12.2 The Calculus of Vector-Valued Functions Calculus of... · 2009. 4. 29. · The Calculus of...

King Abdulaziz University

12.2 The Calculus of Vector-Valued Functions

Dr. Hamed Al-Sulami

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c© 2009 [email protected] http://www.kau.edu.sa/hhaalsalmi

Prepared: April 29, 2009 Presented: April 29, 2009, 2009

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http://www.kau.edu.sa

mailto:[email protected]

http://www.kau.edu.sa/hhaalsalmi

2/15

The Calculus of Vector-Valued Functions

Limits and Continuity

Definition .1: [The Limit of Vector-Valued Function]For a vector-valued function r(t) = f(t) i + g(t) j + h(t)k, the limit of r(t) as t approachesa is given by

limt→a

r(t) = limt→a

[f(t) i + g(t) j + h(t)k] = limt→a

f(t) i + limt→a

g(t) j + limt→a

h(t)k,

provided all of the indicated limits exist. If any of the limits limt→a

f(t), limt→a

g(t), limt→a

h(t) fail

to exist, then limt→a

r(t) does not exist.(DNE)

Note 1: The expression limt→a

r(t) = L means as t approaches a the magnitude of the vector

r(t)−L is approaching 0. Thus ‖r(t) − L‖ → 0 as t → a. The graph❍? illustrate the meaningof this note.

Definition .2: [Continuity of a Vector-Valued Function]A vector-valued function r(t) = f(t) i + g(t) j + h(t)k, is continuous at t = a iflimt→a

r(t) = r(a) (i.e. the limit exist and equal to r(a).) A vector-valued function r iscontinuous on an interval I if it is continuous at every point in I.

❘ ➡ ➡ ➦ � � � � ➥ ➡ �

xy

z

r(t) − L

r(t) − L

L

r(t)

r(t)

The Calculus of Vector-Valued Functions 3/15

Example 1. Find limt→0

[sin t

ti +

√1 − t j + t k

].

Solution: Since limt→0

sin t

t= 1, lim

t→0

√1 − t = 1, and lim

t→0t = 0, then

limt→0

[sin t

ti +

√1 − t j + t k

]= i + j + 0 k❍? .

�

Note 2: A vector-valued function r(t) = 〈f(t), g(t), h(t)〉 , is continuous at t = a if andonly if all of f, g, and h are continuous at t = a.

Example 2. Determine the values of t for which the vector-valued function r(t) =t

t − 2i+

√4 − t j + t k is continuous.

Solution: Sincet

t − 2, is continuous on R \ {2},

√4 − t, is continuous on (−∞, 4], and t, is

continuous every where then

r(t) =t

t − 2i+

√4 − t j+t k is continuous on R\{2}∩(−∞, 4]∩R = (−∞, 2)∪(2, 4]. �

❘ ➡ ➡ ➦ � � � � ➥ ➡ �

x y

z


Differentiation of Vector-Valued Functions

Definition .3: [Derivative of a Vector-Valued Function]The derivative of a vector-valued function r(t), is defined by

r′(t) = limΔt→0

r(t + Δt) − r(t)Δt

for all t for which the limit exists. If r′(a) exists, then r(t)

is differentiable at a. A vector-valued function r is differentiable on an interval I if it isdifferentiable at every point in I.

Note 3: In addition to r′(t), we use the following notations for the derivative of a vector-

valued function Dt[r(t)],d

dt[r(t)], and

dr

dt.

Differentiation of a Vector-Valued Function can be done on a component-by-componentbasis

Theorem .1: [Derivative of a Vector-Valued Function]If r(t) = f(t) i + g(t) j + h(t)k, where f, g, and h are differentiable of t, then

r′(t) = f ′(t) i + g′(t) j + h′(t)k.

