Midterm Exam II
Math 1320 - Engineering Calculus II
April 6, 2015
Answer each question completely in the area below. Show all work and explain your reasoning. If the
work is at all ambiguous, it is considered incorrect. No phones, calculators, or notes are allowed.
Anyone found violating these rules will be asked to leave immediately. Point values are in the square
to the left of the question. If there are any other issues, please ask the instructor.
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agree to abide University Honor Code:
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Signature:
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Solutions
Question
Points
1
15
2
15
3
20
4
20
5
20
6
10
Total:
100
Score
Note: There are 6 questions total on the exam, as well as a challenging bonus question at the end of the
exam that is worth trying if and only if you have extra time.
Midterm Exam 2
Math 1320 - Engineering Calculus II
April 6, 2015
1. Consider the following two vectors:
u = h3, 2i,
5
v = h2, 0i.
(a) Find an expression for the angle ✓ between u and v and draw graphical representation of these
vectors on a standard (x, y ) coordinate system.
Solution: We know that this is just an immediate application of the dot product. Specifically,
u · v = ku| kvk cos ✓,
where ✓ is the desired angle between the two vectors, meaning that
⇢
u·v
1
✓ = cos
,
kuk kvk
thus we just need to compute each of these. Our dot product can be obtained by taking the
sum of the products of the components:
u · v = h3, 2i · h2, 0i = 3 · 2 + 2 · 0 = 6.
Also, the magnitudes can be computed from the distance formula:
p
p
p
kuk = 32 + 22 = 13,
kvk = 22 + 02 = 2.
Thus, our ✓ is:
✓ = cos
1
⇢
u·v
kuk kvk
= cos
1
⇢
6
p
2 13
.
u
✓
v
10
(b) Compute projv u and draw the graphic representation of this operation and its result.
Solution: Recall that projv u means the projection of the vector u onto the vector v. That is,
we want to compute the component of u that is in the direction of v. This corresponds to the
following picture:
The length of the vector we want, denoted compv u is the following:
compv u =
u·v
6
= = 3.
kv k
2
Thus, we want a vector of length compv u in the direction of v, so we just compute the unit
vector in the direction of v:
v
1
= h2, 0i = h1, 0i.
kvk
2
Therefore, our total projection is:
projv u = compv u
2/11
v
= h3, 0i.
kvk
/15 pts
Midterm Exam 2
Math 1320 - Engineering Calculus II
April 6, 2015
Notice, this makes sense. Since v is just in the x direction, the projection of u onto v is just
the x component of u.
u
v
3/11
projv u
/0 pts
Midterm Exam 2
Math 1320 - Engineering Calculus II
April 6, 2015
2. Consider the two following points:
P1 = (1, 2, 3)
5
P2 = ( 1, 0, 4).
(a) Find a vector representation of the line that passes through both P1 , P2 .
Solution: We first need to find the vector that points in the direction of the line through P1
and P2 , which is just the displacement vector between these two points:
!
P1 P2 = hP2
P1 i = h 2, 2, 1i.
Note, this is our v for our line. The equation of a line is then
r = r0 + vt,
where r0 = hP1 i = h1, 2, 3i, so we have:
r = hx, y , z i = h1, 2, 3i + h 2, 2, 1it = h1
10
2t, 2
2t, 3 + ti.
(b) Consider the following vector:
v = h1, 1, 1i.
Find the equation of the plane that contains the line described in part (a) that also is in the
direction of the vector v.
Solution: We want a plane that is in the direction of the line from the previous problem, which
I’ll call v0 = h 2, 2, 1i and now also v = h1, 1, 1i. We need the normal vector to the plane,
which is orthogonal to both of these and therefore can be obtained via the cross product:
i
n = v0 ⇥ v = 2
1
j k
2 1 = h 3, 3, 0i.
