Math 241, Final Exam. 12/8/11.
Name:
• No notes, calculator, or text.
• There are 200 points total. Partial credit may be given.
• Write your full name in the upper right corner of page 1.
• Do problem 1 on p.1, problem 2 on p.2,... (or at least present your solutions in numerical order).
• Circle or otherwise clearly identify your final answer.
1. (5 points): Give the cosine of the angle between the vectors
~u = ~i − ~j + 2~k,
~v = 2~i − ~j + ~k.
Solution: We have
cos θ =
~u · ~v
5
2+1+2
√
= .
=√
k~ukk~vk
6
1+1+4 4+1+1
2. (15 points): Lines. Find parametric equations for the line L through the point (1, 2, 3) that
is parallel to the plane Q : x + y + z = 2 and perpendicular to the line
L1 : x = 1 + 2t; y = 1 − 2t; z = t.
Solution: The vector ~n = h1, 1, 1i is normal to Q; the vector ~v1 = h2, −2, 1i is the direction
vector of L1 . It follows that L has direction vector
~i ~j ~k
~n × ~v1 = 1 1 1 = ~i(1 + 2) − ~j(1 − 2) + ~k(−2 − 2) = 3~i + ~j − 4~k.
2 −2 1 It follows that L has parametric equations
x = 1 + 3t;
y = 2 + t;
z = 3 − 4t, t ∈ R.
3. (15 points): Planes. Find the equation of the plane that passes through the point (1, 2, 3)
and contains the line
L : x = 3t;
y = 1 + t;
z = 2 − t, t ∈ R.
Solution: The line has direction vector ~v = h3, 1, −1i. We set t = 0 in the line to obtain a
second point in the plane, (0, 1, 2); therefore, a second vector in the plane is ~u = (1, 2, 3) −
(0, 1, 2) = h1, 1, 1i. A normal vector to the plane is
~i ~j ~k ~v × ~u = 3 1 −1 = ~i(1 + 1) − ~j(3 + 1) + ~k(3 − 1) = 2~i − 4~j + 2~k.
1 1 1 Hence, an equation for the plane is
2(x − 1) − 4(y − 2) + 2(z − 3) = 0 ⇐⇒ 2x − 2 − 4y + 8 + 2z − 6 = 0
⇐⇒ 2x − 4y + 2z = 0 ⇐⇒ x − 2y + z = 0.
4. (15 points): Tangent plane. Find the equation of the plane tangent to z = x2 y + 2xy 3 at
the point (1, 1, 3).
Solution: With z = f (x, y) = x2 y + 2xy 3 , we compute
fx = 2xy + 2y 3 =⇒ fx (1, 1) = 4;
fy = x2 + 6xy 2 =⇒ fy (1, 1) = 7.
It follows that
4(x − 1) + 7(y − 1) = z − 3 ⇐⇒ 4x − 4 + 7y − 7 = z − 3 ⇐⇒ 4x + 8y − z = 8.
5. (15 points): Chain rule.
2
(a) (10 points): Let z = exy , x = t cos t, y = t sin t. Compute dz/dt. You may leave
your answer in the variables x, y, and t.
Solution: The chain rule gives
dz
∂z dx ∂z dy
2
2
=
+
= (y 2 exy )(−t sin t + cos t) + (2xyexy )(t cos t + sin t).
dt
∂x dt
∂y dt
(b) (5 points): Suppose that
z = F (x, y), x = G(u, v), u = H(t), v = I(t).
Write down the form of the chain rule that you would use to compute dz/dt. (Use a
tree diagram to show variable dependencies.)
Solution: We have
dz
∂z ∂x du ∂z ∂x dv
=
+
.
dt
∂x ∂u dt
∂x ∂v dt
6. (15 points): Gradient and directional derivative. Let F (x, y, z) = 2x3 y − 3y 2 z.
(a) (10 points): Compute the directional derivative of F at (1, −1, 1) in the direction of
the vector ~u = h2, 3, 6i. Is F increasing or decreasing at this point?
