92çB$ø`çü5 (-ç‚)
1. [a](3}) Graph the curves r = 3 sin θ and r = 1 + sin θ.
sol.
(None.)
[b](7}) Find the area of the region that lies inside the curve r = 3 sin θ and outside the curve r = 1 + sin θ.
sol.
Two curves intersect at
3 sin θ = 1 + sin θ
⇒ 2 sin θ = 1
⇒ sin θ = 1 2
⇒ θ = π
6 and 5π 6
Thus
A = Z 5π6
π 6
1
2[(3 sin θ)2− (1 + sin θ)2]dθ
= Z 5π6
π 6
1
2(9 sin2θ − 1 − 2 sin θ − sin2θ)dθ
= Z 5π6
π 6
1
2(8 sin2θ − 1 − 2 sin θ)dθ
= 1 2
Z 5π6
π 6
[8(1 − cos 2θ
2 ) − 1 − 2 sin θ]dθ
= 1
2(3θ − 2 sin(2θ) + 2 cos θ)
5π 6
π 6
= 1 23(4π
6 ) − sin(5π
3 ) + sin(π
3) + cos(5π
3 ) − sin(π 3)
= π +
√3 2 +
√3 2 −
√3 2 −
√3 2
= π
2. (15}) For the curve given by −→r (t) = (t−1, 2 ln t, 2t).Find [a] the unit tangent vector −→
T sol.
−
→r0(t) = (−t2,2t, 2), |−→r0(t)| = q
(−t2)2 + (2t)2+ (2)2 = t−2+ 2. Thus
−
→T = |−−→→rr(t)(t)| = (−t−
2,2t,2)
t−2+2 = (−1,2t,2t1+2t22). [b] the unit normal vector −→N sol.
−
→T0 = (1 + 2t2)(0, 2, 4t) − 4t(−1, 2t, 2t2) (1 + 2t2)2
(4t, 2 − 4t2, 4t)
|−→
T 0| = p(4t)2+ (2 − 4t2)2+ (4t)2 (1 + 2t2)2
=
√4 + 16t2 + 16t4 (1 + 2t2)2
= 2p(1 + 2t2)2 (1 + 2t2)2
= 2
1 + 2t2 Thus
−
→N =
−
→T0
|−→T0|
= (4t, 2 − 4t2, 4t) 2(1 + 2t2) [c] the curvature K
sol.
−
→K = |→− T0|
|−→r0|
= 2/(1 + 2t2) t−2+ 2
= 2t2 (1 + 2t2)2
[d] the osculating plane at the point P (1, 0, 2) sol.−
→B =−→ T ×−→
N |t=1 ⇒
−
→B =
i j k
−13 23 23
2 3 −13
2 3
= (23,23, −13).
Then the osculating plane is
−
→B · (x − 1, y − 0, z − 2) = 0
⇒ 2(x − 1) + 2y − (z − 2)
⇒ 2x − 2 + 2y − z + 2 = 0
⇒ z = 2x + 2y
3. (10}) ì2 f (x, y) =
( x2y
x2+y2 J (x, y) 6= (0, 0) 0 J (x, y) = (0, 0) [a] f (x, y) ÊÞ,µ<õu©/í?
sol.
All points on the plane.
[b] °Fj² −→u = α−→i + β−→j , α2+ β2 = 1, U)j²ûbD−→uf (0, 0)æ
Ê ½ D−→uf (0, 0) = fx(0, 0)α + fy(0, 0)β?
sol.
All −→u with D−→uf (0, 0) = α2β. No!
[c] f (x, y) Ê (0, 0)õª}´? sol.
Not.
4. (15}) ° f (x, y) = (12 − x2+ y2)e1−(x2+y2) íF critical points 1‹J}é
1° f (x, y) Ê x2+ y2 ≤ 1í|×|üM
sol.
saddle point: (0, 0)
local maximum points: (0,q
1
2), (0, −q
1 2) local minimum points: (q
3
2, 0), (−q
3 2, 0) max =√
e, min = −12.
