Math 265
Final Exam
Full Name:
Fall 2012
Instructor/TA:
1. Equation for plane
AB = h4, 1, −2i, AC = h3, 2, −1i
so a normal vector is h3, −2, 5i. An equation is then
3x − 2y + 5z = 4
(or equivalent forms).
b) Plane intersects y-axis at (0, −2, 0).
2. Motion in the plane
v = ht cos t, t sin ti,
a = hcos t − t sin t, sin t + t cos ti.
b) components of acceleration
aT = a.T = 1,
aN = a.N = a.h− sin t, cos ti = t.
3. Tangent plane
Normal to the given surface is n = h2x − 3y − 2, −3x + 5y + 3, −1i. This needs to be
parallel to h2, −4, 1i. So there has to be a number c such that one vector equals c times
the other,
2x − 3y − 2 = 2c
−3x + 5y + 3 = −4c
−1 = 1c
So c = −1, 2x − 3y = 0 and −3x + 5y = 1. This is true only for the point (x, y) = (3, 2).
4. Local Extrema
∇f = h4x3 − 4y − 14x + 4, −4x + 8y − 8i
equals h0, 0i exactly for y = x/2 + 1 and (substitute for y into the other equation)
4x3 − 2x − 4 − 14x + 4 = 4x(x2 − 4) = 0,
so x = 0, ±2. This gives the points (0, 1), (2, 2), (−2, 0).
Classification. We compute
fxx
fyy
fxy
D
=
=
=
=
12x2 − 14
8
−4
8(12x2 − 14) − 16 = 32(x2 − 4)
and get for the points listed above D = −128, 0, 0 (in this order). So (0, 0) is a saddle,
the SPT is inconclusive for the two other points.
Math 265
Final
Page 2
5. Order of integration
The inequalities y 2 ≤ x ≤ 2 − y mean the region is bounded on the left by the parabola
y 2 = x, on the right by the line 2 − y = x. The inequalities 0 ≤ y ≤ 1 say that the region
is bounded on top and bottom by the lines y = 0, 1, but a sketch shows that the line
y = 1 is not actually needed. We find the smallest and biggest x-value
√ as x = 0, 2 (both
happen for y = 0). Then we solve all inequalities for y. Both y ≤ x and y ≤ 2 − x
need to be true, so
√
0 ≤ y ≤ min( x, 2 − x).
So the integral equals
2
Z
Z
√
min( x,2−x)
f (x, y) dy dx.
0
0
6. Spherical / Cylindrical coordinates
Describe solid with inequalities, translate to spherical
ρ ≤ 4 cos φ,
cos phi ≤ sin φ.
This gives limits π/4 ≤ φ ≤ π/2 and
Z π/2 Z 4 cos φ Z
V =
π/4
0
2π
ρ4 sin3 φ dθ dρ dφ.
0
The integral over θ could be in any other position, too.
7. FTC for line integrals / independent of patch / conservative
a) We try to find a potential function f (x, y) such that
fx = y −
so f = xy +
1
x
1
,
x2
+ C(y). Then fy = x + C 0 (y) which means
f (x, y) = xy +
1 1
− + D.
x y
Since F = ∇f , F is conservative.
b) We can use the FTC for line integrals,
Z
1
9
F · dr = f (4, 1) − f (1, 1) = 4 + − 1 − 1 = .
4
4
C
8. Gauss’ Divergence Theorem says that the flux integral in question equals
ZZ
ZZZ
F · n dS =
div F dV.
∂S
S
It turns out that div F is really simple,
div F = 2z.
Math 265
Final
Page 3
For the volume integral, we best use cylindrical coordinates. Note that
r2 ≤ z ≤ 2 − r
gives an upper bound for r (sketch the functions r2 and 2 − r! the biggest possible value
for r happens where their graphs meet, at r = 1). So 0 ≤ r ≤ 1, and
ZZ
Z 1 Z 2−r Z 2π
F · n dS =
2rz dθ dz dr
∂S
0
r2
0
Z 1
r5 − r(2 − r)2 dr
= 2π
Z0 1
3π
r5 − r3 + 4r2 − 4r dr = 2π(1/6 − 1/4 + 4/3 − 2) = − .
= 2π
2
0