MAT137Y1 Lecture : MAT 137Y 2007-08Winter Session, Self Generated Solutions to Problem Set 7.pdf
70
MAT137Y1 Full Course Notes
Verified Note
70 documents
Document Summary
Mat 137y 2007-08 winter session, solutions to problem set 7. 1 (she 3. 7: differentiating, x2 + 4xy + y3 + 5 = 0 = 2x + 4y + 4x dy dx = 0. Differentiating again, we get dx = 0, so dy. At (2, 1) we get (cid:18)dy (cid:19)2 dx dy dx. At (2, 1) we get 2 + 11 d2y. 2 + 4 dx2 = 0, so d2y dx2 = 2. + 3y2 d2y dx2 = 0: for y = 2x, the slope is m1 = 2. For x2 xy + 2y2 = 28, we have. At a point of intersection of the line and the curve we have m2 = 0 since y = 2x. | m1| = 2, so is the angle between 0 and (a) differentiating, x2/3 + y2/3 = a2/3 = 2 (cid:1)1/3. Thus the dx = (cid:0) y slope at (x1,y1) (where x1 (cid:54)= 0) is m = (cid:16) y1.
