458 report — 1878. 



Since the equation of the line is 



x + y cot (p -f(<j>) = o. (1) 



Differentiating with respect to (p we get 



y =->'($> Bin ^ j (2) 



hence irom (1) we get 



*=/ «>)+/' M>) sin <£ cos <£, (3) 



and eliminating $ between (2) and (3) we have the Cartesian equation required. 

 Thus, if a line of constant length slide along two rectangular lines, we have 



v = a cos (p . • . f ((p) = a cos (p. 

 Hence from (2) and (3) 



y = a sin 3 </>, .r = a cos 3 cf> 

 .*.«' + y3 = as, 



3. If we differentiate the values of x and y in equations (2) and (3) we get by 

 squaring and adding, &c, 



(Is 

 ^=2/'(0) cos <£+/'«>) sir. <£, 



Which may be written 



ds d / \ 



df- dti/M *"'* )■ (4) 



sin 

 Thus the equation of the evolute of an ellipse is 



V« r - + b" tan *<£. 

 Hence, from equation (4) we get, putting 



A W>) = N/l-e 2 sin a <£ 

 b 2 1 



(5) 



■■ fl A 3 ((p)' 



Which is the intrinsic equation of the evolute of an ellipse. 



4. From formula (4) we can find conversely the tangential equation from the 

 intrinsic. 



Thus, let S = F(<£) be the intrinsic equation, and then we have from equation (4) 



4( //W SlD S *) =F W sm cf> 

 /($) = / c°sec 2 $| J ¥'{$) sin cpd(p\d(p. 



(6) 



For example, the intrinsic equation of the evolute of the catenary is 



s = e / tan (£d<£ 



,".F(<£)=c/ tan <pd(p. 

 Hence by formula (6) we get the tangential equation of the evolute 



v = c < 1 — log ( sec (p + tan <£ I [• . ^ ' 



tan <p 



5. Let the tangential equation of a curve be v =f((p), then, denoting the radius 

 of curvature by p, we have from equation (4) 



dr.., v (8) 



psin ^ = 4( //(<w siu2< 0' 



