Mathematics 54
Fourth Exam Answer Key
I. ~r(t) = h3 sin 2t, 3 cos 2t, 8ti
2.
1. ~r0 (t) = h6 cos 2t, −6 sin 2t, 8i k~r0 (t)k = 10 ´t s s(t) = 0 10du = 10t =⇒ t = 10 s s 4 ~r(s) = 3 sin , 3 cos , s 5 5 5 ~r(20π) = h0, 3, 16πi =⇒ point (0, 3, 16π) 1 2. T~ (t) = h6 cos 2t, −6 sin 2t, 8i
10 1 h−12 sin 2t, −12 cos 2t, 0i 10 ~ (t) = N 1 (12) 10 = h− sin 2t, − cos 2t, 0i (0, 3, 8π) ⇒ t = π 3 4 T~ (π) = , 0, 5 5
III. 1. f (x, y, z) = (3x + 4y − 5z)6 fx (x, y, z) = 6(3x + 4y − 5z)5 (3) fxy (x, y, z) = 18(5)(3x + 4y − 5z)4 (4) fxyz (x, y, z) = 360(4)(3x + 4y − 5z)3 (−5) fxyz (0, 1, 1) = 7200 2. F (x, y, z) = z 2 +ln(sinh(xy))−exz tan(x+y 3 ) = 0 ∂F/∂y ∂z =− ∂y ∂F/∂z 1 (cosh xy)x − exz sec2 (x + y 3 )(3y 2 ) ∂z sinh xy = − ∂y 2z − exz x tan(x + y 3 )
~ (π) = h0, −1, 0i N 4 3 ~ B(π) = , 0, − 5 5 Osculating plane:
3 4 x − (z − 8π) = 0 5 5
Rectifying plane: −(y − 3) = 0 Normal plane:
3 4 x + (z − 8π) = 0 5 5
k~r0 (π) × ~r00 (π)k k~r0 (π)k3 kh6, 0, 8i × h0, −12, 0ik = kh6, 0, 8ik3 kh96, 0, −72ik = 103 24 kh4, 0, 3ik = 103 3 k= = 0.12 25
3. k(π) =
25 ρ= 3 ~r0 (π) · ~r00 (π) = 0 k~r0 (π)k 0 00 k~r (π) × ~r (π)k = = 12 k~r0 (π)k
4. aT = aN
II. 1.
x3 y + y2
lim
(x,y)→(0,0) 3x6
along y = mx:
mx4 mx2 = lim =0 x→0 3x6 + m2 x2 x→0 3x4 + m2 lim
3
along y = x :
x6 1 = + x6 4 ∴ limit DNE lim
x→0 3x6
x−2y+z e (x − y + z) lim cos−1 x2 − y 2 − z 2 + 2yz (x,y,z)→(0,1,2) 0 e (1) = cos−1 = cos−1 (−1) = π −1
3. w = zex/y ; x = r cos θ, y = r sin θ, z = r2 + θ2 ∂w ∂w ∂x ∂w ∂y ∂w ∂z = + + ∂r ∂x ∂r ∂y ∂r ∂z ∂r ∂w 1 x = zex/y cos θ + zex/y − 2 sin θ + ex/y (2r) ∂r y y IV. 1.
dl dw dA = Al + Aw dt dt dt dl dw =w +l dt dt = 5(−3) + 8(2) dA = 1ft2 /s dt
2. f (3, 4) = 5 x 3 fx (3, 4) = p = 2 2 5 x +y y 4 fy (3, 4) = p = 5 x2 + y 2 3 4 L(x, y) = 5 + (x − 3) + (y − 4) 5 5 p 3 4 (3.01)2 + (3.95)2 ≈ 5 + (0.01) + (−0.05) 5 5 p 2483 2 2 (3.01) + (3.95) ≈ = 4.966 500
3. (use p #2)
x2 + y 2 3 4 dh = (0.2) + (−0.04) 5 5 11 dh = = 0.088 125 h=