❘ ➡ ➡ ➦ � � � � ➥ ➡ �


123456789101112 of 12

r(a)

x

y

z

r(a)

r(a + Δt) − r(a)

r(a + Δt)

x

y

z

r(a)

r(a + Δt) − r(a)

r(a + Δt)

x

y

z

x

y

z

x

y

z

x

y

z

x

y

z

x

y

z

x

y

z

x

y

z

x

y

z

r(a)

r′(a)

x

y

z

The graph of r(t)The graph of r(t)The graph of r(t)The graph of r(t)The graph of r(t)The graph of r(t)The graph of r(t)The graph of r(t)The graph of r(t)The graph of r(t)The graph of r(t)The graph of r(t)

Interpretation of the DerivativeThe figures to the left show thegraphs of r(t), with orientation andthe vectors r(a), r(a + Δt), andr(a + Δt) − r(a). The vector r(a +Δt) − r(a) runs along the secantline joining the terminal points ofr(a + Δt), and r(a). Now, the vec-tor r(t+Δt)−r(t)

Δt points in the direc-tion of increasing t and runs alongthe secant line. As Δt → 0, the se-cant line approaches the tangent lineat the terminal point of r(a), so if the

limit r′(a) = limΔt→0

r(a + Δt) − r(a)Δt

exist and nonzero, it is a vector thatis tangent to the graph of r(t) at thetip of r(a) and point in the directionof increasing t.
























































❘ ➡ ➡ ➦ � � � � ➥ ➡ �


Example 3. plot the curve of r(t) = − cos (2t) i + sin (2t) j and draw the position vectorand the tangent vector at t = 0, t = π

8 , t = π4 .

Solution:

Using the parametric equations we have

x = − cos (2t) and y = sin (2t).x2 +y2 = (− cos (2t))2 +(sin (2t))2 = 1.This is the unite circle. Now, sincer(t) = − cos (2t) i + sin (2t) j, thenr′(t) = 2 sin (2t) i + 2 cos (2t) j.

r(0) = − i + 0 j r′(0) = 0 i + 2 j

r(π

8) =

−1√2

i +1√2

j r′(π

8) =

√2 i +

√2 j

r(π

4) = 0 i + j r′(

π

4) = 2 i + 0 j

To draw the position vector and thetangent vector at any t. For the posi-tion vector draw a line from the point(0, 0) to r(t) and for the tangent vectordraw a line from r(t) to r(t) + r′(t).

1

-1

-2

1-1-2

x

yr′(0)

r(0)

r(π/8)

r′(π/8)

r(π/4)

r′(π/4)

�

❘ ➡ ➡ ➦ � � � � ➥ ➡ �


Example 4. plot the curve of r(t) =√

t i + (2 − t) j and draw the position vector and thetangent vector at t = 1.

Solution:

Using the parametric equations we have

x =√

t and y = 2 − t.x2 = t, and y = 2 − t = 2 − x2 withx =

√t ≥ 0. This is the right part of a

parabola with vertex at (0, 2) and opendawn. Now, since r(t) =

√t i+(2−t) j,

then r′(t) =1

2√

ti − j.

r(1) = i + j r′(1) =12

i − j

To draw the position vector and thetangent vector at any 1. For the posi-tion vector draw a line from the point(0, 0) to (1, 1) and for the tangent vectordraw a line from (1, 1) to r(1)+r′(1) =(3/2, 0).

1

2

-1

-2

1 2-1-2

x

y

r′(1)

r(1)

�

❘ ➡ ➡ ➦ � � � � ➥ ➡ �


Example 5. plot the curve of r(t) = 2 cos t i+sin t j + t k and draw the position vector andthe tangent vector at t = π/4.

Solution:

Using the parametric equations we have x = 2 cos t, y = sin t, andz = t. Thus

x

2= cos t and y = sin t. Hence(x

2

)2

+y2 = cos2 t+sin2 t = 1. Thus the graph lie on the elliptical

cylinder x2 + 4y2 = 4. Now, since r(t) = 2 cos t i + sin t j + t k,then r′(t) = −2 sin t i + cos t j + k.

r(π/4) =√

2 i +1√2

j + π/4 k

r′(π/4) = −√

2 i − 1√2

j + kr(π/4)

r′(π/4)

x

y

z

�

❘ ➡ ➡ ➦ � � � � ➥ ➡ �


Definition .4: [Smooth Curve]The curve given by the vector-valued function r(t) = f(t) i+g(t) j+h(t)k, is called smoothon an open interval I if r′(t) is continuous on I and r′(t) �= 0 for any t in I.

Example 6. Determine whether the vector valued function r(t) is smooth

1. r(t) =⟨t2, t3, t sin t

⟩2. r(t) =

⟨sin2 t, cos2 t

⟩Solution:

1. Since r(t) =⟨t2, t3, t sin t

⟩, then r′(t) =

⟨2t, 3t2, sin t + t cos t

⟩.

r′(t) = 0⇔2t = 0, 3t2 = 0, sin t + t cos t = 0⇔t = 0.

