Solutions to the Second Homework Set (page 22) 1. (a) Parametrization: Let r(t) = hcos t; sin ti and t 2 (0; 2). The t 1 1 value for the point P = ( 2 ; 2 ) can be found from the parametrization: We need to nd t so that r(t) = h 12 ; 12 i so t must satisfy the two equations cos t = p1 ; sin t = p1 2 2 from which we conclude that t = 4 . Tangent vector: r (t) = h, sin t; cos ti. Evaluating r at t = 4 yields r ( 4 ) = h, p12 ; p12 i for the tangent vector. (b) Parametrization: Let r(t) = h1+cos t; ,1+sin ti, t 2 (0; 2). Find t so that r(t) = h1 + 12 ; ,1 + 12 i. So 1 + cos t = 1 + p1 ; ,1 + sin t = ,1 + p1 2 2 . So t = 4 . Tangent vector: r (t) = h, sin t; cos ti and r ( 4 ) = h, 12 ; 12 i. 2 (c) Parametrization: First divide the equation by 6 to get x22 + 2y3 = 1. 2 Now, let x22 = cos2 t and 2y3 = sin2 t, from which we get r3 p x(t) = 2 cos t; y (t) = 2 sin t; and, therefore, r3 p r(t) = h 2 cos t; 2 sin ti: p p p p 0 0 0 p p 0 0 p p Next, we nd t so that r(t) = h 12 ; 2 32 i or p p p Hence, t satises 2 cos t = p1 ; 2 cos t = 12 ; p r3 p p3 : 2 sin t = 2 2 p sin t = 23 ; which implies that t = 3 . q p Tangent vector: r (t) = h, q2 sin t; 32 cos ti which when evaluated at t = 3 yields r ( 3 ) = h, 32 ; 2 32 i. (d) Parametrization: 0 p 0 p 1