Solution: We have ∇F = 6x2 y~i+(2x3 −6yz)~j−3y 2~k; it follows that ∇F (1, −1, 1) =
~u
h2, 3, 6i
h2, 3, 6i
−6~i + 8~j − 3~k. Furthermore, we have
=√
=
. We compute
k~uk
7
4 + 9 + 36
D~u F (1, −1, 1) = h−6, 8, −3i ·
h2, 3, 6i
−12 + 24 − 18
6
=
=− .
7
7
7
Since D~u (1, −1, 1) < 0, the function is decreasing.
2
(b) (5 points): In what direction from (1, −1, 1) is the directional derivative of F a maximum? Your answer need not be a unit vector.
Solution: The direction of maximum increase is ∇F (1, −1, 1) = h−6, 8, −3i.
7. (15 points): Local extrema. Let f (x, y) = x3 + y 3 − 3x − 12y + 20. Find the point(s)
(x, y) at which f (x, y) has a local maximum, minimum, or saddle.
Solution: We have
fx = 3x2 − 3 = 0 =⇒ x = ±1; fy = 3y 2 − 12 = 0 =⇒ y = ±2.
The critical points are {(1, 2), (1, −2), (−1, 2), (−1, −2)}. In order to apply the second
derivative test, we require
6x 0 = 36xy.
fxx = 6x, fyy = 6y, fxy = 0, D = 0 6y We test the points as follows:
(1, 2): fxx > 0 and D > 0 =⇒ (1, 2) is a local minimum.
(1, −2): D < 0 =⇒ (1, −2) is a saddle point.
(−1, 2): D < 0 =⇒ (−1, 2) is a saddle point.
(−1, −2): fxx < 0 and D > 0 =⇒ (−1, −2) is a local maximum.
8. (15 points): Absolute extrema. Find the absolute maximum of f (x, y) = 2x + 3y + 1 on
the triangular region D in the first quadrant bounded by:
L1 : x = 0 (y-axis); L2 : y = 0 (x-axis); L3 : x + y = 2.
Clearly explain your answer.
Solution: Since fx = 2 6= 0 and fy = 3 6= 0, there are no critical points inside D. Hence,
the extrema of f (x, y) lie on the boundary of D.
L1 : We have u1 (y) = f (0, y) = 3y + 1. Since u01 (y) = 3 6= 0, the extrema on L1 occur at
the endpoints: f (0, 0) = 1 and f (0, 2) = 7.
L2 : We have u2 (x) = f (x, 0) = 2x + 1. Since u02 (x) = 2 6= 0, the extrema on L2 occur at
the endpoints: f (0, 0) = 1 and f (2, 0) = 5.
L3 : We have u3 (x) = f (x, 2 − x) = 2x + 3(2 − x) + 1 = 7 − x. Since u03 (x) = −1, the
extrema on L3 occur at the endpoints: f (0, 2) = 7 and f (2, 0) = 5.
It follows that the absolute maximum of f (x, y) on D is f (0, 2) = 7.
3
9. (15 points): Lagrange multipliers. Use Lagrange multipliers to find the point(s) (x, y, z)
with x, y, z ≥ 0 which maximize f (x, y, z) = xyz 2 subject to the constraint x+2y +2z = 6.
Solution: Let g(x, y, z) = x + 2y + 2z. We have
∇f = yz 2~i + xz 2~j + 2xyz~k = λ(~i + 2~j + 2~k) = λ∇g.
Equating components gives
xz 2
= λ, 2xyz = 2λ =⇒ xyz = λ.
2
yz 2 = λ, xz 2 = 2λ =⇒
2
Now, yz 2 = xz2 implies that y = x2 . Similarly, yz 2 = xyz implies that x = z. We substitute
in the constraint equation to obtain
x+2
3
3
3
+ 2x = 4x = 6 =⇒ x = , y = , z = .
