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Wize University Calculus 1 Textbook > Applications of Integration

Arc Length

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Arc Length

While the distance formula can tell us the length of a straight line, we can use integrals to find the length of a curve.


Arc Length of a Curve

Let f(x)f\left(x\right) be a function with continuous derivative function f′(x)f'\left(x\right) defined on the interval [a,b]\left[a,b\right]. The arc length of f(x)f\left(x\right) is given by
L=∫ab1+[f′(x)]2 dx\boxed{L=\int_a^b\sqrt{1+[f'\left(x\right)]^2} \ dx}

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Example: Arc Length

Find the length of the curve made by f(x)=−ln⁡(cos⁡x)f(x)=-\ln(\cos x) on [0,π4][0,\frac{\pi}{4}].

L=∫ab1+f′(x)2dx\displaystyle L=\int_a^b\sqrt{1+f'\left(x\right)^2}dx

f′(x)=−(−sin⁡xcos⁡x)=tan⁡x\displaystyle f'(x)=-\left(\frac{-\sin x}{\cos x}\right)= \tan x

=∫0π41+(tan⁡x)2dx=\displaystyle \int_0^\frac{\pi}{4}\sqrt{1+(\tan x)^2}dx

∫0π4sec⁡2xdx=∫0π4sec⁡xdx\displaystyle \int_0^\frac{\pi}{4} \sqrt{\sec^2 x}dx=\displaystyle \int_0^\frac{\pi}{4} \sec xdx

=ln⁡∣sec⁡x+tan⁡x∣∣0π4=\displaystyle \ln|\sec x + \tan x| \bigg|_0^\frac{\pi}{4}

=ln⁡∣sec⁡π4+tan⁡π4∣−ln⁡∣sec⁡0+tan⁡0∣∣0π4=\displaystyle \ln|\sec \frac{\pi}{4} + \tan \frac{\pi}{4}| -\displaystyle \ln|\sec 0 + \tan 0| \bigg|_0^\frac{\pi}{4}

=ln⁡∣2+1∣−ln⁡∣1+0∣=\displaystyle \ln|\sqrt2 + 1| -\displaystyle \ln|1 + 0|

=ln⁡∣2+1∣=\displaystyle \ln|\sqrt2 + 1|

Practice: Arc Length

Find the arc length of f(x)=12(ex+e−x)f(x)=\frac{1}{2}\left(e^x+e^{-x}\right) on [0,1][0,1].
Extra Practice
Arc Length
Which of the following integrals represent the arc length of the curve r=sin⁡2(θ4)r=\sin^2\left(\frac{\theta}{4}\right)?
Find the arc length of y=2x32y=2x^{\frac{3}{2}} on [0, 53]\left[0,\ \dfrac{5}{3}\right].
Q:\bf{Q:} Find the arc length of y=2x32y=2x^{\frac{3}{2}} on [0, 53]\left[0,\ \dfrac{5}{3}\right].
Find the arc length of the curve y=3+12cosh⁡ 2xy = 3 +\dfrac{1}{2}\cosh\ 2x from x=0x=0 to x=1x=1.
Function from Arc Length
Find a curve through the point (0,1)(0, 1) whose length integral is L=∫121+1y4 dy\displaystyle L=\int_1^2\sqrt{1+\frac{1}{y^4}}\ dy
Arc Length with Partial Fractions
Find the arc length of y=exy=e^x on [0,1][0,1].
Practice: Arc Length with Perfect Square
Q:\textbf{Q:} Consider the curve y=f(x)=x33+14xy=f(x)=\dfrac{x^3}{3}+\dfrac{1}{4x}. Find the arc length of this curve over the interval [1,3][1, 3]. Find a function giving the arc length of the curve over the interval [1,x].[1, x].
Find the arc length of y=2x32y=2x^{\frac{3}{2}} on [0, 53]\left[0,\ \dfrac{5}{3}\right].
(a) 525\sqrt2 (b) 23\frac23 (c) 143\frac{14}{3} (d) 737\sqrt3 (e) 227\frac{2}{27}
Q:\bf{Q:} Find the arc length of y=2x32y=2x^{\frac{3}{2}} on [0, 53]\left[0,\ \dfrac{5}{3}\right].
Arc Length
Which of the following integrals represent the arc length of the curve r=sin⁡2(θ4)r=\sin^2\left(\frac{\theta}{4}\right)?
Practice Question: Arc Length
Find the arc length function for the curve y=2x3/2y=2x^{3/2} with starting point P0=(0,0)P_0=\left(0,0\right).
Q:\bf{Q:} Find the arc length of y=2x32y=2x^{\frac{3}{2}} on [0, 53]\left[0,\ \dfrac{5}{3}\right].
Find the arc length of the curve y=3+12cosh⁡ 2xy = 3 +\dfrac{1}{2}\cosh\ 2x from x=0x=0 to x=1x=1.
Compute the arc length of the curve f(x)=∫1xt5−1dt f\left(x\right)=\int_1^x\sqrt{t^5-1}dt\ for  1≤x≤9\ 1\le x\le9.
Hint: use the Fundamental Theorem of Calculus: ddx∫cxf(t)dt=f(x)\frac{d}{dx}\int_c^x f(t)dt=f(x)
Arc Length with Partial Fractions
Find the arc length of y=exy=e^x on [0,1][0,1].
Function from Arc Length
Find a curve through the point (0,1)(0, 1) whose length integral is L=∫121+1y4 dy\displaystyle L=\int_1^2\sqrt{1+\frac{1}{y^4}}\ dy
Practice: Arc Length with Perfect Square
Q:\textbf{Q:} Consider the curve y=f(x)=x33+14xy=f(x)=\dfrac{x^3}{3}+\dfrac{1}{4x}. Find the arc length of this curve over the interval [1,3][1, 3]. Find a function giving the arc length of the curve over the interval [1,x].[1, x].