19 Calculus of Vector Valued Functions.pdf

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 the}limit 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 , lim_t→1− tan(πt)

t₋1

À

=

¿

lim

t→1−

−(t−1)

t₋1 , limt→1− sin(πt)

t2₋₁ , lim_t→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

µ_sin(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

)

=

­

,

t

+

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 , lim_t→2(t+1), lim_t→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_∆1_tis 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, 2e}2t®

. 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)32_ˆı+(3+t)32_ˆ+kˆ

´ dt. Solution. ¿Z 1 0 ³

(3−t)32

´

dt,

Z 1

0

(3+t)32_dt,

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))2_dt

=

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+πˆ+ln(t+cost)ˆkfrom the point (1, 0, 0) to point