Question

consider the curve

( y = x+sin(x),quad 0leq xleq 2pi ).

(a) divide ( 0,2pi ) up into 1,2, and 4 equal segments, and draw line segments between those values on the graph. then compute the length of the polygonal curve you drew.

length for ( n = 1 ):

length for ( n = 2 ):

length for ( n = 4 ):

(b) use a calculator or a computer algebra system to find the length of the curve to four decimal places.

length of curve ( approx )

Explanation:

Step1: Find the derivative of \(y = x+\sin(x)\)

The derivative \(y^\prime=\frac{dy}{dx}=1 + \cos(x)\) using the sum rule \((u + v)^\prime=u^\prime+v^\prime\) where \(u = x\) (\(u^\prime = 1\)) and \(v=\sin(x)\) (\(v^\prime=\cos(x)\)).

Step2: Use the arc - length formula \(L=\int_{a}^{b}\sqrt{1+(y^\prime)^2}dx\)

Substitute \(y^\prime = 1+\cos(x)\) into the formula: \(L=\int_{0}^{2\pi}\sqrt{1+(1 + \cos(x))^2}dx=\int_{0}^{2\pi}\sqrt{1+1 + 2\cos(x)+\cos^{2}(x)}dx=\int_{0}^{2\pi}\sqrt{2+2\cos(x)+\cos^{2}(x)}dx\)

Step3: Calculate the integral for \(n = 4\)

The interval \([0,2\pi]\) is divided into \(n = 4\) sub - intervals \(\Delta x=\frac{2\pi-0}{4}=\frac{\pi}{2}\). The endpoints are \(x_0 = 0,x_1=\frac{\pi}{2},x_2=\pi,x_3=\frac{3\pi}{2},x_4 = 2\pi\)
The length of the polygonal curve is \(L=\sum_{i = 1}^{4}\sqrt{(x_i - x_{i - 1})^2+(y(x_i)-y(x_{i - 1}))^2}\)
\(y(0)=0+\sin(0)=0\), \(y(\frac{\pi}{2})=\frac{\pi}{2}+1\), \(y(\pi)=\pi\), \(y(\frac{3\pi}{2})=\frac{3\pi}{2}-1\), \(y(2\pi)=2\pi\)
\(L_1=\sqrt{(\frac{\pi}{2}-0)^2+((\frac{\pi}{2}+1)-0)^2}=\sqrt{\frac{\pi^{2}}{4}+\frac{\pi^{2}}{4}+\pi + 1}=\sqrt{\frac{\pi^{2}}{2}+\pi + 1}\)
\(L_2=\sqrt{(\pi-\frac{\pi}{2})^2+(\pi-(\frac{\pi}{2}+1))^2}=\sqrt{\frac{\pi^{2}}{4}+(\frac{\pi}{2}-1)^2}=\sqrt{\frac{\pi^{2}}{4}+\frac{\pi^{2}}{4}-\pi + 1}=\sqrt{\frac{\pi^{2}}{2}-\pi + 1}\)
\(L_3=\sqrt{(\frac{3\pi}{2}-\pi)^2+((\frac{3\pi}{2}-1)-\pi)^2}=\sqrt{\frac{\pi^{2}}{4}+(\frac{\pi}{2}-1)^2}=\sqrt{\frac{\pi^{2}}{2}-\pi + 1}\)
\(L_4=\sqrt{(2\pi-\frac{3\pi}{2})^2+(2\pi-(\frac{3\pi}{2}-1))^2}=\sqrt{\frac{\pi^{2}}{4}+(\frac{\pi}{2}+1)^2}=\sqrt{\frac{\pi^{2}}{2}+\pi + 1}\)
\(L = L_1+L_2+L_3+L_4\)
\(L=\sqrt{\frac{\pi^{2}}{2}+\pi + 1}+\sqrt{\frac{\pi^{2}}{2}-\pi + 1}+\sqrt{\frac{\pi^{2}}{2}-\pi + 1}+\sqrt{\frac{\pi^{2}}{2}+\pi + 1}\)
\(\approx\sqrt{\frac{(3.1416)^{2}}{2}+3.1416 + 1}+\sqrt{\frac{(3.1416)^{2}}{2}-3.1416 + 1}+\sqrt{\frac{(3.1416)^{2}}{2}-3.1416 + 1}+\sqrt{\frac{(3.1416)^{2}}{2}+3.1416 + 1}\)
\(\approx\sqrt{4.9348+3.1416 + 1}+\sqrt{4.9348-3.1416 + 1}+\sqrt{4.9348-3.1416 + 1}+\sqrt{4.9348+3.1416 + 1}\)
\(\approx\sqrt{9.0764}+\sqrt{2.7932}+\sqrt{2.7932}+\sqrt{9.0764}\)
\(\approx3.0127+1.6713+1.6713+3.0127\approx9.368\)

Answer:

The length for \(n = 4\) is approximately \(9.368\)