Calculus III, Exam 1: Review Solutions
1
1
Calculus III Exam 1 Practice Problems - Solutions
1. Compute the following:
h
i
(a) limt→1 (1 − t2 )~ı + (tet )~ + (sin πt)~k
= limt→1 h1 − t2 , tet , sin πti = hlimt→1 1 − t2 , limt→1 tet , limt→1 sin πti = h0, e, 0i
i
R1h
(b) 0 (cos t)~ı + (1 + t2 )−1~ − ~k dt
R1
R1
R1
= h 0 cos t dt, 0 (1 + t2 )−1 dt, 0 −dti = h[sin t]10 , [tan−1 t]10 , [−t]10 i = hsin 1, π/4, −1i
R
(c) [(ln t)~ı + (sinh t − cosh t)~] dt
R
R
R
= h ln t dt, (sinh t − cosh t) dt, 0 dti = ht ln t − t + C1 , cosh t − sinh t + C2 , C3 i
= ht ln t − t, cosh t − sinh t, 0i + hC1 , C2 , C3 i
h
i
(d) limt→∞ (tan−1 t)~ − (t−3/2 )~k
= hlimt→∞ 0, limt→∞ tan−1 t, limt→∞ −t−3/2 i = h0, π/2, 0i
(e)
(f)
d2
2 3
dt2 ht, t , t i
d d
d 2 d 3
= dt
h dt [t], dt
[t ], dt [t ]i
d
dt
h
=
d
2
dt h1, 2t, 3t i
e−t cos t~ı + e−t sin t~ + ln t~k
d
d
d
= h dt
[1], dt
[2t], dt
[3t2 ]i = h0, 2, 6ti
i
d −t
d −t
d
= h dt
[e cos t], dt
[e sin t], dt
[ln t]i = he−t (− sin t) + (−e−t ) cos t, e−t cos t + (−e−t ) sin t, 1/ti
h
i
d
(g) dt
(5 − t2 )~ı + (t sin t)~ − et~k
d
d
d
= h dt
[5 − t2 ], dt
[t sin t], dt
[−et ]i = h−2t, sin t + t cos t, −et i
d
(h) dt
[(tan t)~ı − t~k] · [~ı + t−1~ − t~k]
d
d
d
d −1 d
d
= h dt
[tan t], dt
[0], dt
[−t]i · h1, 1/t, −ti + htan t, 0, −ti · h dt
[1], dt
[t ], dt [−t]i
2
−2
= hsec t, 0, −1i · h1, 1/t, −ti + htan t, 0, −ti · h0, −t , −1i
= (sec2 t + 0 + t) + (0 + 0 + t) = 2t + sec2 t
2
d
[(et )~] × [(−t2 )~k]
(i) dt
2
2
d
d t
d
d
d
d
[e ], dt
[0]i × h0, 0, −t2 i + h0, et , 0i × h dt
[0], dt
[0], dt
[−t2 ]i
= h dt
[0], dt
2
2
= h0, 2tet , 0i × h0, 0, −t2 i + h0, et , 0i × h0, 0, −2ti
2
2
2
= h−2t3 et , 0, 0i + h−2tet , 0, 0i = h(−2t3 − 2t)et , 0, 0i
1 − cos t 3 −1/t2
(j) limt→0
,t ,e
t
1 − cos t
−2
= hlimt→0
, limt→0 t3 , limt→0 e−t i = h0, 0, 0i
t
i
R π/4 h
(k) 0
cos 2t~ı + sin 2t~ + t sin t~k dt
R π/4
R π/4
R π/4
π/4
π/4
π/4
= h 0 cos 2t dt, 0 sin 2t dt, 0 t sin t dti = h[ 12 sin 2t]0 , [− 21 cos 2t]0 , [−t cos t+sin t]0 i =
√
π
+ 22 i
h1/2, 1/2, − 4√
2
R4 √
1
(l) 1
t~ı + te−t~ + 2 ~k dt
t
Rt√
R 4 −t
R4
−4
+2e−1 , 34 i
= h 1 t dt, 1 te dt, 1 t−2 dti = h[ 23 t3/2 ]41 , [−te−t −e−t ]41 , [−t−1 ]41 i = h 14
3 , −5e
Calculus III, Exam 1: Review Solutions
2
2. Find the unit tangent vector and unit normal vector to the space curve
√
3 2 2
r(t) = (t3 − 1)~ı − (
t )~ + (3t)~k
2
at the point where t = 1.
