Math 2433 - Exam 1 Review KEY. Problem 1. (a) (i) AB = (2, 4, 1), AC = (-3, 3, -6),
BC = (-5, -1, -7). AB · AC = 0 ⇒ AB+AC, A,B,C are vertices of a rt. triangle.
Math 2433 Exam 1 Review Key Problem 1.
√ ~ =3 6 (a) ||AC|| (b) 3x − y − 2z − 3 = 0 (c) (x − 2)2 + y 2 + (z − 32 )2 =
21 4
Problem 2. (a) ub =
√2 i 14
(b) compb a =
+
√3 j 14
−
√1 k. 14
−
9 14 k
√5 14
(c) projb c = 97 i + (d) θ = arccos √570
27 14 j
Problem 3. (a) P and L are parallel since the normal for P is perpendicular to the direction vector of L (dot product = 0). (b) x(t) = −3 − 2t, y(t) = 2t, z(t) = 5 + 2t (c) x(t) = −3 + 3t, y(t) = −t, z(t) = 5 + 4t
Problem 4. (a) x + 4y − z = 2 (b) P1 and P2 are not parallel. (c) cos θ =
x 7
=
y− 24 7 −5
=
z− 26 7 −13
1 √ 2 7
Problem 5. (a) lim f(t) = 7i + j. t→3
(b) 4t − 1 + 2π cos(πt) sin(πt)
Problem 6. (a) v(t) = r0 (t) = (−e−t sin t + e−t cos t)i + (−e−t cos t − e−t sin t)j + 3k √ (b) speed : ||r0 (t)|| = 2e−2t + 9. (c) a(t) = v0 (t) = r00 (t) = −2e−t cos ti + 2e−t sin tj.
1
Problem 7. (a) 9, 61 3
R3√
(b) L(C) =
0
t2 + t4 dt =
R3 √ t 1 + t2 dt = 13 (103/2 − 1) 0
Problem 8. (a) x = 6 + 2t, y = 9 + 6t, z = 9 + 9t R3√ R3 (b) L(C) = 0 4 + 4t2 + t4 dt = 0 (2 + t2 ) dt = 15 (c) T(1) = 32 i + 23 j + 13 k and N(1) = − 32 i + 13 j + 23 k
Problem 9. (a) T(t) =
1 2
cos ti +
1 2
√
sin tj +
3 2 k
; N(t) = − sin ti + cos tj
1 4t (c) aT = 2; aN = t; a(t) = 2T + tN (b) κ =
Problem 10. dom(f ) = {(x, y)|x2 + y 2 < 9}
Problem 11. (a) ellipses
(b) parabolas
Problem 12. 0 = lim 0 = 0 x2 + 0 x→0 2y (b) Along the y-axis, all points are (0, y) so we have lim = lim 2 = 2 y→0 0 + y y→0 (a) Along the x-axis, all points are (x, 0) so we have lim
x→0
(c) Along y = 3x2 , all points are (x, 3x2 ) so we have lim
x→0 x2
(d)
lim
6x2 6 3 = lim = 2 x→0 4 + 3x 2
f (x, y) = DNE since all paths do not yield the same answer
(x,y)→(0,0)
Problem 13. fx = y 3 exy + y1 , fy = 2yexy +xy 2 exy − yx2 , fxy = fyx = 3y 2 exy +xy 3 exy − y12 , fxx = y 4 exy , fyy = 2exy +4xyexy +x2 y 2 exy + 2x y3
Problem 14. (a) (i) ∇f = (x2 yex−1 + 2xyex−1 + 2y 2 )i + (x2 ex−1 + 4xy)j.
√ (ii) ∇f (1, −1) = −i − 3j; direction of most rapid decrease: −∇f (1, −1) = i + 3j; most rapid decr: −||∇f || = − 10
(b) (i) ∇F = (2x + 4y)i + (3z + 4x)j + 3yk 0
(ii) Fu (1, 1, −5) = (6i − 11j + 3k) ·
( 12 i
+
3 4j
−
1 4
√
√ −21 − 3 3 3k) = 4 2
