MATH100 Quiz: MATH100 Midterm 2007 Winter Solutions

• School: University of Alberta
• Department: Mathematics
• Course: MATH100
• Professor: All

Document Summary

Solutions for math 100 2007 midterm: each limit exists and is given by (a) (b) lim x!1 x 1. = (cid:0: in order that the two curves be orthogonal it is necessary that at the point of intersection the slopes be the negative reciprocals of each other. Thus y = k (cid:0) x =) y0 = (cid:0)1 and for y = x2 =) y0 = 2x: Hence, it is necessary that y0jline (cid:2) y0jparabola = ((cid:0)1) (cid:2) (2x) = (cid:0)1 () x = 4: the derivatives are given by (a) y0 = 1 + 1 + e2x ; (b) y0 = (1 + sin (2x)) sech2(cid:0)x + sin2 (x)(cid:1) ; (c) (cid:0)y0 sin (y) = (1 + y0) cos (x + y) =) y0 = (cid:0) (d) ln (2) 2yy0 = By the fundamental properties of polynomials and trigonometric functions it follows that f (x) is continuous for all x 2.