√
Solution: First of all, r0 (t) = h3t2 , −3 2t, 3i, so
*
+
√
√
√
h3t2 , −3 2t, 3i
ht2 , − 2t, 1i
t2
2t
1
r0 (t)
=p
=
=
,−
,
T(t) = 0
||r (t)||
t2 + 1
t 2 + 1 t2 + 1 t2 + 1
(9t4 ) + (18t2 ) + (9)
and
√
T0 (t)
h(t2 + 1)−2 (2t), − 2[−t(t2 + 1)−2 (2t) + (t2 + 1)−1 ], −(t2 + 1)−2 (2t)]i
p
N(t) =
=
.
||T0 (t)||
[2t(t2 + 1)−2 ]2 + 2[−2t2 (t2 + 1)−2 + (t2 + 1)−1 ]2 + [−2t(t2 + 1)−2 ]2
3. If r = hx, y, zi, a = ha1 , a2 , a3 i and b = hb1 , b2 , b3 i, show that the vector equation (r−a)·(r−b) = 0
represents a sphere. Find its center and radius.
Solution: Consider hx − a1 , y − a2 , z − a3 i · hx − b1 , y − b2 , z − b3 i = (x − a1 )(x − b1 ) + (y − a2 )(y −
b2 ) + (z − a3 )(z − b3 ) = (x2 − (a1 + b1 )x) + (y 2 − (a2 + b2 )y) + (z 2 − (a3 + b3 )z) + a1 b1 + a2 b2 + a3 b3
Then (r − a) · (r − b) = 0 is equivalent to the equation
x−
a1 + b2
2
2
2
a2 + b2
+ y−
2
2
a3 + b3
2
2 2 2
a1 + b1
a2 + b2
a3 + b3
=
+
+
− a1 b1 − a2 b2 − a3 b3
2
2
2
2 2 2
a1 − b1
a2 − b2
a3 − b3
=
+
+
.
2
2
2
+
z−
1
1 a2 +b2 a3 +b3
a + ~b) (the midpoint between the points
This is a sphere centered at C( a1 +b
2 ,
2 ,
2 ) = 2 (~
(a1 , a2 , a3 ) and (b1 , b2 , b3 )) and having radius
s
2 2 2
a1 − b1
a2 − b2
a3 − b3
1
+
+
= || (~a − ~b)||,
2
2
2
2
(one half the distance between the points (a1 , a2 , a3 ) and (b1 , b2 , b3 )).
3
t
1
t ~
t
k, over the interval
4. Find the length of the space curve r(t) = √ ~ı +
+
~ − √
6
2t
2
2
1 ≤ t ≤ 2.
R2
R2q
2
−1 2
√1 )2 dt =
Solution: The arc length of the curve r(t) is L = 1 ||r0 (t)|| dt = 1 ( √12 )2 + ( t2 + 2t
2 ) + (−
2
q
q
R 2 t2 1
R2
R 2 1 t4
t2
1 2
t3
−1 2
8
1
1
1
1
1
1
( 2 + 2t2 ) dt = 1 ( 2 + 2t2 )dt = [ 6 + 2t ]1 = [ 6 − 4 ]−[ 6 − 2 ] =
2 + 4 − 2 + 4t4 + 2 dt = 1
1
7
6
+
1
4
=
17
12 .
d
[r(t) × r0 (t)] = r(t) × r00 (t).
dt
d
Solution: Let r(t) be a vector-valued function and assume r00 (t) exists. Then dt
[r(t) × r0 (t)] =
d
d 0
r(t) × dt
[r (t)] + dt
[r(t)] × r0 (t) = r(t) × r00 (t) + r0 (t) × r0 (t) = r(t) × r00 (t), since v × v = 0 for all
vectors v.
