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

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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