1 1
For a plane, we need a point and a normal vector, so we can just choose a point on the line,
say P1 = (1, 2, 3), which gives us that our plane equation is:
!
n · P P1 = 0,
which is the same as:
h 3, 3, 0i · hx
1, y
2, z
3i =
4/11
3(x
1) + 3(y
2) + 0(z
3) = 0.
/15 pts
Midterm Exam 2
Math 1320 - Engineering Calculus II
April 6, 2015
3. Match the following equations to one of the surfaces on the next page. If you explain/show work, you
may get extra credit even for a wrong answer.
Note I : not all of the equations are given in rectangular coordinates.
Note II: not all of the surfaces have corresponding equations, but every equation has a corresponding
surface.
5
(a)
x2
y 2 + z 2 = 1.
Solution: The easiest way to identify this one is that notice due to the negative signs, you
obtain hyperbolas. Specifically, when z = 0, no solutions exist. This is clearly (v).
5
(b) z = r .
Solution: We’ve talked many times about this example in cylindrical coordinates. It’s the classical cone, (iii).
5
(c) x = z 2 + y 2 .
Solution: Again, taking slices can yield the answer, but notice that it’s effectively something
that grows in the x-direction, the only of which, is (viii).
5
(d) ⇢ cos
= 0.
Solution: This is probably the trickiest of the bunch, but if you recall your coordinate changes,
we know that z = ⇢ cos , so this is just z = 0, the plane described by (i).
5/11
/20 pts
Midterm Exam 2
Math 1320 - Engineering Calculus II
(i)
(v)
(ii)
(vi)
(iii)
(vii)
(iv)
(viii)
6/11
April 6, 2015
/0 pts
Midterm Exam 2
Math 1320 - Engineering Calculus II
April 6, 2015
4. Consider two bees flying around in a beehive, where the first bee’s position at time t is described by
r1 (t) = ht, t, 1i,
and the second bee’s trajectory is
r2 (t) = ht 2 , t
10
3, 5 + ti.
(a) Colliding would be catastrophic for both of the bees. Do they ever have to worry about this?
That is, do the two bees ever collide? If so, at what time?
Solution: To check if the trajectories intersect, we simply have to see if the x, y , z components
are all the same at any given time. Thus, if we first examine the x component, we check
t = t 2 =) t 2 t = 0, which only occurs at t = 0 and t = 1, two possible candidates.
Now, if we check the y component, which, for the bees to collide, requires t = t 3 =)
2t = 3, which only occurs at t = 3/2, we see that the bees never collide. We do not even
need to check the z component.
10
(b) Suppose the hive is a sphere of radius 3, centered at the origin, which we can describe by
x 2 + y 2 + z 2 = 32 .
Does the first bee ever escape the sphere? That is, does r1 ever intersect the sphere? If so,
when?
Solution: Similar to a lab question, to see if the first bee intersects the sphere, we simply check
if the position of the bee r1 ever satisfies the equation of the sphere. To do so, we plug in each
of the x, y , z coordinates for x, y , z in the sphere equation:
(t)2 + ( t)2 + 12 = 32 ,
which after simplifying, yields
2t 2 + 1 = 9 =) t = ±2.
Clearly, we do not care about t =
2, so the bee leaves the hive at t = 2.
7/11
/20 pts
Midterm Exam 2
Math 1320 - Engineering Calculus II
April 6, 2015
5. Consider the following vector equation:
r(t) = 3ti + 2t 2 j.
5
(a) Compute T(t), the unit tangent equation.
Solution: We’ve defined T to be the following:
T(t) =
r0 (t)
,
kr0 (t)k
so we clearly just need to take the derivative to find r0 (t), which is, in this case
r0 (t) = h3, 4ti.
Computing the magnitude of this, we see
p
p
kr0 (t)k = 32 + (4t)2 = 9 + 16t 2 ,
meaning that we have our unit tangent vector
T(t) =
5
r0 (t)
1
=
h3, 4ti.
0
kr (t)k
9 + 16t 2
(b) Compute N(t), the unit normal equation.