Hence r(t) =⟨t2, t3, t sin t

⟩is not smooth on any interval contains See the 0.❍?

2. Since r(t) =⟨sin2 t, cos2 t

⟩, then r′(t) = 〈2 sin t cos t,−2 cos t sin t〉.

r′(t) = 0⇔2 sin t cos t = 0, −2 sin t cos t = 0⇔ sin t = 0, cos t = 0

⇔t = nπ, or t = (2n + 1)π

2n ∈ Z.

Hence r(t) =⟨sin2 t, cos2 t

⟩is not smooth on any interval contains nπ, or (2n + 1)π

2 .❍?

�

❘ ➡ ➡ ➦ � � � � ➥ ➡ �

1

-1

-2

1-1-2

x

y

r′(t)

r(t)

r(t)

r′(t)

x

y

z


Differentiation Rules

Theorem .2: [Properties of the Derivative]Suppose that r(t) and s(t) are differentiable vector-valued functions, f(t) is a differentiablereal-valued function, and c is any scalar.

1.d

dt[r(t) ± s(t)] = r′(t) ± s′(t)

2.d

dt[cr(t)] = cr′(t)

3.d

dt[f(t)r(t)] = f(t)r′(t) + f ′(t)r(t)

4.d

dt[r(t) · s(t)] = r′(t) · s(t) + r(t) · s′(t)

5.d

dt[r(t) × s(t)] = r′(t) × s(t) + r(t) × s′(t)

6.d

dt[r(f(t))] = f ′(t)r′(f(t))

Note 4: Two vectors a and b are orthogonal if a · b = 0. Also note that ‖a‖2 = a · a.

❘ ➡ ➡ ➦ � � � � ➥ ➡ �


Theorem .3: [Magnitude and Orthogonality]Let r(t) be a vector-valued function. ‖r(t)‖ is constant if and only if r(t) and r′(t) areorthogonal for all t.

Proof:

‖r(t)‖ = c, for some c ≥ 0 ⇔ ‖r(t)‖2 = c2

⇔ r(t) · r(t) = c2 differentiate both sides⇔ r(t) · r′(t) + r′(t) · r(t) = 0⇔ 2r(t) · r′(t) = 0 divide both sides by 2⇔ r(t) · r′(t) = 0⇔ r(t), r′(t) are orthogonal.

�

Note 5: If ‖r(t)‖ = c, then the graph of r(t) lies on a sphere( circle in 2-dimension) centerat the origin of radius c. The above theorem says the graph of r(t) lies on a sphere center atthe origin if and only if the tangent vector is orthogonal to the position vector at every point

on the graph. The vector-valued function r(t) =cos t√

3i +

sin t√3

j +√

2√3

k❍? is an example

of such vector-valued function.

❘ ➡ ➡ ➦ � � � � ➥ ➡ �

r(t)

r′(t)

x

y

z


Arc Length

Theorem .4: [Arc Length]If C is a smooth curve given by r(t) = f(t) i + g(t) j + h(t)k, on an interval [a, b], then the

arc length of C on the interval is s =

b∫a

‖r′(t)‖ dt =

b∫a

√(f ′(t))2 + (g′(t))2 + (h′(t))2 dt.

Example 7. Find the arc length of the curve given by r(t) = 2t i + ln t j + t2 k❍?

for 1 ≤ t ≤ e.

Solution: Since r′(t) = 2 i +1t

j + 2t k, then

‖r′(t)‖ =

√(2)2 +

(1t

)2

+ (2t)2 =

√4t2 + 1 + 4t4

t2=

√(2t2 + 1)2

t2=

∣∣∣∣2t2 + 1t

∣∣∣∣ =2t2 + 1

t,

since t ≥ 1. Hencee∫

1

‖r′(t)‖ dt =

e∫1

2t2 + 1t

dt =

e∫1

[2t +

1t

]dt =

[t2 + ln t

]e

1= [(e2 + ln e)− (1 + ln 1)] = e2.

�

❘ ➡ ➡ ➦ � � � � ➥ ➡ �

x y

z


Example 8. Find the arc length of the curve given by r(t) =t√2

i +t4

8j +

14t2

k❍? for

−2 ≤ t ≤ −1.

