Arc length is used to find the distance along a curved path over a given interval. In calculus, this usually involves using a derivative inside a square root and evaluating a definite integral. Many arc length integrals require a calculator or numerical approximation, but the setup comes directly from the formula \(\text{Arc Length}=\displaystyle\int_a^b\sqrt{1+\left(f'(x)\right)^2}\,dx\).
Notes
Practice Problems
Find the arc length over the given interval.
(Calculator needed for integrals)
\(\textbf{1)}\) \( f(x)=4(x+1)^{3/2} \,\,\,\, [2,5] \)
\(\,\,\,\text{Arc Length} = \displaystyle\int_{a}^{b}\sqrt{1+ \left( f'(x) \right)^2 } \, dx\)
\(\,\,\,\,\,\,\,\,f'(x)=4\cdot \frac{3}{2} \left( x + 1 \right)^{1/2}=6 \left( x + 1 \right)^{1/2}\)
\(\,\,\,\displaystyle\int_{2}^{5}\sqrt{1+ \left( 6 \left( x + 1 \right)^{1/2} \right)^2 } \, dx\)
\(\,\,\,\displaystyle\int_{2}^{5}\sqrt{1+ 36 \left( x + 1 \right) } \, dx\)
\(\,\,\,\displaystyle\int_{2}^{5}\sqrt{36x+37} \, dx\)
\(\,\,\,\,\,\,\,\,u=36x+37\)
\(\,\,\,\,\,\,\,\,du=36 \,dx\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{du}{36}=dx\)
\(\,\,\,\,\,\,\,\,x=2 \rightarrow u=109\)
\(\,\,\,\,\,\,\,\,x=5 \rightarrow u=217\)
\(\,\,\,\displaystyle \frac{1}{36} \int_{109}^{217}\sqrt{u} \, du\)
\(\,\,\,\displaystyle\frac{1}{36} \cdot \frac{2}{3}u^{3/2} \Big|_{109}^{217}\)
\(\,\,\,\displaystyle\frac{1}{54}(217)^{3/2}-\frac{1}{54}(109)^{3/2}\)
\(\,\,\,59.196-21.074\)
\(\,\,\,\)The arc length is \( \approx 38.12\)
\(\textbf{2)}\) \( y=\sin x \,\,\,\, [0,\pi] \)
\(\,\,\,\text{Arc Length}=\displaystyle\int_a^b\sqrt{1+\left(y’\right)^2}\,dx\)
\(\,\,\,y=\sin x\)
\(\,\,\,y’=\cos x\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^\pi\sqrt{1+\cos^2 x}\,dx\)
\(\,\,\,\text{Use a calculator to evaluate the integral.}\)
\(\,\,\,\)The arc length is \( \approx 3.8202 \)
\(\textbf{3)}\) \( y=x^2 \,\,\,\, [0,4] \)
\(\,\,\,\text{Arc Length}=\displaystyle\int_a^b\sqrt{1+\left(y’\right)^2}\,dx\)
\(\,\,\,y=x^2\)
\(\,\,\,y’=2x\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^4\sqrt{1+(2x)^2}\,dx\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^4\sqrt{1+4x^2}\,dx\)
\(\,\,\,\text{Use a calculator to evaluate the integral.}\)
\(\,\,\,\)The arc length is \( \approx 16.8186 \)
\(\textbf{4)}\) \( y=\frac{\sqrt[3]{x}}{2} \,\,\,\, [1,3] \)
\(\,\,\,y=\frac{1}{2}x^{1/3}\)
\(\,\,\,y’=\frac{1}{2}\cdot\frac{1}{3}x^{-2/3}\)
\(\,\,\,y’=\frac{1}{6x^{2/3}}\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_1^3\sqrt{1+\left(\frac{1}{6x^{2/3}}\right)^2}\,dx\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_1^3\sqrt{1+\frac{1}{36x^{4/3}}}\,dx\)
\(\,\,\,\text{Use a calculator to evaluate the integral.}\)
\(\,\,\,\)The arc length is \( \approx 2.0127 \)
\(\textbf{5)}\) \( y=\frac{1}{x} \,\,\,\, [1,5] \)
\(\,\,\,y=x^{-1}\)
\(\,\,\,y’=-x^{-2}\)
\(\,\,\,y’=-\frac{1}{x^2}\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_1^5\sqrt{1+\left(-\frac{1}{x^2}\right)^2}\,dx\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_1^5\sqrt{1+\frac{1}{x^4}}\,dx\)
\(\,\,\,\text{Use a calculator to evaluate the integral.}\)
\(\,\,\,\)The answer is \( \approx 4.1515 \)
\(\textbf{6)}\) \( y=\frac{1}{x} \,\,\,\, [.1,1] \)
\(\,\,\,y=x^{-1}\)
\(\,\,\,y’=-x^{-2}\)
\(\,\,\,y’=-\frac{1}{x^2}\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_{0.1}^{1}\sqrt{1+\left(-\frac{1}{x^2}\right)^2}\,dx\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_{0.1}^{1}\sqrt{1+\frac{1}{x^4}}\,dx\)
\(\,\,\,\text{Use a calculator to evaluate the integral.}\)
\(\,\,\,\)The arc length is \( \approx 9.1526 \)
\(\textbf{7)}\) \( y=3x+1 \,\,\,\, [0,4] \)
\(\,\,\,y=3x+1\)