5. (10}) øÀ˘$,íÅ}0uT (x, y) = −e−2ycos x. øó¥Ê$,W, w
Ê©øõí«j²u•OÅÓ‹|×íj²‡ª J¤ó¥¦¬õ (x = π4, y = 0),
°øƒb f (x) U)Ç$y = f (x) /u¤ó¥í«*
sol.
f0(x) = TTxy = 2cos xsin x, so f (x) = 2 ln(√
2 sin x).
sol.
(∂w∂y)z = 5, ∂y∂z∂2w = 5.
7. (10}) Let f (x) = Rx
1 ey2dy. Find the average valus of f on the interval [0, 1].
sol.
f = 1 1 − 0
Z 1 0
f (x)dx
= Z 1
0
Z x 1
ey2dydx
= − Z 1
0
Z 1 x
ey2dydx
= − Z 1
0
Z y 0
ey2dxdy (by Fubini Thm)
= − Z 1
0
yey2dy
= − 1 2ey2
1
0
= 1
2(1 − e)
8. (10}) Let D be the lamina enclosed by x-axis, y = sin x, and 0 ≤ x ≤ π.
The density at (x, y) is given by ρ(x, y) = y. Find the coordinates of the center of mass and the moment of inertial about the y-axis.
sol.
âú˚4ø x = π2. m =Rπ
0
Rsin x
0 ydydx = π4. Mx =Rπ
0
Rsin x
0 y2dydx = 49.
∴y = Mmx = 9π16
Iy = Z π
0
Z sin x 0
x2ydydx
= Z π
0
1
2x2sin2xdx
= 1 2
Z π 0
x2· 1 − cos 2x
2 dx
= 1 4
Z π 0
x2dx − 1 4
Z π 0
x2cos 2xdx
= π3 12− 1
8 Z π
0
x2d(sin 2x)
Z π 0
x2d(sin 2x) = (x2sin 2x)|π0 − Z π
0
(sin 2x)2xdx
= 0 + Z π
0
xd cos 2x
= (x cos 2x)|π0 − Z π
0
(cos 2x)dx
= π − (1
2sin 2x)|π0
= π − 0 = π Iy = π123 −18π.
9. (10}) vÆ6ñ x2+ y2 ≤ ψ D7ñ4x2+ 4y2+ z2 ≤ bψ u°¶}íñ sol.
2 = ±pbψ − 4x2− 4y2, D = {(x, y) : x2+ y2 ≤ ψ}
V = Z Z
D
pbψ − 4x2− 4y2− (−pbψ − 4x2− 4y2)dA
= 2 Z Z
D
pbψ − 4x2− 4y2dA
= 2 Z 2π
0
Z 2 0
2√
1b − r2rdrdθ
= ψ Z 2π
0
Z 2
0 r(1b − r2)12drdθ
= ψ · 2π ·
−1
3(1b − r2)32
2
0
= 8π
3 (bψ − 24√ 3)
10. (10}) Find the volume of the largest rectangular box with edges parallel to the axes that can be inscribed in the ellipsoid 9x2+ 36y2+ 4z2 = 36.
sol.
f (x, y) = xyz, 9x2+36y2+4z2 = 36, 18x+8zzx = 0 and 72y +8zzy = 0.
fx = yz + xyzx, fy = xz + xyzy
fx = 0 ⇒ yz2+ xyzzx = 0 ⇒ y36−9x42−36y2 + xy(−18x)8 = 0 fy = 0 ⇒ xz2 + xyzzy = 0 ⇒ x36−9x24−36y2 + xy(−72x)8 = 0
⇒ x2+ 2y2 = 2, x2+ 8y2 = 4
⇒ x2 = 43, y2 = 13, z2 = 93 ⇒ xyz = √23.
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