Limits and Continuity Derivatives Integrals Arclength Exercises
Calculus of Vector-Valued Functions
Mathematics 54 - Elementary Analysis 2
Institute of Mathematics
University of the Philippines-Diliman
1 / 17
Limits and Continuity Derivatives Integrals Arclength Exercises
Definition Examples
Limits and Continuity of Vector Functions
Definition
­
®
Given ~
R (t) = f (t) , g (t) , h (t) .
1
We define the limit of ~
R as t approaches a by
D
E
R (t) = limf (t) , limg (t) , limh (t) ,
lim~
t→a
t→a
t→a
t→a
provided that lim f (t), lim g (t), and lim h (t) exist.
t→a
2
t→a
t→a
The function ~
R (t) is continuous at t = a if
~
R (a) exists;
lim~
R (t) exists;
t→a
~
R (a) = lim~
R (t).
t→a
2 / 17
Limits and Continuity Derivatives Integrals Arclength Exercises
Definition Examples
Limits of Vector-Valued Functions
Example 1.
Evaluate lim~
R (t) where
t→2
À
¿
t 2 − 4 sin (2t − 4)
~
,
.
R (t) = t + 1,
t −2
t −2
Solution.
¿
À
t2 − 4
sin (2t − 4)
R (t) = lim (t + 1) , lim
lim~
, lim
t→2
t→2 t − 2 t→2
t→2
t −2
¿
À
2 cos (2t − 4)
= 3, lim (t + 2) , lim
t→2
t→2
1
(L’Hopital’s Rule)
= 〈3, 4, 2〉
3 / 17
Limits and Continuity Derivatives Integrals Arclength Exercises
Definition Examples
Limits of Vector-Valued Functions
Example 2.
Evaluate lim−~
R (t) where
t→1
~
R (t) =
¿
À
|t − 1| sin(πt) tan(πt)
,
,
.
t − 1 t2 − 1
t −1
Solution. Note that for t → 1− , t < 1. We have
¿
À
|t − 1|
sin(πt)
tan(πt)
~
lim R (t) = lim−
, lim
, lim
t→1−
t→1 t − 1 t→1− t 2 − 1 t→1− t − 1
¿
À
sin(πt)
− (t − 1)
tan(πt)
= lim−
, lim− 2
, lim−
t→1
t − 1 t→1 t − 1 t→1 t − 1
¿
À
π cos(πt)
π sec2 (πt)
= −1, lim−
, lim−
(L’Hopital’s Rule)
t→1
t→1
2t
1
D
π E
= −1, − , π
2
4 / 17
Limits and Continuity Derivatives Integrals Arclength Exercises
Definition Examples
Continuity of Vector-Valued Functions
Example 3.
Determine if the function
À
¿
 sin(t) , t − 1, et
, t 6= 0
~
t
R (t) =

ı̂ − 2 ̂ + k̂
,t = 0
is continuous at t = 0.
Solution.
~
R (0) exists. ~
R (0) = ı̂ − 2 ̂ + k̂.
¿
µ
¶
À
sin(t)
t
lim~
R (t) = lim
, lim(t − 1), lim e = 〈1, −1, 1〉
t→0
t→0
t→0
t→0
t
~
~
R (0) = 〈1, −2, 1〉 6= 〈1, −1, 1〉 = lim R (t)
t→0
Hence, ~
R(t) is discontinuous at t = 0.
5 / 17
Limits and Continuity Derivatives Integrals Arclength Exercises
Definition Examples
Derivatives of Vector-Valued Functions
Definition
The derivative of a vector-valued function ~
R is defined by
~
R (t + ∆t) −~
R (t)
,
t→∆t
∆t
~
R 0 (t) = lim
if this limit exists.
~
~
R(t)
The vectors R(t+∆t)−
and
∆t
~
R (t + ∆t) −~
R (t) are parallel.
~
R (t + ∆t) −~
R (t)
As ∆t → 0, the vector
∆t
approaches a vector tangent to the
graph of ~
R (t).
Thus, ~
R 0 (t) is a vector tangent to the
graph of ~
R (t) in the direction of the
curve.
6 / 17
Limits and Continuity Derivatives Integrals Arclength Exercises
Definition Examples
Derivatives of Vector-Valued Functions
Let ~
R(t) = 〈f (t), g(t), h(t)〉, where f , g and h are differentiable scalar
functions of t.
