Why differentiating vector
function is important?
How to differentiate vector
functions?
What is the result of
differentiating vector
functions?
DESTACAMENTO X CALINGIN |
DIFFERENTIATING VECTOR FUNCTIONS
Why differentiating
vector function is
important?
10/1/2021
DESTACAMENTO X CALINGIN |
DIFFERENTIATING VECTOR FUNCTIONS
3
y
𝑟Ԧ 𝑡 𝑖𝑠 a position vector
at point P
P
C(t)
x
-x
Since this is a vector
function, we can use
the derivative to
describe the motion
of the object at
specific point in time.
-y
10/1/2021
DESTACAMENTO X CALINGIN |
DIFFERENTIATING VECTOR FUNCTIONS
4
The derivative of a vector-valued function can be understood to be an
instantaneous rate of change as well; for example, when the function
represents the position of an object at a given point in time, the derivative
represents its velocity at that same point in time
• 𝐫 (𝒕) is position vector at any point in time with components:
• 𝐫Ԧ t = x t 𝑖Ƹ + y t 𝐽መ + z t ෡𝑘
• The derivative r′(t) of vector rԦ is the velocity vector function
(always tangent to the curve) with components
• 𝐫Ԧ ′ t = x′ t 𝑖Ƹ + y′ t 𝐽መ + z′ t ෡𝑘
• The second derivative 𝐫Ԧ ′′ (𝒕) of vector rԦ is the acceleration
vector function with components
• 𝐫Ԧ ′′ t = x′′ t 𝑖Ƹ + y′′ t 𝐽መ + z′′ t ෡𝑘
Source: Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license
10/1/2021
DESTACAMENTO X CALINGIN |
DIFFERENTIATING VECTOR FUNCTIONS
6
Retrieved from https://instruct.math.lsa.umich.edu/lecturedemos/ma215/docs/13_2/examples.html
10/1/2021
DESTACAMENTO X CALINGIN |
DIFFERENTIATING VECTOR FUNCTIONS
7
10/1/2021
DESTACAMENTO X CALINGIN |
DIFFERENTIATING VECTOR FUNCTIONS
8
Review: Common derivatives
𝒚=𝑓 𝑥
constant "C"
𝑘𝑥
ⅆ𝑦
= 𝑓′ 𝑥
ⅆ𝑥
0
𝑛𝑥 𝑛−1
1
𝑥
𝑥
𝑒𝑥
𝑒 𝑎𝑥
10/1/2021
cos 𝑥
sin x
tan x
𝑠𝑒𝑐 2 𝑥
sec x
sec x tan x
cot x
𝑐𝑠𝑐 2 𝑥
csc x
- csc x cot x
ln 𝑥
𝑘
(where “k” is constant)
𝑥𝑛
ⅆ𝑦
= 𝑓′ 𝑥
ⅆ𝑥
1
𝑥
− sin 𝑥
cos 𝑥
𝒚=𝑓 𝑥
−
1
𝑥2
1
2 𝑥
𝑒𝑥
𝑎𝑒 𝑎𝑥
DESTACAMENTO X CALINGIN |
DIFFERENTIATING VECTOR FUNCTIONS
9
Properties
If f(t) and g(t) are differentiable vector functions with respect to t, and A