2
2
4
2
x
10. (15 points): Polar coordinates. Set up an integral in polar coordinates for the volume of
the solid E that lies below the plane z = 6 and above the paraboloid z = 3x2 + 3y 2 in the
first octant. (How do the plane and paraboloid intersect?)
Solution; We have
ZZ
V =
(6 − 3x2 + 3y 2 ) dA =
Z
√
π
2
Z
(6 − 3r2 )r dr dθ = 3
Z
0
0
D
2
0
π
2
√
Z
2
(2 − r2 ) dr dθ.
0
11. (15 points): Rectangular coordinates. Suppose that E is the tetrahedron (a polyhedron
with four vertices and four triangular faces, three of which meet at each vertex) bounded by
the planes
x = y, x + y = 4, y = z, z = 0 (xy-plane).
ZZZ
Express the integral
dV as an iterated integral with dV = dy dx dz. Do not evaluate.
E
(Draw a quick sketch. What is the shadow (projection) of E on the xz-plane?)
Solution: We have
ZZZ
Z
2
Z
4−z
Z
dV =
dy dx dz.
0
E
4−x
z
x
12. (15 points): Cylindrical coordinates. Convert the integral
Z
3
√
Z
9−x2
Z
9−x2 −y 2
y
−3
0
0
4
p
x2 + y 2 dz dy dx
from rectangular to cylindrical coordinates. Do not evaluate.
Solution: We have
√
3
Z
Z
−3
0
π
Z
9−x2
3
Z
9−x2 −y 2
Z
0
9−r2
Z
0
π
3
Z
9−r2
Z
r sin θ · r · r dz dr dθ
0
0
0
r3 sin θ dz dr dθ.
=
0
Z
p
y x2 + y 2 dz dy dx =
0
13. (15 points): Spherical coordinates. Use spherical coordinates to evaluate the triple integral
ZZZ
xz dV
E
2
2
2
where
pE is the solid region that lies within the sphere x + y + z = 4 and below the cone
z = x2 + y 2 in the first octant.
Solution: We have
Z
π
2
π
2
Z
Z
2
π
2
Z
2
π
2
Z
Z
(ρ sin φ cos θ)(ρ cos φ)(ρ sin φ) dρ dφ dθ =
0
π
4
0
0
! Z
π
2
Z
=
cos θ dθ
!
3 1
u
3
! Z
√
2
2
32
5
=
2
sin2 φ cos φ dφ
π
4
0
=
π
2
ρ4 dρ
32
15
√
(1 − ( 2/2)3 ) =
0
=
sin θ
0
32
15
ρ4 sin2 φ cos φ cos θ dρ dφ dθ
π2 ! Z
0
π
4
2
!
1
√
u= 22
u2 du
ρ5
5
R
over the square R with vertices {(0, 2), (1, 1), (2, 2), (1, 3)}. Use a suitable change of
variable.
It follows that
∂(x, y) 1 1
∂(u, v) = 2 = 2 .
5
0
√ !
√
√
2
32 8 2
8
1−
=
−
= (4 − 2).
4
15
15
15
14. (15 points): Change of variable in a double integral. Compute the double integral
ZZ
x−y
dA
x+y
Solution: Let u = x − y and v = x + y. We have
∂(u, v) 1 −1
=
= 2.
∂(x, y) 1 1 2 !
Therefore, we have
ZZ
x−y
1
dA =
x+y
2
ZZ
R
u
du dv.
v
S
It remains to determine S and to do the evaluation. Since the transformation is linear, it will
map the square to a quadrilateral whose vertices are:
(0, 2) 7→ (−2, 2); (1, 1) 7→ (0, 2); (2, 2) 7→ (0, 4); (1, 3) 7→ (−2, 4).
The quadrilateral S is in fact, a square. We obtain
ZZ
R
x−y
1
dA =
x+y
2
Z
2
4
Z
0
−2
u
1
du dv =
v
2
Z
2
4
u2
2v
0 !
Z
dv = −
−2
= −(ln 4 − ln 2) = −(2 ln 2 − ln 2) = − ln 2.
6
2
4
dv
v