~
R (t + ∆t) −~
R (t)
∆t→0
∆t
­
® ­
®
f (t + ∆t) , g (t + ∆t) , h (t + ∆t) − f (t) , g (t) , h (t)
= lim
∆t→0
∆t
¿
À
f (t + ∆t) − f (t)
g (t + ∆t) − g (t)
h (t + ∆t) − h (t)
= lim
, lim
, lim
∆t→0
∆t→0
∆t→0
∆t
∆t
∆t
®
­ 0
0
0
= f (t) , g (t) , h (t)
~
R 0 (t) = lim
Theorem
®
­
®
­
Given ~
R (t) = f (t) , g (t) , h (t) . Then ~
R 0 (t) = f 0 (t) , g 0 (t) , h 0 (t)
provided that f 0 (t), g 0 (t) and h 0 (t) exist.
7 / 17
Limits and Continuity Derivatives Integrals Arclength Exercises
Definition Examples
Derivatives of Vector-Valued Functions
Example 4.
Determine ~
R 0 (0) and ~
R 00 (0) if
­
®
~
R (t) = e2t , cos t, − tan−1 t .
Solution.
~
R 0 (t)
~
R 00 (t)
¿
À
−1
2e2t , − sin t,
=⇒ ~
R 0 (0) = 〈2, 0, −1〉
1 + t2
¿
À
2t
=
4e2t , − cos t,
=⇒ ~
R 00 (0) = 〈4, −1, 0〉
(1 + t 2 )2
=
8 / 17
Limits and Continuity Derivatives Integrals Arclength Exercises
Definition Examples
Tangent Line
Example 5.
Determine a vector equation of the tangent line to the curve
p
®
­
~
R (t) = ln t, 5 − 4t, t 3 − 6 at the point corresponding to t = 1.
Solution.
point of tangency: terminal of the vector ~
R(1), which is (0, 1, −5)
R 0 (1)
direction vector of the tangent line : tangent vector ~
~
R 0 (t) =
¿
1
2
, −p
, 3t 2
t
5 − 4t
À
=⇒ ~
R 0 (1) = 〈1, −2, 3〉
Hence, the vector equation of the tangent line is
~L(t) = 〈t, 1 − 2t, −5 + 3t〉 .
9 / 17
Limits and Continuity Derivatives Integrals Arclength Exercises
Definition Examples
Theorems on Differentiation
Let ~
F ) and ~
G be vector functions and f be a real-valued function.
1
2
3
4
5
(~
F ±~
G)0 (t) = ~
F 0 (t) ± ~
G 0 (t)
¡
¢ 0
¡
¢
(f ~
F )0 (t) = f (t) ~
F (t) + f 0 (t) ~
F (t)
(~
F ·~
G)0 (t) = ~
F (t) · ~
G 0 (t) + ~
G (t) · ~
F 0 (t)
(~
F ×~
G)0 (t) = ~
F (t) × ~
G 0 (t) + ~
F 0 (t) × ~
G(t)
¡
¢
¡
¢
(~
F ◦ f )0 (t) = ~
F 0 f (t) f 0 (t)
10 / 17
Limits and Continuity Derivatives Integrals Arclength Exercises
Definition Examples
Theorems on Differentiation
Example 6.
­
®
Let ~
F (t) = t − 1, t 3 , cos t , ~
G (t) = 〈ln t, sinh t, −4〉 and f (t) = e−t .
0
0
Find (~
F ·~
G) (t) and (~
G ◦ f ) (t).
Solution.
À
1
, cosh t, 0 +
t
­
®
〈ln t, sinh t, −4〉 · 1, 3t 2 , − sin t
t −1
=
+ ln t + t 3 cosh t + 3t 2 sinh t + 4 sin t
t
¡
¢0
~
F ·~
G (t) =
­
®
t − 1, t 3 , cos t ·
¿
11 / 17
Limits and Continuity Derivatives Integrals Arclength Exercises
Definition Examples
Theorems on Differentiation
Example 6.
­
®
Let ~
F (t) = t − 1, t 3 , cos t , ~
G (t) = 〈ln t, sinh t, −4〉 and f (t) = e−t .
0
0
Find (~
F ·~
G) (t) and (~
G ◦ f ) (t).
Solution.
