Math 3C — Exam #1 Sample

Math 3C — Exam #1 Sample
Laney College, Spring 2011
Fred Bourgoin
1. (14 points) Consider the function f (x, y) = x2 + 3y 2 + 1.
(a) Sketch a contour diagram for f with at least 3 level curves.
(b) Sketch the graph of f .
(c) Find a function g(x, y, z) such that the graph of f is a level surface of g.
(Don’t forget to indicate which level surface of g it is.)
2. (12 points) Consider the points P = (3, 2, −5) and Q = (−1, 3, −2) in R3 .
−−→
(a) Write the vector P Q in terms of ~i, ~j, and ~k.
−−→
(b) Find the magnitude of QP .
(c) What is the distance between P and Q?
3. (10 points) Suppose P and Q are as in the preceding problem, and let R = (2, 6, 4).
(a) What is the area of the triangle △P QR?
(b) Find an equation for the plane through P , Q, and R.
4. (8 points) Find the angle between the planes z = 2x − y + 3 and x − 3y + 2z = 5.
p
5. (8 points) Sketch the domain of f (x, y) = 1 + x − y 2 .
6. (10 points) Shortly after takeoff, a plane is climbing northwest through still air at
an airspeed of 200 km/hr, and rising at a rate of 300 m/min. Resolve its velocity
vector into components. The x-axis points east, the y-axis points north, and the
z-axis points up.
7. (10 points) The following table gives values of a function h(x, y) at 20 points.
1
3
x
5
7
9
−3
y
−2
−4
−2
−3
−5
−6
−7
−1
0
1
3
0
2
−3
−1
1
−5
−3
−1
−1
−4
−2
0
(a) Could h be a linear function? Why, or why not?
(b) Find a possible expression for h.
1
8. (6 points) Describe the level surfaces of f (x, y, z) = p
1
x2
+ 2y 2 + z 2
.
9. (16 points) Let ~u = 2~i + ~j − 3~k, ~v = ~i − ~j + 2~k, and w
~ = ~i − 5~j − ~k.
(a) Compute 2~u − 3~v .
(b) Are two of the vectors perpendicular? Which ones? Justify.
~ 2 such that w
~ =w
~1 + w
~ 2, w
~ 1 is parallel to ~v , and w
~2
(c) Find vectors w
~ 1 and w
is perpendicular to ~v .
10. (6 points) Find the volume of the parallelepiped defined by the vectors
~u = 2~i + ~j − 3~k ,
~v = ~i − ~j + 2~k ,
and
w
~ = ~i − 5~j − ~k
of the preceding problem.
EC. True/False (No justification required)
(a) If ~u · ~v < 0, then the angle between ~u and ~v is greater than π/2.
(b) If the contours of g(x, y) are concentric circles, then the graph of g is a cone.
(c) It is never true that ~v × w
~ =w
~ × ~v .
(d) The sphere x2 + y 2 + z 2 = 10 intersects the plane x = 10.
(e) For any vectors ~u and ~v , we have (~u + ~v ) · (~u − ~v ) = k~uk2 − k~v k2 .
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Math 3C — Exam #1 Sample Solutions
Laney College, Spring 2011
Fred Bourgoin
1. (14 points) Consider the function f (x, y) = x2 + 3y 2 + 1.
(a) Sketch a contour diagram for f with at least 3 level curves.
Solution. Setting f (x, y) = c yields
x2
c−1
+ y2 =
,
3
3
so your contour diagram should consist of concentric ellipses, closer to one
another as you move away from the origin. Shown here are the contours
c = 2, 3, 4, 5, 6 (from the inside out).
(b) Sketch the graph of f .
Solution. The graph is a paraboloid with vertex (0, 0, 1).
3
(c) Find a function g(x, y, z) such that the graph of f is a level surface of g.
(Don’t forget to indicate which level surface of g it is.)
Solution. Let g(x, y, z) = z − (x2 + 3y 2 + 1). Then the graph of f (x, y) is
also the graph of the level surface g(x, y, z) = c = 0.
2. (12 points) Consider the points P = (3, 2, −5) and Q = (−1, 3, −2) in R3 .
−−→
(a) Write the vector P Q in terms of ~i, ~j, and ~k.
−−→
Solution. P Q = (−1 − 3)~i + (3 − 2)~j + (−2 + 5)~k = −4~i + ~j + 3~k.
−−→
(b) Find the magnitude of QP .
p
√
−−→
−−→
Solution. kQP k = kP Qk = (−4)2 + (1)2 + (3)2 = 26.
(c) What is the distance between P and Q?
√
−−→
Solution. The distance between P and Q is the magnitude of P Q, so 26.
3. (10 points) Suppose P and Q are as in the preceding problem, and let R = (2, 6, 4).
(a) What is the area of the triangle △P QR?
−→
Solution. Since P R = −~i + 4~j + 9~k, we have
~i ~j ~k −−→ −→ P Q × P R = −4 1 3 = −3~i + 33~j − 15~k .
−1 4 9 −−→
The area of △P QR is half the area of the parallelogram with edges P Q and
−→
P R, so
√
√
1323
21
3
1p
=
.
(−3)2 + (33)2 + (−15)2 =
2
2
2
(b) Find an equation for the plane through P , Q, and R.
−−→ −→
Solution. The vector P Q × P R is normal to the plane, so the equation is of
the form −3x + 33y − 15z = c. To find c, plug in one of the points; you get
−3x + 33y − 15z = 132 .
