Limits and Continuity Derivatives Integrals Arclength Exercises
Calculus of Vector-Valued Functions
Mathematics 54 - Elementary Analysis 2
Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Limits and Continuity of Vector Functions
Definition.
Given~R(t)=
x(t) ,y(t) ,z(t)®
.
1 We define thelimit of~Rastapproachesaby
lim
t→a~R(t)=
D
lim
t→ax(t) , limt→ay(t) , limt→az(t)
E
,
provided that lim
t→ax(t), limt→ay(t), and limt→az(t) exist.
2 The function~R(t) iscontinuousatt=aif
~R(a) exists; lim
t→a~R(t) exists;
Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Limits of Vector Functions
Example
Evaluate the following limits:
1 lim
t→2~R(t) where~R(t)=
¿
t+1,t 2−4
t−2,
sin (2t−4)
t−2
À
.
2 lim
t→1−~R(t) where~R(t)=
¿
|t−1|
t−1 , sin(πt)
t2−1 , tan(πt)
t−1
À
.
1 We have
lim
t→2~R(t) =
¿
lim
t→2(t+1) , limt→2
t2−4
t−2, limt→2
sin (2t−4)
t−2
À
=
¿
3, lim
t→2(t+2) , limt→2
2 cos (2t−4) 1
À
(L’Hopital’s Rule)
Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Limits of Vector Functions
2 Note that fort→1−,t<1. We have
lim
t→1−~R(t) =
¿
lim
t→1−
|t−1|
t−1, limt→1− sin(πt)
t2−1 , limt→1− tan(πt)
t−1
À
=
¿
lim
t→1−
−(t−1)
t−1 , limt→1− sin(πt)
t2−1 , limt→1− tan(πt)
t−1
À
=
¿
−1, lim
t→1−
−πcos(πt) 2t , limt→1−
πsec2(πt) 1
À
(L’Hopital’s Rule)
=D−1,π 2,π
E
Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Continuity of Vector Functions
Example
Determine whether the function
~R(t)=
¿
sin(t)
t ,t−1,e t
À
, t6=0 ˆ
ı−2 ˆ+kˆ, t=0
is continuous att=0.
Solution:
a. ~R(0) exists.~R(0)=ˆı−2 ˆ+kˆ.
b. lim
t→0~R(t)=
¿
lim
t→0
µsin(t)
t
¶
, lim
t→0(t−1), limt→0e
t
À
= 〈1,−1, 1〉
c. ~R(0)= 〈1,−2, 1〉 6= 〈1,−1, 1〉 =lim
Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Continuity of Vector Functions
Example
Find value/s oftsuch that
~R(t)=
¿sin(t
−2) 2−t ,t+1,
1
et−1
À
is continuous.
Solution:
Possible discontinuities att=0, 2.
Whent=0. Note that~R(2) is undefined. Also
lim
t→0+ 1
et−1= +∞ and tlim→0− 1
et−1= −∞
Therefore, lim
Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
~
R
(
t
)
=
sin(2−t−t2),
t
+
1,
et1−1®
Whent=2. Note that~R(2) is also undefined but observe that
lim
t→2~R(t) =limt→2
D
sin(t−2) 2−t ,t+1,
1
et−1
E
=
¿
lim
t→2 sin(t−2)
2−t , limt→2(t+1), limt→2
1
et−1
À
= ¿
1, 3, 1
e2−1 À
.
Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Derivative of
~
R
(
t
)
=
x
(
t
) ,
y
(
t
) ,
z
(
t
)
®
Derivative of~Ris defined by~R0(t)= lim
t→∆t
~R(t+∆t)−~R(t)
∆t , if this limit exists.
Since∆1tis a scalar, the vectors ~R(t+∆t)−~R(t)
∆t and~R(t+∆t)−~R(t) are parallel.
As∆t→0, the vector
Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Derivative of
~
R
(
t
)
=
x
(
t
) ,
y
(
t
) ,
z
(
t
)
®
~R0(t) = lim
∆t→0
~R(t+∆t)−~R(t) ∆t
= lim
∆t→0
x(t+∆t) ,y(t+∆t) ,z(t+∆t)®−x(t) ,y(t) ,z(t)® ∆t
=
¿
lim
∆t→0
x(t+∆t)−x(t) ∆t , lim∆t→0
y(t+∆t)−y(t) ∆t , lim∆t→0
z(t+∆t)−z(t) ∆t
À
=
x0(t) ,y0(t) ,z0(t)®
Theorem
Given~R(t)=
x(t) ,y(t) ,z(t)®
. Then~R0(t)=
x0(t) ,y0(t) ,z0(t)®
provided that
Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Derivatives of Vector Functions
Example
Determine~R0(t) and~R00(t) if
~R(t)=
lnt, sect,−tan−1t®
.
Solution.
We have
~R0(t) =
¿1
t, secttant, −1 1+t2
À
~R00(t) =
¿
−t12, sec3t+secttan2t, 2t (1+t2)2
Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Tangent Line
Example
Determine a vector equation of the tangent line to~R(t)=
lnt,e−t,t3®
at the point corresponding tot=2.
Solution.