Solution: This calculation is a little messier, but at the end of the day, still just basic calculus.
We’ve defined the unit normal to be
N(t) =
T(t)
,
kT0 (t)k
so we need to take our answer from the previous question and differentiate it. Note, for the y
component, you do have to use the quotient rule, but what we find is
⌧
48t
36
12
0
T (t) =
,
=
h 4t, 3i .
2
3/2
2
3/2
(9 + 16t )
(9 + 16t )
(9 + 16t 2 )3/2
Therefore, we can now compute the magnitude noticing that we can pull the scalar out in front
out, to see that we get
kT0 (t)k =
p
12
12(9 + 16t 2 )1/2
12
2 + (3)2 =
(
4t)
=
.
9 + 16t 2
(9 + 16t 2 )3/2
(9 + 16t 2 )3/2
Therefore, we now have our unit normal vector
N(t) =
T(t)
1
=p
h 4t, 3i.
0
kT (t)k
9 + 16t 2
8/11
/10 pts
Midterm Exam 2
10
Math 1320 - Engineering Calculus II
April 6, 2015
(c) Compute the curvature, (t).
Solution: For this problem, you could feasibly use the definition
(t) =
kT0 (t)k
,
kr0 (t)k
since we already have both of these, but we can also use the other definition, which is an easy
calculation in this case. The other definition we’ve used is
(t) =
kr0 (t) ⇥ r00 (t)k
.
kr0 (t)k3
We know what r0 (t) is, so we can take another derivative to find that
r00 (t) = h0, 4i.
Thus, we now have the following:
i j k
r (t) ⇥ r (t) = 3 4t 0 = h0, 0, 12.
0 4 0
0
00
Thus, our final answer is simply
(t) =
kr0 (t) ⇥ r00 (t)k
12
12
=p
3 = (9 + 16t 2 )3/2 .
kr0 (t)k3
2
9 + 16t
9/11
/10 pts
Midterm Exam 2
10
Math 1320 - Engineering Calculus II
April 6, 2015
6. In class, we proved that if |r(t)| = c, where c is a constant, then r(t) ? r0 (t) for all t. Prove the
converse. That is, prove the following statement:
If r0 (t) ? r(t) (that is, the vectors are orthogonal) for all t, then |r(t)| = c, a constant.
For this problem, it’s sufficient to consider r(t) to have the form r(t) = hf (t), g(t), h(t)i.
Hint: What does it mean for two vectors to be orthogonal? Write this down and then integrate.
Solution: Let’s consider the form provided in the problem, that is r(t) = hf (t), g(t), h(t)i, then
r0 (t) = hf 0 (t), g 0 (t), h0 (t)i.
For two vectors to be orthogonal, this means that their dot product is zero, in other words:
0 = r(t) · r0 (t) = f (t)f 0 (t) + g(t)g 0 (t) + h(t)h0 (t).
On the left side if integrate, we just get a constant, c:
Z
0 dt = c.
The right side of the equation is screaming to be integrated, yielding, by the chain rule:
Z
Z
Z
f (t)2 g(t)2 h(t)2
f (t)f 0 (t) dt + g(t)g 0 (t) dt + h(t)h0 (t) dt =
+
+
.
2
2
2
Thus, we’re left with:
2c = f (t)2 + g(t)2 + h(t)2 ,
which means that
p
2c =
p
f (t)2 + g(t)2 + h(t)2 ,
but what is this object on the right hand side? It’s exactly the magnitude of r(t), that is:
p
kr(t)k = f (t)2 + g(t)2 + h(t)2 ,
p
so we just have that some constant 2c, which we can now call a different constant C:
p
2c = C = kr(t)k,
which is exactly what we hoped to prove.
10/11
/10 pts
Midterm Exam 2
Math 1320 - Engineering Calculus II
April 6, 2015
7. (Extra credit) Prove that a circle of radius a has constant curvature.
Solution: This just Example 3 in section 10.3 of the book, found on page 710.
11/11
/0 pts