\(\,\,\,y’=3\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^4\sqrt{1+3^2}\,dx\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^4\sqrt{10}\,dx\)
\(\,\,\,\text{Arc Length}=4\sqrt{10}\)
\(\,\,\,\)The arc length is \(\approx 12.6491\)
\(\textbf{8)}\) \( y=x^3 \,\,\,\, [0,2] \)
\(\,\,\,y=x^3\)
\(\,\,\,y’=3x^2\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^2\sqrt{1+(3x^2)^2}\,dx\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^2\sqrt{1+9x^4}\,dx\)
\(\,\,\,\text{Use a calculator to evaluate the integral.}\)
\(\,\,\,\)The arc length is \(\approx 8.6303\)
\(\textbf{9)}\) \( y=\sqrt{x} \,\,\,\, [1,9] \)
\(\,\,\,y=x^{1/2}\)
\(\,\,\,y’=\frac{1}{2}x^{-1/2}\)
\(\,\,\,y’=\frac{1}{2\sqrt{x}}\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_1^9\sqrt{1+\left(\frac{1}{2\sqrt{x}}\right)^2}\,dx\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_1^9\sqrt{1+\frac{1}{4x}}\,dx\)
\(\,\,\,\text{Use a calculator to evaluate the integral.}\)
\(\,\,\,\)The arc length is \(\approx 8.2681\)
\(\textbf{10)}\) \( y=\ln x \,\,\,\, [1,4] \)
\(\,\,\,y=\ln x\)
\(\,\,\,y’=\frac{1}{x}\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_1^4\sqrt{1+\left(\frac{1}{x}\right)^2}\,dx\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_1^4\sqrt{1+\frac{1}{x^2}}\,dx\)
\(\,\,\,\text{Use a calculator to evaluate the integral.}\)
\(\,\,\,\)The arc length is \(\approx 3.3428\)
\(\textbf{11)}\) \( y=e^x \,\,\,\, [0,1] \)
\(\,\,\,y=e^x\)
\(\,\,\,y’=e^x\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^1\sqrt{1+\left(e^x\right)^2}\,dx\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^1\sqrt{1+e^{2x}}\,dx\)
\(\,\,\,\text{Use a calculator to evaluate the integral.}\)
\(\,\,\,\)The arc length is \(\approx 2.0035\)
\(\textbf{12)}\) \( y=\cos x \,\,\,\, [0,\pi] \)
\(\,\,\,y=\cos x\)
\(\,\,\,y’=-\sin x\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^\pi\sqrt{1+\left(-\sin x\right)^2}\,dx\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^\pi\sqrt{1+\sin^2 x}\,dx\)
\(\,\,\,\text{Use a calculator to evaluate the integral.}\)
\(\,\,\,\)The arc length is \(\approx 3.8202\)
\(\textbf{13)}\) \( y=\frac{2}{3}x^{3/2} \,\,\,\, [0,4] \)
\(\,\,\,y=\frac{2}{3}x^{3/2}\)
\(\,\,\,y’=\frac{2}{3}\cdot\frac{3}{2}x^{1/2}\)
\(\,\,\,y’=\sqrt{x}\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^4\sqrt{1+\left(\sqrt{x}\right)^2}\,dx\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^4\sqrt{x+1}\,dx\)
\(\,\,\,\text{Arc Length}=\displaystyle\left[\frac{2}{3}(x+1)^{3/2}\right]_0^4\)
\(\,\,\,\text{Arc Length}=\frac{2}{3}\left(5^{3/2}-1\right)\)
\(\,\,\,\)The arc length is \(\approx 6.7869\)
\(\textbf{14)}\) \( y=\frac{1}{3}\left(x^2+2\right)^{3/2} \,\,\,\, [0,2] \)
\(\,\,\,y=\frac{1}{3}\left(x^2+2\right)^{3/2}\)
\(\,\,\,y’=\frac{1}{3}\cdot\frac{3}{2}\left(x^2+2\right)^{1/2}\cdot 2x\)
\(\,\,\,y’=x\sqrt{x^2+2}\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^2\sqrt{1+\left(x\sqrt{x^2+2}\right)^2}\,dx\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^2\sqrt{1+x^2(x^2+2)}\,dx\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^2\sqrt{x^4+2x^2+1}\,dx\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^2(x^2+1)\,dx\)
\(\,\,\,\text{Arc Length}=\left[\frac{x^3}{3}+x\right]_0^2\)
\(\,\,\,\)The arc length is \(\frac{14}{3}\approx 4.6667\)
\(\textbf{15)}\) \( x=y^2 \,\,\,\, [0,3] \)
\(\,\,\,\text{When }x\text{ is a function of }y\text{, use }\displaystyle\int_c^d\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy\)
\(\,\,\,x=y^2\)
\(\,\,\,\frac{dx}{dy}=2y\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^3\sqrt{1+(2y)^2}\,dy\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^3\sqrt{1+4y^2}\,dy\)
