By Thomas R. Kane

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Solution: Let a be the position vector of A relative to B, and b a unit vector parallel to edge BC, and use primes to denote differentiation with respect to z in R. Then SEC. 4 52 [Chapter 2 (A) From Fig. 3, where y is the distance from B to the X-axis. As the distance between A and B is fixed at 7 ft, Q2 so that = 9 + yl + Z2 = 72 = 49 y = (40 - z2)1'2 Hence α and = - 3 η ! - (40 - z2)1/2n2 + zn3 a' = ^(40 - z 2 )- 1 ' 2 ^ + n3 Thus a'|z=2 = « n2 + n3 (B) As b is perpendicular to both a and n3, it can be expressed in terms of the cross product of these vectors : , b α X n3 |a X n3| - ( 4 0 - ζψ^ + 3n2 (49 - z2)1'2 Differentiating, b' = (49 - ζ2)*/2(40 - z2)-1/2zni + [ - ( 4 0 - z2)1/2ni + 3n2](49 - z 2 )" 1 ^ 49 - z2 Thus b i _ ~ 6 n i + 3n2 =2 1/2 .

3) d< k X t k X n J d * "d* Hence Ä 'dn v , Ä'd ' | * χ . , . - * χ . , χ . ( | ) · di Χ 1 ( | ) · while Thus 'f-»x-)-*x-)-*x-)S-S di R R It follows from Sec. 2 that '

P ds2 This follows from Sees. 4, together with the facts that Rd2p/ds2 is perpendicular to Rdp/ds and Rdp/ds is a unit vector. SECS. 1 Diß. 4 The plane passing through P and normal to v is called the rectifying plane of C at P . 1 The position vector p of the center of curvature of a space curve C at a point P of C, relative to P , is called the vector radius of curvature of C at P , and is given by p= (^T ( p ' X p " ) X p ' w here p is the position vector of P relative to a point 0 fixed in a reference frame Ä in which C is fixed and primes denote differentiation with respect to zin R, z being any scalar variable such that the position of P on C depends on z.

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