Q. 78 - Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1. Given Information

Circle of radius 5.

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

$$\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} 5 is: 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}$$