{4}\) from \(x=0\) to \(x=8\). Find its length.

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\(\sqrt{1+\dfrac{9}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac{5}{4}\)

\(8\cdot \dfrac{5}{4}\)

\(10\)

Answer: \(10\)

Function \(f\) has length 7 on \([0,3]\). Function \(g(x)=f(x)+4\) is a vertical shift. What is the length of \(g\) on the same interval?

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\(g'=f'\), so the integrand does not change.

The length is still 7.

Answer: \(7\)

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