BRIEF ANSWERS FOR TEST 2

• T or F?
  • F. The speed for A and B is the same constant value. Therefore they both experience only centripetal acceleration, which is directly proportional to curvature and to the square of speed. Since they have the same speed, only the curvature matters. The curve travelled by A has twice the curvature, so A experiences TWICE (not four times) the acceleration that B experiences.
  • F. Torsion = -(dB/ds) dot N, so if the binormal turns toward N then torsion is negative.
• Let p(t) = (e^t*cos(t), e^t*sin(t), e^t). For simplicity, we will occasionally abbreviate the functions e^t, cos(t) and sin(t) by: e, c, s respectively. Thus p(t) = (ec,es,e). Now v(t) = p'(t) = (ec-es,ec+es,e). Therefore |v| = square_root((ec-es)^2 + (ec+es)^2 + e^2) = (square_root(3))*e^t. Since we want to find the arc length of the curve between (1,0,1) (at t=0) and (e^(2*Pi),0,e^(2*Pi)) (at t=1), we must integrate (square_root(3))*e^t from t=0 to t=1. The answer is (square_root(3))*(e^(2*Pi) - 1).
• We need to find the TNB frame at t=0 for a particle travelling according to p(t) = (ec,es,e). From the last problem: T(t) = v/|v| = (ec-es,ec+es,e)/(square_root(3)*e) = (1/square_root(3))*(c-s,c+s,1), so T(0) = (1/square_root(3))*(1,1,1). Next, N(t) = T'(t)/|T'(t)| = [(1/square_root(3))(-c-s,c-s,0)]/(square_root(2/3)), so N(0) = (1/square_root(2))*(-1,1,0). B(0) = T(0) x N(0) = (1/square_root(6))*(-1,-1,2).
• Again p(t) = (ec,es,e).
  • This curve consists of the points on the cone z^2 = x^2 + y^2 which lie above a logarithmic spiral in the xy-plane.
  • kappa = |T'|/|v|. We computed T' in the previous problem and |v| in the problem before that, so kappa = |(1/square_root(3))(-c-s,c-s,0)|/[square_root(3)*e^t] = (square_root(2)/3)*e^(-t).
  • There are many ways to express the answer in the previous part so that we can see that the answer depends on the position only, and not on time. For example: if the position is (x,y,z), then kappa = (square_root(2)/3)*z^(-1) is one such way.
• H(x,y) = 2*x*y/(x+y)
  • The limit of H(x,y) as (x,y) goes to (0,1) exists (and equals H(0,1) = 0), since H(x,y) is continuous at (0,1). This is because it is a ratio of polynomials and the denominator is not zero at (0,1).
  • The limit as (x,y) goes to (0,0) does not exist since if we approach (0,0) along the line y = x we get lim_(x goes to 0) 2*x*x/(x+x) = lim_(x goes to 0) x = 0. However, if we travel along y = -x+x^2 we get lim_(x goes to 0) 2*x*(-x+x^2)/(x-x+x^2) = lim_(x goes to 0) 2(-1+x) = -2. Since different paths give different answers, there is no limit.