Note that~R0(t) gives the direction of the line tangent to the curve~R(t). The point of tangency is the terminal point of the vector~R(2) which is (ln 2,e−2, 8).
Recall that line passing through the point (x0,y0,z0) and is parallel to the vector〈a,b,c〉has the vector equation
~L(t)=
x0+at,y0+bt,z0+ct
®
.
~R0(t)=1
t,−e−t, 3t2
®
⇒ ~R0(2)=1
2,−e−2, 12
®
∴L(t)=~R(2)+t~R0(2)=
ln 2+2t,e−2−e−2t, 8+12t®
Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Theorems on Differentiation
Let~F(t) and~G(t) be vector functions andf(t) be a real-valued function.
1 (~F+~G)0(t)=~F0(t)+~G0(t)
2 (f~F)0(t)=¡f(t)¢~F0(t)+~F(t)¡f0(t)¢
Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Theorems on Differentiation
Example Let~F(t)=
t−1,t3, cost®
,~G(t)= 〈lnt, sinht,−4〉andf(t)=e−t. Evaluate:
1 (~F·~G)0(t) 2 (~G◦f)0(t)
Solution.
1 ¡~F·~G¢0(t)
= t−1,t3, cost®
·
¿1
t, cosht, 0
À
+ 〈lnt, sinht,−4〉 ·
1, 3t2,−sint®
= t−t1+lnt+t3cosht+3t2sinht+4 sint
2 (~G◦f)0(t) =
¿ 1
e−t, coshe− t, 0
À ¡
−e−t¢
=
Limits and Continuity Derivatives Integrals Arclength Exercises
Integrals of Vector Functions
Given~R(t)=
x(t) ,y(t) ,z(t)®
.
indefinite integral:
Z
~R(t)dt=
¿Z
x(t)dt,
Z
y(t)dt,
Z
z(t)dt
À
definite integral:
Zb
a
~R(t)dt=
¿Z b
a
x(t)dt,
Z b
a
y(t)dt,
Zb
a z(t)dt
À
Example
Given~R(t)=
t2−1, cos 2t, 2e2t®
. EvaluateR~
R(t)dt.
Z
~R(t)dt =
¿Z ¡
t2−1¢
dt,
Z
cos 2t dt,
Z
2e2tdt
À
=
¿1
3t
3−t+C1,1
2sin 2t+C2,e 2t+C
3
À
C1,C2,C3constants
= ¿1
t3−t,1sin 2t,e2t À
Limits and Continuity Derivatives Integrals Arclength Exercises
Integrals of Vector Functions
Example
Evaluate
Z 1
0
³
(3−t)32ˆı+(3+t)32ˆ+kˆ
´ dt. Solution. ¿Z 1 0 ³
(3−t)32
´
dt,
Z 1
0
(3+t)32dt,
Z 1 0 dt À = ¿
−25(3−t) 5 2 ¯ ¯ ¯ 1 0, 2 5(3+t)
5 2 ¯ ¯ ¯ 1
0,t
¯ ¯ ¯ 1 0 À = ¿ −2 5 ¡
25/2−35/2¢
,2 5
¡
25−35/2¢
, 1
Limits and Continuity Derivatives Integrals Arclength Exercises
Arclength of a Space Curve
Recall of Length of Arc of a Plane Curve
Given a plane curve with parametric equationsx=x(t),y=y(t),aÉtÉb. The length of the curve is given by
s=
Z b
a
q
(x0(t))2
+¡y0(t)¢2
dt.
The length of a space curve is defined in the same way.
Given~R(t)=
x(t) ,y(t) ,z(t)®
,t∈[a,b] The length of the graph of~Ris given by
s =
Zb
a
q
(x0(t))2
+¡
y0(t)¢2
+(z0(t))2dt
=
Z b
a
° °~R0(t)
Limits and Continuity Derivatives Integrals Arclength Exercises
Arclength of a Vector Function
Example
Find the length of the curve given by~R(t)=
D
4t32,−3sint, 3cost
E
where
t∈[0, 2].
Note that
~R0(t)=D6t12,−3cost,−3sint
E
.
s =
Z2
0
r ³
6t12
´2
+(−3cost)2+(−3sint)2dt
=
Z 2
0
p
36t+9dt
=3
1 6(4t+1)
3 2
t=2
t=0
=27−1
Limits and Continuity Derivatives Integrals Arclength Exercises
Exercises
1 Evaluate lim
t→∞
¿
tan−1t,e−2t, lnt
t−1
À
.
2 Find the derivative of~R(t)= ¿
sin2(3t+1), ln(t2−1), t
t2+1
À
.
3 Find the vector equation of the tangent line to the graph of
~R(t)=
1+2pt,t3−t,t3+t®
at the point (3, 0, 2).
4 Evaluate the integral
Z π
2
0
sin2tcost, costsin2t, tan2t®
dt.
5 Find the length of the graph of~R(t)= ¿
2t,t2,2 3t
3
À
,t∈[0, 1].
6 Set-up the definite integral that represents the arc length of the curve
~R(t)=e3tˆı− sint
2t+πˆ+ln(t+cost)ˆkfrom the point (1, 0, 0) to point