Solution: Since r′(t) =1√2

i +t3

2j +

−12t3

k, then

‖r′(t)‖ =

√(1√2

)2

+(

t3

2

)2

+(−12t3

)2

=

√t6

4+

12

+1

4t6=

√(t3

2+

12t3

)2

=∣∣∣∣ t32 +

12t3

∣∣∣∣ =∣∣∣∣ t6 + 1

2t3

∣∣∣∣ =t6 + 1−2t3

since −2 ≤ t ≤ −1. Hence−1∫

−2

‖r′(t)‖ dt =

−1∫−2

t6 + 1−2t3

dt =

−1∫−2

[−t3

2− t−3

2

]dt

=[−t4

8+

t−2

4

]−1

−2

=[(

−18

+14

)−

(−168

+116

)]

=3316

.

�

❘ ➡ ➡ ➦ � � � � ➥ ➡ �

x

y

z


Integration of Vector-Valued Functions

Definition .5: [Antiderivative]The vector-valued function R(t), is an antiderivative of the vector-valued function r(t) ifR′(t) = r(t).

Definition .6: [Indefinite Integral]If R(t), is any antiderivative of r(t), the indefinite integral of r(t) is defined to be∫

r(t) dt = R(t) + C where C is an arbitrary constant vector. Which mean if r(t) =

f(t) i + g(t) j + h(t)k, then∫

r(t) dt =[∫

f(t) dt

]i +

[∫g(t) dt

]j +

[∫h(t) dt

]k.

Example 9. Evaluate∫ ⟨

cos (2t),− sin (2t),1

1 + t2

⟩dt

Solution:∫ ⟨cos (2t),− sin (2t),

11 + t2

⟩dt =

⟨∫cos (2t) dt,−

∫sin (2t) dt,

∫1

1 + t2dt

⟩

=⟨

12

sin (2t) + C1,12

cos (2t) + C2, tan−1 t + C3

⟩

=12

sin (2t) i +12

cos (2t) j + tan−1 t k + C

�

❘ ➡ ➡ ➦ � � � � ➥ ➡ �


Definition .7: [Definite Integral]If R(t), is any antiderivative of r(t), on [a, b]the definite integral of r(t) is defined to be

b∫a

r(t) dt = R(b) − R(a). Which mean if r(t) = f(t) i + g(t) j + h(t)k, then

b∫a

r(t) dt =

⎡⎣ b∫

a

f(t) dt

⎤⎦ i +

⎡⎣ b∫

a

g(t) dt

⎤⎦ j +

⎡⎣ b∫

a

h(t) dt

⎤⎦ k.

Example 10. Evaluate

1∫0

⟨e−3t, t2 sin (t3), t cos t

⟩dt

Solution:1∫

0

⟨e−3t, t2 sin (t3), t cos t

⟩dt =

⟨ 1∫0

e−3t dt,

1∫0

t2 sin (t3) dt,

1∫0

t cos t dt

⟩

=

⟨−1

3

1∫0

e−3t−3 dt,1

3

1∫0

sin (t3)3 t2dt,

1∫0

t d(sin t)

⟩

=

⟨1 − e−3

3❍? ,

1 − cos (1)

3❍? , cos (1) + sin (1) − 1❍?

⟩

�

❘ ➡ ➡ ➦ � � � � ➥ ➡ �

1∫0

t cos t dt =

1∫0

t d(sin t) or use u = t, dv = cos t dt

= t sin t|10 −1∫

0

sin t dt Integration by part du = dt, v = sin t

= t sin t|10 − [− cos t]10

∫udv = uv −

∫vdu

= [sin 1 − 0 sin 0] + [cos 1 − cos 0]= cos (1) + sin (1) − 1

1∫0

t2 sin (t3) dt =13

1∫0

sin (t3)3 t2dt

=13

[− cos (t3)

]10

=13

[− cos (1) + 1]

=1 − cos (1)

3

1∫0

e−3t dt =−13

1∫0

e−3t−3 dt

=−13

[e−3t

]1

0

=−13

[e−3 − 1

]=

1 − e−3

3

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12.2 The Calculus of Vector-Valued Functions Calculus of... · 2009. 4. 29. · The Calculus of...

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