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Limits and Continuity Derivatives Integrals Arclength Exercises
Calculus of Vector-Valued FunctionsMathematics 54 - Elementary Analysis 2
Institute of MathematicsUniversity of the Philippines-Diliman
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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 the limit of~R as t approaches a by
limta
~R (t)=
limtax (t) , limtay (t) , limtaz (t)
,
provided that limta x (t), limta y (t), and limta z (t) exist.
2 The function~R (t) is continuous at t = a if~R (a) exists;
limta
~R (t) exists;
~R (a)= limta
~R (t).
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Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Limits of Vector Functions
Example
Evaluate the following limits:
1 limt2
~R (t) where~R (t)=
t+1, t24t2 ,
sin(2t4)t2
.
2 limt1
~R (t) where~R (t)= |t1|
t1 ,sin(pit)
t21 ,tan(pit)
t1
.
1 We have
limt2
~R (t) =
limt2 (t+1) , limt2
t24t2 , limt2
sin(2t4)t2
=
3, limt2 (t+2) , limt2
2cos(2t4)1
(LHopitals Rule)
= 3,4,2 .
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Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Limits of Vector Functions
2 Note that for t 1, t < 1. We have
limt1
~R (t) =
limt1
|t1|t1 , limt1
sin(pit)
t21 , limt1tan(pit)
t1
=
limt1
(t1)t1 , limt1
sin(pit)
t21 , limt1tan(pit)
t1
=1, lim
t1picos(pit)
2t, lim
t1pisec2(pit)
1
(LHopitals Rule)
=1,pi
2,pi
.
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Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Continuity of Vector Functions
Example
Determine whether the function
~R (t)=
sin(t)
t, t1,et
, t 6= 0
2 + k, t = 0
is continuous at t = 0.Solution:
a. ~R (0) exists.~R (0)= 2 + k.b. lim
t0~R (t)=
limt0
(sin(t)
t
), lim
t0(t1), limt0 et= 1,1,1
c. ~R (0)= 1,2,1 6= 1,1,1 = limt0
~R (t)
Hence,~R(t) is discontinuous at t = 0.
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Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Continuity of Vector Functions
Example
Find value/s of t such that
~R (t)=
sin(t2)2 t , t+1,
1
et 1
is continuous.
Solution:Possible discontinuities at t = 0,2.
When t = 0. Note that~R (0) is undefined. Also
limt0+
1
et 1 =+ and limt01
et 1 =
Therefore, limt0
~R (t) does not exist.
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Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
~R (t)= sin(t2)2t , t+1, 1et1
When t = 2. Note that~R (2) is also undefined but observe that
limt2
~R (t) = limt2
sin(t2)
2t , t+1, 1et1
=
limt2
sin(t2)2t , limt2(t+1), limt2
1et1
=1,3, 1
e21
.
Hence,~R is continuous at every t R\{0,2}.
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Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Derivative of~R (t)= x (t) ,y (t) ,z (t)Derivative of~R is defined by~R (t)= lim
tt~R(t+t)~R(t)
t , if this limit exists.
Since 1t is a scalar, the vectors~R(t+t)~R(t)
t and~R (t+t)~R (t)
are parallel.
As t 0, the vector~R (t+t)~R (t) becomes a vectortangent to the graph of~R (t).
So~R (t) is a vector having one ofits representations to be tangentto the graph of~R (t) in thedirection of the increasingparameter t.
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Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Derivative of~R (t)= x (t) ,y (t) ,z (t)
~R (t) = limt0
~R (t+t)~R (t)t
= limt0
x (t+t) ,y (t+t) ,z (t+t)x (t) ,y (t) ,z (t)
t
=
limt0
x (t+t)x (t)t
, limt0
y (t+t)y (t)t
, limt0
z (t+t)z (t)t
= x (t) ,y (t) ,z (t)
Theorem
Given~R (t)= x (t) ,y (t) ,z (t). Then~R (t)= x (t) ,y (t) ,z (t) provided thatx (t) ,y (t) and z (t) exist.
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Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Derivatives of Vector Functions
Example
Determine~R (t) and~R (t) if
~R (t)= ln t, sec t, tan1 t .Solution.We have
~R (t) =
1
t, sec t tan t,
11+ t2
~R (t) = 1
t2, sec3 t+ sec t tan2 t, 2t
(1+ t2)2
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Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Tangent Line
Example
Determine a vector equation of the tangent line to~R (t)= ln t,et , t3 atthe point corresponding to t = 2.
