Math 1172 - Midterm 3 - Form A - Page 2
Problem 1
[18 pts]
True or False. Give a brief explanation or example to justify your answer.
a) [3 pts]
For any vectors u and v in R3 , u · (v × u) = 0.
True: The cross product of u and v is orthogonal to both u. Thus u · (v × u) = 0
b) [3 pts] For any vectors u, v, and w in R3 , if v and w are both orthogonal to
u then v and w are parallel.
False: Consider the three vectors u = h1, 0, 0i, v = h0, 1, 0i, and w = h0, 0, 1i. v and w are
both orthogonal to u, but v and w are not parallel (they are also orthogonal).
c) [3 pts] The curves r(t) =< 3 + t, 4 − 2t, −1 + 2t > and s(t) =< t2 , 8 cos(t − 1), t − 6 >
intersect at the point (1, 8, −5).
True: The two curves will intersect at (1, 8, −5) if there exists times t1 and t2 such that
r(t1 ) = s(t2 ) = (1, 8, −5). In particular, we need for equality to hold in the third coordinate.
If intersection occurs,
−1 + 2t1 = −5
and
t2 − 6 = −5.
Thus t1 = −2 and t2 = 1. Evaluating r(−2) and s(1) we find r(−2) = s(1) = (1, 8, −5).
d) [3 pts]
The curve r(t) =< t, 2t, t2 > is parameterized with respect to arc length.
False: If the speed of the curve is constantly 1 the curve will be parameterized by arc length.
Let us calculate the speed of the particle.
r0 (t) =< 1, 2, 2t >
p
√
s(t) = |r0 (t)| = 12 + 22 + (2t)2 = 5 + 4t2 6≡ 1
e) [3 pts]
The function f (x, y) = tan(x2 + y 2 ) is continuous at all points in R2 .
p
False: Tangent is undefined at π/2. Thus f (x, y) is undefined at ( π/2, 0) and hence is not
continuous.
x2 + xy
1
f ) [3 pts]
lim
=
2
2
(x,y)→(1,−1) x − y
2
True:
x(x + y)
x
1
1
x2 + xy
=
lim
=
lim
=
=
(x,y)→(1,−1) (x − y)(x + y)
(x,y)→(1,−1) (x − y)
(x,y)→(1,−1) x2 − y 2
(1 − (−1))
2
lim
Math 1172 - Midterm 3 - Form A - Page 3
Problem 2
[21 pts]
Consider the vectors v = 3i − 3k and w = −5i + 4j + 7k.
a) [4 pts]
Compute v · w.
Solution:
v · w = 3 · (−5) + 0 · 4 + (−3) · 7 = −36
b) [4 pts]
Compute v × w.
Solution:
i j k v × w = 3 0 −3 = (0 · 7 − (−3) · 4)i − (3 · 7 − (−3) · (−5))j + (3 · 4 − 0 · (−5))k
−5 4 7 = h12, −6, 12i
c) [4 pts]
Find projv w.
Solution: v · v = 3 · 3 + 0 · 0 + (−3) · (−3) = 18
projv w =
d) [4 pts]
−36
v·w
v=
h3, 0, −3i = h−6, 0, 6i
v·v
18
Find a unit vector that is orthogonal to both v and w.
Solution: v × w is orthogonal to both v and w. Thus we just need to rescale the vector so
that it is a unit vector
v×w
h12, −6, 12i
2 −1 2
=p
=h ,
, i
2
2
2
|v × w|
3 3 3
12 + (−6) + 12
e) [5 pts]
Find vectors a and b such that a + b = v and a − b = w.
Solution: Adding the equations a + b = v and a − b = w together we find that
2a = v + w = h−2, 4, 4i.
Thus a = h−1, 2, 2i. Using the first given equation
b = v − a = h4, −2, −5i
Math 1172 - Midterm 3 - Form A - Page 4
Problem 3
[23 pts] Consider a particle that has initial position r(0) =< 0, 1, 0 > and initial velocity
v(0) =< 1, 1, 0 >. Moreover, the particle has acceleration given by a(t) =< 2, 0, cos t >
for t ≥ 0.
a) [5 pts]
Find the velocity, v(t).
Solution:
Z
a(t) = h2t, 0, sin ti + C1 .
v(t) =
Let us solve for C1 using the initial conditions.
h1, 1, 0i = v(0) = h2 · 0, 0, sin 0i + C1 = C1 .