\(\,\,\,\text{Use a calculator to evaluate the integral.}\)
\(\,\,\,\)The arc length is \(\approx 9.7471\)
\(\textbf{16)}\) \( x=\ln y \,\,\,\, [1,e] \)
\(\,\,\,x=\ln y\)
\(\,\,\,\frac{dx}{dy}=\frac{1}{y}\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_1^e\sqrt{1+\left(\frac{1}{y}\right)^2}\,dy\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_1^e\sqrt{1+\frac{1}{y^2}}\,dy\)
\(\,\,\,\text{Use a calculator to evaluate the integral.}\)
\(\,\,\,\)The arc length is \(\approx 2.0035\)
\(\textbf{17)}\) Find the arc length of the parametric curve \(x=t\), \(y=t^2\) from \(t=0\) to \(t=2\).
\(\,\,\,\text{For parametric curves, }\text{Arc Length}=\displaystyle\int_a^b\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\)
\(\,\,\,x=t\)
\(\,\,\,y=t^2\)
\(\,\,\,\frac{dx}{dt}=1\)
\(\,\,\,\frac{dy}{dt}=2t\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^2\sqrt{1^2+(2t)^2}\,dt\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^2\sqrt{1+4t^2}\,dt\)
\(\,\,\,\text{Use a calculator to evaluate the integral.}\)
\(\,\,\,\)The arc length is \(\approx 4.6468\)
\(\textbf{18)}\) Find the arc length of the parametric curve \(x=t^2\), \(y=t^3\) from \(t=0\) to \(t=1\).
\(\,\,\,\text{Arc Length}=\displaystyle\int_a^b\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\)
\(\,\,\,x=t^2\)
\(\,\,\,y=t^3\)
\(\,\,\,\frac{dx}{dt}=2t\)
\(\,\,\,\frac{dy}{dt}=3t^2\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^1\sqrt{(2t)^2+(3t^2)^2}\,dt\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^1\sqrt{4t^2+9t^4}\,dt\)
\(\,\,\,\text{Use a calculator to evaluate the integral.}\)
\(\,\,\,\)The arc length is \(\approx 1.4397\)
\(\textbf{19)}\) Find the arc length of the polar curve \(r=2\) from \(\theta=0\) to \(\theta=\pi\).
\(\,\,\,\text{For polar curves, }\text{Arc Length}=\displaystyle\int_a^b\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta\)
\(\,\,\,r=2\)
\(\,\,\,\frac{dr}{d\theta}=0\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^\pi\sqrt{2^2+0^2}\,d\theta\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^\pi 2\,d\theta\)
\(\,\,\,\text{Arc Length}=2\pi\)
\(\,\,\,\)The arc length is \(2\pi\approx 6.2832\)
\(\textbf{20)}\) Find the arc length of the polar curve \(r=\theta\) from \(\theta=0\) to \(\theta=\pi\).
\(\,\,\,\text{For polar curves, }\text{Arc Length}=\displaystyle\int_a^b\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta\)
\(\,\,\,r=\theta\)
\(\,\,\,\frac{dr}{d\theta}=1\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^\pi\sqrt{\theta^2+1^2}\,d\theta\)
\(\,\,\,\text{Arc Length}=\displaystyle\int_0^\pi\sqrt{\theta^2+1}\,d\theta\)
\(\,\,\,\text{Use a calculator to evaluate the integral.}\)
\(\,\,\,\)The arc length is \(\approx 6.1099\)
See Related Pages\(\)
In Summary
In calculus, arc length refers to the distance along a curved path. It is a measure of the distance between two points along a curve, rather than a straight line. Arc length is an important concept in calculus because it allows us to measure and understand the properties of curved objects and shapes. It is an advanced topic that builds upon the concepts of limits and derivatives, which are introduced in a first-year calculus course.
Arc length is related to several other concepts in calculus, including integration, limits, and derivatives. It is also closely related to the concept of curvature. Other related topics include circular arc length, the circumference of a circle, and the arc length of a curve that is not straight. These concepts are important in fields such as physics, engineering, and geometry.