~
G 0 (t) =
¿
1
, cosh t, 0
t
À
and f 0 (t) = −e−t
À
¡
¢
1
−t
, cosh e , 0 −e−t
−t
e
­
®
= −1, −e−t cosh e−t , 0
(~
G ◦ f )0 (t) =
¿
12 / 17
Limits and Continuity Derivatives Integrals Arclength Exercises
Integrals of Vector-Valued Functions
­
®
Given ~
R (t) = f (t) , g (t) , h (t) .
Z
indefinite integral :
b
Z
definite integral :
a
~
R (t) dt =
~
R (t) dt =
¿Z
Z
f (t) dt,
¿Z
b
À
Z
g (t) dt,
b
Z
f (t) dt,
h (t) dt
b
Z
g (t) dt,
a
a
À
h (t) dt
a
Example 7.
Z 1
Evaluate
〈3t 2 − 5, cos t, 2t〉 dt.
0
Solution.
Z 1
〈3t 2 − 5, cos t, 2t〉 dt
0
À
Z 1
cos tdt,
2tdt
0
0
¿ 0
¯1
¯1
¯1 À
¯
¯
¯
(t 3 − 5t)¯ , sin t ¯ , t 2 ¯
= 〈−4, sin 1, 1〉
=
¿Z
=
1
(3t 2 − 5)dt,
0
Z
1
0
0
13 / 17
Limits and Continuity Derivatives Integrals Arclength Exercises
Integrals of Vector-Valued Functions
Example 8.
­
®
Given ~
R 0 (t) = t 2 − 1, sin t, 2e2t and ~
R(0) = 〈2, −1, 0〉. Find ~
R(t).
Solution.
Z
~
R 0 (t) dt
¿Z
=
~
R(t) =
¿
¡ 2
¢
t − 1 dt,
Z
Z
sin t dt,
2e2t dt
À
À
1 3
t − t + C1 , − cos t + C2 , e2t + C3 , C1 , C2 , C3 ∈ R
3
~
R(0) = 〈C1 , −1 + C2 , 1 + C3 〉 = 〈2, −1, 0〉
=⇒ C1 = 2, C2 = 0, C3 = −1
¿
À
1 3
~
R(t) =
t − t + 2, − cos t, e2t − 1
3
14 / 17
Limits and Continuity Derivatives Integrals Arclength Exercises
Arclength of a Space Curve
Recall that the length of a plane curve with parametric equations
x = f (t), y = y(t), a É t É b, is given by
Z
s=
a
b q¡
¢2 ¡
¢2
f 0 (t) + g 0 (t) dt.
The length of a space curve is defined in the same way.
Theorem
­
®
The length of the graph of ~
R (t) = f (t) , g (t) , h (t) , where t ∈ [a, b], is
given by
Z bq
¢2
¡
¢2 ¡
s =
f 0 (t) + g 0 (t) + (h0 (t))2 dt
Za b
° 0 °
°~
=
R (t)° dt.
a
15 / 17
Limits and Continuity Derivatives Integrals Arclength Exercises
Arclength of a Space Curve
Example 9.
D 3
E
Find the length of the curve defined by ~
R (t) = 4t 2 , −3sint, 3cost ,
where t ∈ [0, 2].
Solution.
D 1
E
~
R0 (t) = 6t 2 , −3cost, −3sint
Z 2 r³
´
1 2
s =
6t 2 + (−3cost)2 + (−3sint)2 dt
0
Z 2p
=
36t + 9dt
0
¯
1
27 − 1
3 ¯2
= (4t + 1) 2 ¯ =
= 13.
0
2
2
16 / 17
Limits and Continuity Derivatives Integrals Arclength Exercises
Exercises
1
2
3
4
¿
À
−1
−2t ln t
Evaluate lim tan t, e ,
.
t→∞
t −1
¿
Find the derivative of ~
R (t) = sin2 (3t + 1), ln(t 2 − 1),
À
t
.
t2 + 1
Find the vector equation of the tangent line to the graph of
®
­
p
~
R (t) = 1 + 2 t, t 3 − t, t 3 + t at the point (3, 0, 2).
Evaluate the integral
Z π
®
2 ­
sin2 t cos t, cos t sin2 t, tan2 t dt.
0
5
6
¿
À
2 2 3
~
Find the length of the graph of R (t) = 2t, t , t , t ∈ [0, 1].
3
Set-up the definite integral that represents the arc length of the
sin t
curve ~
R(t) = e3t ı̂ −
̂ + ln(t + cos t)k̂ from the point (1, 0, 0)
2t + π
3π
to point (e , 0, ln(π − 1)).
17 / 17