Thus
v(t) = h2t + 1, 1, sin ti.
b) [4 pts]
Find the speed of the particle at time t = 2π.
Solution: Recall that speed is the magnitude of the velocity. Thus
p
s(t) = (2t + 1)2 + 12 + sin(t)2 , and
s(2π) =
c) [5 pts]
p
(4π + 1)2 + 12 + sin(2pi)2 =
p
(4π + 1)2 + 12
Find the position function r(t).
Solution:
Z
r(t) =
v(t) = ht2 + t, t, − cos ti + C2 .
Let us solve for C2 using the initial conditions.
h0, 1, 0i = r(0) = h02 + 0, 0, − cos 0i + C2 = h0, 0, −1i + C2 .
Thus C2 = h0, 1, 1i, and
r(t) = ht2 + t, t + 1, 1 − cos ti.
d) [5 pts]
Find an equation of the line tangent to the curve r(t) at time t = 2π.
Solution:
v(2π) = h2(2π) + 1, 1, sin(2π)i = h4π + 1, 1, 0i.
r(2π) = h(2π)2 + (2π), (2π) + 1, 1 − cos(2π)i = h4π 2 + 2π, 2π + 1, 0i.
Thus an equation of the line tangent to the curve r(t) at time t = 2π is
R(t) = h4π 2 + 2π, 2π + 1, 0i + th4π + 1, 1, 0i.
e) [4 pts] Set up an integral that represents the distance traveled by the particle along
the curve over the interval 0 ≤ t ≤ 2π. DO NOT EVALUATE THE INTEGRAL.
Solution:
Z 2π
Z 2π p
s(t)dt =
(2t + 1)2 + 12 + sin(t)2 dt.
0
0
Math 1172 - Midterm 3 - Form A - Page 5
Problem 4
r
[16 pts]
Consider the function f (x, y) =
x2
y
.
+1
a) [4 pts] Find the domain of f . Sketch the domain in the xy-plane.
Solution: The function will be undefined when x2 + 1 = 0 or when x2y+1 < 0. x2 ≥ 0 so
x2 + 1 ≥ 1. Thus the first problem does not occur. Since x2 + 1 > 0, the sine of x2y+1 will be
determined by the sine of y. Hence the function is defined when y ≥ 0. In other words the
domain is
D = {(x, y) : y ≥ 0} .
b) [4 pts] At what points of R2 is the function continuous?
Solution: Rational functions and roots are continuous on their domains. Hence so is the
composition of such functions. Thus
Points of continuity = {(x, y) : y ≥ 0} .
c) [6 pts] Sketch the the level curves (in the xy-plane) corresponding to z0 = 0, z0 = 1,
and z0 = 2.
Solution: A generic level curve at level z = c is given by f (x, y) = c. Solving this equation
for y we have that
y = c2 (x2 + 1).
So when c = 0, the level curve is y = 0, when c = 1,the level curve is y = x2 + 1, and when
c = 2, the level curve is y = 4x2 + 4.
d) [2 pts]
Sketch the yz-trace.
Solution: The yz-trace will be formed by setting x = 0 in the initial equation. Thus the
√
yz-trace is z = y.
Math 1172 - Midterm 3 - Form A - Page 6
Problem 5
[22 pts]
Limits and Partial Derivatives
a) [10 pts]
If f (x, y) = x2 + sin(xy 2 ), find fx and fy .
Solution:
fx = 2x + y 2 cos(xy 2 )
fy = 2xy cos(xy 2 ).
b) [6 pts]
Use a tree diagram and write the required Chain Rule formula for
w = f (x, y, z), x = g(u, v), y = h(u, v), and z = k(u).
∂w
where
∂v
Solution:
So
c) [6 pts]
∂w
∂w ∂x ∂w ∂y
=
·
+
·
.
∂v
∂x ∂v
∂y ∂v
Use the Two-Path Test to prove that
lim
(x,y)→(0,0) x2
2xy
− 5y 2
does not exist.
Solution: Let us approach (0, 0) on the line y = cx. Then
lim
(x,cx)→(0,0) x2
2xcx
2c
2c
2xy
= lim 2
= lim
=
2
2
2
x→0 x − 5(cx)
x→0 1 − 5c
− 5y
1 − 5c2
Along the path y = 0 the limit is 0 and along the path y = x the limit is − 21 . Thus the two
path test tells us the limit does not exists