Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/184



Find the radius of curvature at any point of the catenary $$y = \tfrac{a}{2} (e^{\tfrac{x}{a}} + e^{-\tfrac{x}{a}})$$.




 * Solution.
 * style="text-align: right;"|$$\frac{dy}{dx}$$
 * $$= \frac{1}{2} (e^{\frac{x}{a}} - e^{-\frac{x}{a}}); \frac{d^2 y}{dx^2} = \frac{1}{2a} (e^{\frac{x}{a}} - e^{-\frac{x}{a}})$$.
 * colspan="3"|Substituting in (42),
 * colspan="2" style="text-align: right;"|$$R = \frac{\left[ 1 + \left( \frac{e^{\frac{x}{a}} - e^{-\frac{x}{a}}}{2} \right)^2 \right]^{\frac{3}{2}} }{\frac{ e^{\frac{x}{a}} - e^{-\frac{x}{a}} }{2a}}$$
 * $$ = \frac{\left( \frac{ e^{\frac{x}{a}} - e^{-\frac{x}{a}} }{2} \right)^3}{ \frac{ e^{\frac{x}{a}} - e^{-\frac{x}{a}} }{2a}} = \frac{a (e^{\frac{x}{a}} - e^{-\frac{x}{a}})^2}{4} = \frac{y^2}{a}$$. Ans.
 * colspan="3"|If the equation of the curve is given in parametric form, find the first and second derivatives of y with respect to x from (A) and (B), &sect;97, namely:
 * style="text-align: right;"|$$\frac{dy}{dx}$$
 * $$= \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }$$, and
 * style="text-align: right;"|$$\frac{d^2 y}{dx^2}$$
 * $$= \frac{ \frac{dx}{dt} \frac{d^2 y}{dt^2} - \frac{dy}{dt} \frac{d^2 x}{dt^2} }{ \left( \frac{dx}{dt} \right)^3 }$$;
 * }
 * style="text-align: right;"|$$\frac{dy}{dx}$$
 * $$= \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }$$, and
 * style="text-align: right;"|$$\frac{d^2 y}{dx^2}$$
 * $$= \frac{ \frac{dx}{dt} \frac{d^2 y}{dt^2} - \frac{dy}{dt} \frac{d^2 x}{dt^2} }{ \left( \frac{dx}{dt} \right)^3 }$$;
 * }
 * $$= \frac{ \frac{dx}{dt} \frac{d^2 y}{dt^2} - \frac{dy}{dt} \frac{d^2 x}{dt^2} }{ \left( \frac{dx}{dt} \right)^3 }$$;
 * }

and then substitute the results in (42).

Find the radius of curvature of the cycloid




 * Solution.
 * style="text-align: center;"|$$\frac{dx}{dt} = a(1 - \cos t)$$,
 * style="text-align: center;"|$$\frac{dy}{dt} = a \sin t$$;
 * style="text-align: center;"|$$\frac{d^2 x}{dt^2} = a \sin t$$,
 * style="text-align: center;"|$$\frac{d^2 y}{dt^2} = a \cos t$$,
 * }
 * style="text-align: center;"|$$\frac{d^2 y}{dt^2} = a \cos t$$,
 * }


 * Substituting in (G) and (H), and then in (42), we get