Advanced Calculus fi..

128 Advanced Calculus, Fifth Edition6. Let I' move in space with spccd I as in Problcm 5. Thc c.lrr~~:trrr-c, of the path is definedas K. whc1.cThus if A- # 0. T and (I/~)rlT/rls = N arc perpendicular unit vcctors: T is tangcnt tothc curvc at P. N is normal to the curvc at P and is termed thc prirrciptrl normal. Therrrrliu.~ cfc-u~~trture is p = I/K. The vcctor B = T x N is called the hirrorn~ttl. Establishthc following relations:a) B is a unit vcctor and (T. N. B) is a positive triple of unit vcctorsb) $ . ~ = O a n d g . ~ = OC) thcre is a scalar -s such that 2 = -rN; r i4 known as thc mrsinr~d) 2 = -KT + r B. The equationsare known as the Frrneffortnlt1tl.v. For further properties of curves, see the book by Struiklisted at the end of the chapter.7. Let a point P move in space at speed v = ti.r/dt = ivl # 0. Show:a) a = 2 = $T+ $N (p # 0,. [Hint: Sct v =vT, differentiate. and use the formulacfT/ds = KN = (l/p)N of Prohlcm 0.1b) K = LI-~~Vx al.C) If w = rla/dt and v x a # 0, then r = Iv x al--'v x a. w.d) The path is a straight line if and only if K = 0.e) The path lies in a plane if and only iT s = 0.Note.point.Wherc v = 0, the curvature K is defincd to bc 0. If v = 0, thc path reduces to a8. For each of the following surfaccs, find the tangent plane and normal linc at thc pointindicated. verifying that the point is in the surface:a) .r' + y' + ,-' = 9 at (2.2, I )b) r,:t b2 - -2 -, -Oat(O,O, 1)c) .r3-.ry'+!:'-z3=0at(l. 1, 1)d) .u' + ?.' -:' = 0 at (0, 0. 0)Why docs the procedure break down in (d)? Show by graphing that a solution isimpossible.e) .rJ -: =Oat(.v~,y~.;~).wherc.r~y~f) xy+vz+.\z = I at(.rl.y~.z~).whcrc.r~v~ + VJ:~ +.r1;1 = 19. Show that thc tangcnt plane at (.ul. y ~. :I) to a surl'acc given by an cquation z = .f(x. y)is as follows:=:IObtain the cquatinns of thc normal linc.