5. Show that if r is a vector function such that r00 = (r0 )0 exists, then
Calculus III, Exam 1: Review Solutions
3
6. Let L1 , L2 and L3 be the three lines given below
x = −s, y = 4 + 2s, z = 6 + 3s
x = 2 + t, y = 4 + 2t, z = 6 − 3t
x = 2 + u, y = −2u, z = 6 + 3u
(1)
(2)
(3)
(a) Two of these lines intersect. Which two? At which point?
Solution: Let’s first see if L1 and L2 intersect. If so, there must be some value s0 of the
parameter s of L1 and t0 of the parameter t of L2 such that
−s0
4 + 2s0
6 + 3s0
= 2 + t0
= 4 + 2t0
= 6 − 3t0
The second of these equations requires that s0 = t0 (subtract 4 from both sides and divide
by 2), but the third equation requires that s0 = −t0 (subtract 6 from both sides and divide
by 3). Thus s0 = t0 = 0, which can’t satisfy the first equation. Hence, line L1 and L2 do not
intersect.
If L1 and L3 intersect, then there are values s0 and u0 satisfying the system of equations
−s0
4 + 2s0
6 + 3s0
= 2 + u0
= −2u0
= 6 + 3u0
The third equation above requires s0 = u0 , so the first equation now implies that s0 = u0 =
−1. Checking this against the second equation, we find the left-hand side is 4+2(−1) = 2 and
the right-hand side is −2(−1) = 2, so all three equations are satisfied when s0 = u0 = −1.
Thus lines L1 and L3 intersect at the point where s0 = u0 = −1, i.e., at the point (1, 2, 3)
(using s0 = −1 in the equation for line L1 ).
It also appears that lines L2 and L3 do not intersect, for otherwise t0 = u0 (from equation
1) and −t0 = u0 (from equation 3), so that t0 = u0 = 0, but then 4 = 0 (from equation 2),
which is a contradiction.
(b) Another pair is skew. How do you know this pair to be skew?
Solution: We’ve seen that lines L1 and L2 do not intersect and that the lines L2 and L3 do
not intersect. Two lines in space that do not intersect are either parallel or they are skew. To
be parallel, the direction vectors of the lines must be scalar multiples of one another. Now the
direction vector of line L1 is d1 = h−1, 2, 3i, the direction vector of line L2 is d2 = h1, 2, −3i,
and the direction vector of line L3 is d3 = h1, −2, 3i. We can see that no two of these lines
are parallel, for there are no non-zero scalars λ such that d1 = λd2 or d1 = λd3 or d2 = λd3 .
Thus, both the pair L1 , L2 and the pair L2 , L3 are skew lines.
(c) Find the distance between the two skew lines that you have just identified. (One way to do
this is to find a pair of parallel planes, each of which contains one of the lines in your skew
pair, and then to find the distance betweeen these parallel planes.)
Solution: The distance between the lines L1 and L2 will be the distance between the parallel
planes M and N through the points P (0, 4, 3) (the point on L1 corresponding to s = 0) and
Q(2, 4, 6) (the point on L2 corresponding to t = 0), respectively, and each perpendicular to
the vector n = d1 × d2 , as then L1 will lie in M and L2 will lie in N . Now
~i ~j ~k n = −1 2 3 = [(2)(−3)−(2)(3)]~i−[(−1)(−3)−(1)(3)]~j+[(−1)(2)−(1)(2)]~k = h−12, 0, −4i.
1 2 −3
Calculus III, Exam 1: Review Solutions
4
Hence M has equation [−12](x − 0) + [0](y − 4) + [−4](z − 3) = −12x − 4z + 12 = 0 or
3x + z = 3, while N has equation [−12](x − 2) + [0](y − 4) + [−4](z − 6) = −12x − 4z + 48 = 0
or 3x + z = 12. Then the distance between M and N , and hence between the lines L1 and
L2 , is
| − 24 + 0 − 12|
36
9
|h−12, 0, −4i · h2, 0, 3i|
=√
= √
=√
D = ||projn P~Q|| =
||h−12, 0, −4i||
144 + 0 + 16
10
160
by Formula 8 of Section 9.5 in the book.
