420 Mr DA VIES on the Equations of Loci 



whose tangent was given. The problem, as we have resolved it for the 

 tangent, is in a form more general than it was proposed ; and resembles 

 those cases in common algebra where irrelevant answers are mixed up with 

 those required by the peculiar terms of the question. The same is true of 

 all the secondary branches of spherical curves that result from solving the 

 questions for some trigonometrical functions of an arc instead of solving 

 them for the arc itself; viz. those which we have denominated cylindrical 

 branches given by means of sines, cosines, versed-sines, secants, and cose- 

 cants, with the conical ones given by means of tangents and cotangents. 



(f t ). To find the ultimate inclination of the tangent to the equator, we 

 have, by NAPIER, 



1 "-= cot QES 



sin<p 

 Now when = we have cot QES = = infinity, or QES = 0. 



AI U 



Hence ultimately the curve coincides with the equator. This agrees with 

 the last determination, that the curve does not cross the equator, but 

 coalesces with it at the infinite value of 6. 



(g,). To find the ultimate inclination to the radius-vector. We have 

 by NAPIER, 



cos <p = tan a cot ESQ, or 



tan ESQ = tan a sec (p. 



But when tp = 0, sec = 1 , and tan ESQ = tan a, or 



ESQ = a, or ESQ = it + a, as we should expect. 



3. We shall next take (6'), namely tan ( <) e " . 



(a /7 ). Let <j) Tr; then tan ( $) = 0, or log tan = ; and hence since 



e.^1, all its powers are greater than 1, and consequently 6 = -^ . The 

 curve, therefore, winds round P' as a polar asymptote. 



(ft,,). Let = y- Then log tan ( <p) - + -1 = infinity ; or d = + i . 



That is, the curve approximates continually towards the equator, and only 

 meets it when 6 has obtained an infinite positive value. 