Solution.
Note that~R(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(ln2,e2,8).Recall that line passing through the point (x0,y0,z0) and is parallel tothe vector a,b,c has the vector equation~L(t)= x0+at,y0+bt,z0+ ct.
~R(t)= 1t , et , 3t2 ~R(2)= 12 ,e2,12 L(t)=~R(2)+ t~R(2)= ln2+ t2 ,e2e2t,8+12t .
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Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Theorems on Differentiation
Let~F (t) and~G (t) be vector functions and f (t) be a real-valued function.
1 (~F +~G) (t)=~F (t)+~G (t)2 (f~F) (t)= (f (t))~F (t)+~F (t)(f (t))3 (~F ~G) (t)=~F (t) ~G (t)+~G (t) ~F (t)4 (~F ~G) (t)=~F (t)~G (t)+~F (t)~G (t)5 (~F f ) (t)=~F (f (t))(f (t))
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Limits and Continuity Derivatives Integrals Arclength Exercises Definition Examples
Theorems on Differentiation
Example
Let~F (t)= t1, t3, cos t,~G (t)= ln t, sinh t,4 and f (t)= et . Evaluate:1 (~F ~G) (t)2 (~G f ) (t)
Solution.
1(~F ~G) (t)= t1, t3, cos t 1
t, cosh t, 0
+ln t, sinh t,4 1, 3t2,sin t
= t1t
+ ln t+ t3 cosh t+3t2 sinh t+4sin t
2 (~G f ) (t) =
1
et, coshet , 0
(et)= 1,et coshet ,0
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Limits and Continuity Derivatives Integrals Arclength Exercises
Integrals of Vector Functions
Given~R (t)= x (t) ,y (t) ,z (t).indefinite integral :
~R (t)dt =
x (t)dt,
y (t)dt,
z (t)dt
definite integral : b
a~R (t)dt =
ba
x (t)dt, b
ay (t)dt,
ba
z (t)dt
Example
Given~R (t)= t21,cos2t,2e2t. Evaluate ~R (t)dt.~R (t)dt =
(t21) dt, cos2t dt, 2e2t dt
=
1
3t3 t+C1, 1
2sin2t+C2,e2t +C3
C1,C2,C3 constants
=
1
3t3 t, 1
2sin2t,e2t
+~C, where~C = C1,C2,C3
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Limits and Continuity Derivatives Integrals Arclength Exercises
Integrals of Vector Functions
Example
Evaluate 1
0
((3 t) 32 + (3+ t) 32 + k
)dt.
Solution. 10
((3 t) 32
)dt, 1
0(3+ t) 32 dt,
10
dt
= 25 (3 t)
52
10
, 25 (3+ t)52
10
, t10
=2
5
(25/235/2) , 2
5
(2535/2) ,1
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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 equations x= x(t), y = y(t), a t b.The length of the curve is given by
s= b
a
(x (t))2+ (y (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~R is given by
s = b
a
(x (t))2+ (y (t))2+ (z (t))2 dt
= b
a
~R (t)dt.16 / 18
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Limits and Continuity Derivatives Integrals Arclength Exercises
Arclength of a Vector Function
Example
Find the length of the curve given by~R (t)=
4t32 ,3sint,3cost
where
t [0,2].
Note that~R (t)=
6t
12 ,3cost,3sint
.
s = 2
0
(6t
12
)2+ (3cost)2+ (3sint)2dt= 2
0
p36t+9dt
= 31
6(4t+1)
3
2
t=2t=0
= 2712
= 13.
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Limits and Continuity Derivatives Integrals Arclength Exercises
Exercises
1 Evaluate limt
tan1 t,e2t ,
ln t
t1
.
2 Find the derivative of~R (t)=
sin2(3t+1), ln(t21), tt2+1
.
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 pi2
0
sin2 t cos t,cos2 t sin t, tan2 t
dt.
5 Find the length of the graph of~R (t)=
2t, t2,2
3t3
, t [0,1].
6 Set-up the definite integral that represents the arc length of the curve
~R(t)= e3t sin t2t+pi + ln(t+cos t)k from the point (1,0,0) to point
(e3pi,0, ln(pi1)).18 / 18
Limits and ContinuityDefinitionExamples
DerivativesDefinitionExamples
IntegralsArclengthExercises
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