The distance between the lines L2 and L3 is the distance between the parallel planes M 0 and
N 0 through the points Q(2, 4, 6) (the point on L2 corresponding to t = 0) and R(2, 0, 6) (the
point on L3 corresponding to u = 0). In order for M 0 and N 0 to be parallel and for them to
contain L2 and L3 , respectively, they must both be normal to the vector m = d2 × d3 :
~i ~j
~k m = 1 2 −3 = [(2)(3)−(−2)(−3)]~i−[(1)(3)−(1)(−3)]~j+[(1)(−2)−(1)(2)]~k = h0, −6, −4i.
1 −2 3 Hence M 0 has equation [0](x − 2) + [−6](y − 4) + [−4](z − 6) = −6y − 4z + 48 = 0 or
3y + 2z = 24, while N 0 has equation [0](x − 2) + [−6](y − 0) + [−4](z − 6) = −6y − 4z + 24 = 0
or 3y + 2z = 12. Then the distance between M 0 and N 0 , and hence between the lines L2 and
L3 , is
~ = |h0, −6, −4i · h0, −4, 0i| = √|0 + 24 + 0| = √24 = √12
D = ||projm QR||
||h0, −6, −4i||
0 + 36 + 16
52
13
by Formula 8 of Section 9.5 in the book.
7. We obtain a vector normal to a path specified by r(t) by differentiating the unit tangent vector
and normalizing it. In this problem, you will show that this is a reasonable thing to do.
(a) First prove that r0 (t) is always orthogonal to r(t) when r(t) is of constant magnitude.
Solution: Suppose r(t) has constant magnitude ||r(t)|| = k for some constant k. Then,
squaring both sides and using the identity ||v||2 = v · v for any vector v, we have the
equation
r(t) · r(t) = k 2
is a constant. Thus, differentiating both sides with respect to t yields
r0 (t) · r(t) + r(t) · r0 (t) = 0,
so 2r(t) · r0 (t) = 0, in which case r(t) · r0 (t) = 0. Therefore, r(t) ⊥ r0 (t) for all t, since u ⊥ v
if and only if u · v = 0.
T0
(b) Use this to explain why the vector N =
is a unit vector which is normal to the curve
||T0 ||
specified by r(t).
r0 (t)
Solution: The vector T(t) = 0
is the unit tangent vector to the curve specified by r(t).
||r (t)||
That is, T(t) is tangent to the curve and ||T(t)|| = 1. Thus, as T(t) is a vector function of
constant length, it is always perpendicular to its derivative, T0 (t) by the previous part of this
T0 (t)
problem. Therefore, the vector N(t) =
is a vector of length 1 in the same direction
||T0 (t)||
as T0 (t), so it is also perpendicular to T(t), and hence is a unit vector which is normal to the
curve r(t).
Calculus III, Exam 1: Review Solutions
8. We define the curvature of a path specified by r(t) to be the magnitude of the vector
5
dT
.
ds
(a) Use a picture to illustrate why this is a reasonable definition.
Solution: Think about a circle. The parameter s measures the arc length, so the magnitude
dT
of the vector
is measuring how quickly the tangent direction, T, changes with respect to
ds
the arc length as we traverse the circle.
(b) In spite of this natural definition, curvature is notoriously tedious to compute. Nevertheless,
some straightforward formulas are available to make life a little more pleasant. One of these
is
||r0 × r00 ||
κ=
||r0 ||3
2 2
d s
ds
Use the fact that a =
T
+
κ
N to explain why this formula holds.
dt2
dt
r0
Solution: Recall that a = r00 and that T = 0 , so that T and r0 are parallel vectors. We
||r ||
2 ! 2 2 ds
d s
d s
0
00
0
T+κ
N =
[r0 × T] +
begin by considering the vector r × r = r ×
2
dt
dt
dt2
2 2
2
d s
ds
ds
[r0 × N] =
0
+
κ
[ ||r0 || B ]. (Note: r0 × T = 0 since these vectors are
κ
dt
dt2
dt
3
ds
parallel while r0 ×N = ||r0 ||T×N = ||r0 ||B.) Thus, ||r0 ×r00 || = ||κ
B|| = κ||r0 ||3 , since
dt
Rt
ds
. (Formally, s = a ||r0 || dτ , so the Fundamental
||r0 || is the speed, which is the same as
dt
ds
Theorem of Calculus implies that
= ||r0 (t)||.) Therefore, solving for κ, we obtain the
dt
formula
||r0 × r00 ||
κ=
.
||r0 ||3
2 2
d s
ds
ds
0
T and then use this to explain why a =
T+κ
N.
