Question

Use Theorem 6.7 to prove that a circle of radius 5 has circumference$10\mathrm{\pi }$.

Short answer

Circle of radius 5 has circumference $10\mathrm{\pi }$.

Step-by-step solution

Step 1. Given Information 

Circle of radius 5.

Step 2. Proving circumference of circle of radius 5 to be 10π.

$$\begin{gathered} \mathrm{Then}\mathrm{the}\mathrm{arc}\mathrm{length}\mathrm{of}f\left(x\right)\mathrm{from}x=a\mathrm{to}x=b\mathrm{can}\mathrm{be}\mathrm{represented}\mathrm{by}\mathrm{the} \\ \mathrm{definite}\mathrm{integral}:{\int }_{\mathrm{a}}^{\mathrm{b}}\sqrt{1+(\mathrm{f}\text{'}{\left(\mathrm{x}\right))}^2}d\mathrm{x} \\ \mathrm{Equation}\mathrm{of}\mathrm{circle}\mathrm{of}\mathrm{radius}5is:x^2+y^2=5^2 \\ \mathrm{Therefore},y=\sqrt{5^2-x^2}=f\left(x\right) \\ \mathrm{and}f\text{'}\left(x\right)=-\frac{x}{\sqrt{5^2-x^2}} \\ \mathrm{Circumference}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{is}\mathrm{twice}\mathrm{the}\mathrm{arc}\mathrm{length}\mathrm{from}x=-5\mathrm{to}x=5\mathrm{of}f\left(x\right) \\ \mathrm{Therefore},C=2{\int }_{-5}^5\sqrt{1+(\mathrm{f}\text{'}{\left(\mathrm{x}\right))}^2}d\mathrm{x} \\ \mathrm{C}=2{\int }_{-5}^5\sqrt{1+{\left(-\frac{\mathrm{x}}{\sqrt{5^2-{\mathrm{x}}^2}}\right)}^2}d\mathrm{x} \\ \mathrm{C}=2{\int }_{-5}^5\sqrt{1+\left(\frac{{\mathrm{x}}^2}{5^2-{\mathrm{x}}^2}\right)}d\mathrm{x} \\ \mathrm{C}=2{\int }_{-5}^5\sqrt{{\frac{5^2-{\mathrm{x}}^2+\mathrm{x}}{5^2-{\mathrm{x}}^2}}^2}d\mathrm{x} \\ \mathrm{C}=2{\int }_{-5}^5\frac{5}{\sqrt{5^2-{\mathrm{x}}^2}}d\mathrm{x} \\ \mathrm{Put},\mathrm{x}=5\sin \mathrm{t},\mathrm{dx}=5\cos \mathrm{t}\mathrm{dt} \\ \mathrm{Limits}\mathrm{of}\mathrm{integration}\mathrm{change}\mathrm{to}-\frac{\mathrm{\pi }}{2}\mathrm{to}\frac{\mathrm{\pi }}{2} \\ \mathrm{C}=2{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{5\times 5\times \cos \mathrm{t}}{\sqrt{5^2-{(5\sin \mathrm{t})}^2}}\mathrm{dt} \\ \mathrm{C}=2{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{5\times 5\times \cos \mathrm{t}}{\sqrt{5^2-{(5\sin \mathrm{t})}^2}}\mathrm{dt} \\ \mathrm{C}=10{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{5\times \cos \mathrm{t}}{5\times \cos \mathrm{t}}\mathrm{dt} \\ \mathrm{C}=10{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}1\mathrm{dt} \\ \mathrm{C}=10{\overline{)\mathrm{t}}}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \\ \mathrm{C}=10\left[\frac{\mathrm{\pi }}{2}-\left(-\frac{\mathrm{\pi }}{2}\right)\right] \\ \mathrm{C}=10\pi \end{gathered}$$