4. (8 points) Find the angle between the planes z = 2x − y + 3 and x − 3y + 2z = 5.
Solution. The angle between the planes is the same as the angle between vectors
that are normal to them. Rewriting the first equation as 2x − y − z = −3, we see
that ~u = 2~i − ~j − ~k is a normal vector to the first plane. A vector that is normal
to the second plane is ~v = ~i − 3~j + 2~k. The angle between ~u and ~v is
~u · ~v
3
3
−1
−1
−1
√ √
√
θ = cos
= cos
= cos
≈ 1.24 rad.
k~uk k~v k
6 14
2 21
p
5. (8 points) Sketch the domain of f (x, y) = 1 + x − y 2 .
Solution. We need to have 1 + x − y 2 ≥ 0; that is, x ≥ y 2 − 1. This is the region
“inside” the sideways parabola x = y 2 − 1 (and the parabola itself).
4
6. (10 points) Shortly after takeoff, a plane is climbing northwest through still air at
an airspeed of 200 km/hr, and rising at a rate of 300 m/min. Resolve its velocity
vector into components. The x-axis points east, the y-axis points north, and the
z-axis points up.
Solution. 300 m/min is 18 km/hr. The plane’s ground speed is
p
p
2002 − 182 = 39,676 ≈ 199 km/hr.
Since it is flying 45◦ north of west, its velocity is
√
√
p
2~ p
2~
~v = − 39,676
i + 39,676
j + 18 ~k
2
2
p
p
= − 19,838~i + 19,838 ~j + 18 ~k .
7. (10 points) The following table gives values of a function h(x, y) at 20 points.
1
3
x
5
7
9
−3
y
−2
−4
−2
−3
−5
−6
−7
−1
0
1
3
0
2
−3
−1
1
−5
−3
−1
−1
−4
−2
0
(a) Could h be a linear function? Why, or why not?
Solution. Yes, since both m =
∆z
∆x
(b) Find a possible expression for h.
= − 12 and n =
∆z
∆y
= 2 are constant.
Solution. If the function is linear, then it is of the form h(x, y) = − 12 x +
2y + c. To find c, notice that f (1, 0) = 3; so c = 72 . Hence, a possible
expression for h is h(x, y) − 12 x + 2y + 72 .
8. (6 points) Describe the level surfaces of f (x, y, z) = p
Solution. Setting f (x, y, z) = c yields
x2
z2
1
+ y2 +
= 2.
2
2
2c
5
1
x2 + 2y 2 + z 2
.
These are ellipsoids.
9. (16 points) Let ~u = 2~i + ~j − 3~k, ~v = ~i − ~j + 2~k, and w
~ = ~i − 5~j − ~k.
(a) Compute 2~u − 3~v .
Solution. 2~u − 3~v = 2(2~i + ~j − 3~k) − 3(~i − ~j + 2~k) = ~i + 5~j.
(b) Are two of the vectors perpendicular? Which ones? Justify.
Solution. Yes, ~u and w
~ are perpendicular since ~u · w
~ = 2 − 5 + 3 = 0.
(c) Find vectors w
~ 1 and w
~ 2 such that w
~ =w
~1 + w
~ 2, w
~ 1 is parallel to ~v , and w
~2
is perpendicular to ~v .
Solution. First replace ~v with a unit vector in the same direction:
~v ′ =
~v
1
= √ (~i − ~j + 2~k) .
k~v k
6
Then
2 ~ ~
(i − j + 2~k)
3
1
and w
~2 = w
~ −w
~ 1 = (~i − 13~j − 7~k) .
3
~ · ~v ′ ) ~v ′ =
w
~ 1 = (w
10. (6 points) Find the volume of the parallelepiped defined by the vectors
~u = 2~i + ~j − 3~k ,
~v = ~i − ~j + 2~k ,
and
w
~ = ~i − 5~j − ~k
of the preceding problem.
Solution. The volume is |(~u × ~v ) · w|.
~
~i ~j
~k ~u × ~v = 2 1 −3 = −~i − 7~j − 3~k
1 −1 2 so |(~u × ~v ) · w|
~ = |(−~i − 7~j − 3~k) · (~i − 5~j − ~k)| = 37 .
EC. True/False (No justification required)
(a) If ~u · ~v < 0, then the angle between ~u and ~v is greater than π/2.
Solution. True.
~u · ~v = k~uk k~v k cos θ < 0
=⇒
cos θ < 0
=⇒
π
< θ < π.
2
(b) If the contours of g(x, y) are concentric circles, then the graph of g is a cone.
Solution. False. Let g(x, y) = x2 + y 2 . Its graph is not a cone even though
its level curves are concentric circles.
(c) It is never true that ~v × w
~ =w
~ × ~v .
Solution. False. If ~v = w
~ = ~0, then ~v × w
~ =w
~ × ~v .
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(d) The sphere x2 + y 2 + z 2 = 10 intersects the plane x = 10.
Solution. False. To look for the intersection, plug 10 in the equation of the
sphere for x:
102 + y 2 + z 2 = 10
=⇒
y 2 + z 2 = −90 ,
which is impossible; they don’t intersect.
(e) For any vectors ~u and ~v , we have (~u + ~v ) · (~u − ~v ) = k~uk2 − k~v k2 .
Solution. True. Using the properties of the dot product,
(~u + ~v ) · (~u − ~v ) = ~u · (~u − ~v ) + ~v · (~u − ~v )
= ~u · ~u − ~u · ~v + ~v · ~u − ~v · ~v
= k~uk2 − k~v k2 .
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