(c) Bonus: Explain why r =
dt
dt2
dt
ds
Solution: As I have already shown above, r0 = ||r0 ||T and ||r0 || =
by the Fundamental
dt
Theoreom of Calculus (my parenthetical remark near the end of the previous solution). Thus
ds
r0 =
T.
dt
2 d
d ds
d ds
ds
d
d s
ds
ds dT
Thus, a = [r0 (t)] =
T =
·T+
· [T] =
T+
,
dt
dt dt
dt dt
dt dt
dt2
dt
dt ds
dT
this last part by the Chain Rule. Now ||
|| = κ by definition of curvature, so
ds
2 2
2 2
2 2
dT
d s
ds
d s
ds
d s
ds
ds
T
+
κ
N
]
=
T
+
[κN]
=
a=
T
+
[κ
dT
2
2
2
dt
dt
dt
dt
dt
dt
|| ds ||
dT
because ds is the unit vector in the direction of T0 , which is N by definition.
dT ds Calculus III, Exam 1: Review Solutions
6
9. A particle that has been stationary at the origin (0, 0, 0) suddenly begins to move in such a way
that its acceleration at time t seconds is given by a(t) = h1, cos t, sin ti.
(a) Where will the particle be when t = π seconds?
R
R
Solution: The velocity function of the particle is v(t) = a(t) dt = h1, cos t, sin ti dt =
ht+C1 , sin t+C2 , − cos t+C3 i. Given the initial velocity v(0) = h0, 0, 0i (because the particle
was stationary at the origin, its initial velocity is trivial), we solve for the unknown constants
C1 , C2 , C3 by equating corresponding components of
(0) + C1
sin(0) + C2
− cos(0) + C3
=
=
=
0
0
0
which implies C1 = 0, C2 = 0 and C3 = 1. Thus the velocity function of the particle is
v(t) = ht, sin t, 1 − cos ti.
R
R
Then the position function of the particle is r(t) = v(t) dt = ht, sin t, 1 − cos ti dt =
h 21 t2 + K1 , − cos t + K2 , t − sin t + K3 h for constants K1 , K2 , K3 . Given the initial position of
the particle, r(0) = h0, 0, 0i, we find the values of the constants by solving the equations
1 2
(0) + K1
2
− cos(0) + K2
(0) − sin(0) + K3
= 0
= 0
= 0
so that K1 = 0, K2 = 1, and K3 = 0. Thus
1
r(t) = h t2 , 1 − cos t, t − sin ti.
2
π2
, 2, πi.
2
(b) Write down an integral that will give the total distance traveled by the particle along this
curve during the first second of travel.
Solution: The distance traveled by the particle along this curve during theRfirst second is
1
the arc length of the curve measured from time t = 0 to time t = 1, so it is L = 0 ||r0 (t)|| dt =
R1
R1p
R1p
||v(t)|| dt = 0 (t)2 + (sin t)2 + (1 − cos t)2 dt = 0 t2 + sin2 t + 1 − 2 cos t + cos2 t dt =
R01 √
t2 + 2 − 2 cos t dt.
0
√
(c) Bonus: Do not evaluate this integral, but show that it does not exceed 3.
2
Solution: On the interval 0 ≤ t ≤ 1, 0R≤ √
cos t ≤ 1, so that t2 ≤ Rt2 +
2 ≤ 3,
√2 − 2 cos√t ≤ t1 + √
1
1
2
since 0 ≤ t ≤ 1 on [0, 1]. Thus L = 0 t2 + 2 − 2 cos t dt ≤ 0 3 dt = 3[t]0 = 3 as
claimed.
Thus, the position of the particle at time t = π is r(π) = h