is differentiable scalar functions of t, then:
C is constant
•
ⅆ
ⅆ𝑡
𝑐𝑓(𝑡) = 𝑐 𝑓 ′ 𝑡
Sum/Difference
Rule
•
ⅆ
ⅆ𝑡
𝑓 𝑡 ±𝑔 𝑡
Product Rule
(Dot)
•
ⅆ
ⅆ𝑡
𝑓 𝑡 ∙𝑔 𝑡
Product Rule
(Cross)
•
ⅆ
ⅆ𝑡
𝑓 𝑡 ×𝑔 𝑡
Product Rule
(Vector and
Scalar Functions)
•
ⅆ
ⅆ𝑡
𝐴𝑓 𝑡
10/1/2021
DESTACAMENTO X CALINGIN |
= 𝑓 ′ 𝑡 ± 𝑔′(𝑡)
= [𝑓 𝑡 ∙ 𝑔′ 𝑡 ] + [𝑔 𝑡 ∙ 𝑓 ′ 𝑡 ]
= [𝑓 𝑡 × 𝑔′ 𝑡 ] + [𝑔 𝑡 × 𝑓 ′ 𝑡 ]
= [𝐴𝑓 ′ 𝑡 ] + [𝐴′𝑓 𝑡 ]
DIFFERENTIATING VECTOR FUNCTIONS
10
Evaluate:
Solution:
ⅆ
𝑟Ԧ
ⅆ𝑡
ⅆ
𝑟Ԧ
ⅆ𝑡
ⅆ
𝑟Ԧ
ⅆ𝑡
𝑡
;
𝑡 =
𝑡 3 𝑖Ƹ
𝑐
𝑘𝑥
𝑘
𝑥𝑛
𝑛𝑥 𝑛−1
𝑓 𝑥
10/1/2021
+
2𝑡 4 𝑗Ƹ
−
+ 2𝑡 4 𝑗Ƹ
−
1
𝑘෠
𝑡
1
෠
𝑘
𝑡
𝑑 3
𝑑 1
𝑑
4
𝑡 = (𝑡 )𝑖Ƹ +
(2𝑡 )𝑗Ƹ − ( )𝑘෠
𝑑𝑡
𝑑𝑡 𝑡
𝑑𝑡
Remember:
ⅆ𝒇(𝒙)
= 𝑓′ 𝑥
ⅆ𝑥
0
𝑟Ԧ t =
𝑡 3 𝑖Ƹ
= 3𝑡 3−1 𝑖Ƹ + 2(4)𝑡 4−1 𝑗Ƹ − −1 𝑡 −1−1 𝑘෠
𝑟′
Ԧ 𝑡
DESTACAMENTO X CALINGIN |
=
3𝑡 2
=
3𝑡 2
3 + 𝑡 −2 𝑘
෠
+
8𝑡
𝑗Ƹ
𝑖Ƹ
𝑖Ƹ +
8𝑡 3 𝑗Ƹ
DIFFERENTIATING VECTOR FUNCTIONS
1
+ 2 𝑘෠
𝑡
11
Evaluate:
ⅆ
ⅆ𝑡
𝑟Ԧ t ∙ 𝑢 t ;
𝑟Ԧ t = cos 𝑡 𝑖Ƹ
𝑟Ԧ ′ 𝑡 = −sin(t)𝑖Ƹ
Solution:
ⅆ
ⅆ𝑡
𝑟Ԧ t ∙ 𝑢 t
=
2 𝑐𝑜𝑠 2
𝑡
𝑢 t = 2 sin 𝑡 𝑖Ƹ + cos 𝑡 𝑗Ƹ
𝑢′ 𝑡 = 2cos t 𝑖Ƹ +(−𝑠𝑖𝑛 𝑡 )𝑗Ƹ
= 𝑟Ԧ t ∙ 𝑢′ t + 𝑟′
Ԧ t ∙𝑢 t
Ƹ
= {cos 𝑡 𝑖Ƹ ∙ (2cos t 𝑖Ƹ + (−𝑠𝑖𝑛 𝑡 )𝑗)}
= 2 𝑐𝑜𝑠 2 𝑡 𝑖Ƹ ∙ 𝑖Ƹ
&
Ƹ
+ {−sin(t)𝑖Ƹ ∙ (2 sin 𝑡 𝑖Ƹ + cos 𝑡 𝑗)}
−2cos t 𝑠𝑖𝑛 𝑡 𝑖Ƹ ∙ 𝑗Ƹ −2sin2 t 𝑖Ƹ ∙ 𝑖Ƹ −(2 𝑐𝑜𝑠(𝑡)sin 𝑡 𝑖Ƹ ∙ 𝑗Ƹ
2
Remember:
−2sin t
= 2(𝑐𝑜𝑠 2 𝑡 − sin2 t )
10/1/2021
DESTACAMENTO X CALINGIN |
DIFFERENTIATING VECTOR FUNCTIONS
𝑓 𝑥
𝑓′ 𝑥
cos 𝑥
− sin 𝑥
sin x
cos 𝑥
12
10/1/2021
DESTACAMENTO X CALINGIN |
DIFFERENTIATING VECTOR FUNCTIONS
13
❑ 𝐫 (𝒕) is position vector at any point in time with components:
✓ 𝐫Ԧ t = x t 𝑖Ƹ + y t 𝐽መ + z t ෡𝑘
➢ Component form: 𝐫Ԧ t = < x t , y t , z t >
❑ The derivative r′(t) of vector rԦ is the velocity vector function
(always tangent to the curve) with components
✓ 𝐫Ԧ ′ (t) = 𝐯 𝐭 = x′ t 𝑖Ƹ + y′ t 𝐽መ + z′ t ෡𝑘
➢ Component form: 𝐯 𝐭 =< x ′ t , y ′ t , z ′ t >
❑ The speed is the magnitude of 𝐯 𝐭
′
✓ ||Ԧ𝐫 t || = | 𝐯 𝐭 | =
10/1/2021
DESTACAMENTO X CALINGIN |
x′ t
DIFFERENTIATING VECTOR FUNCTIONS
2
+ y′ t
2
+ z′ t
2
14
❑ The derivative r′(t) of vector rԦ is the velocity vector function
(always tangent to the curve) with components
✓ 𝐫Ԧ ′ (t) = 𝐯 𝐭 = x′ t 𝑖Ƹ + y′ t 𝐽መ + z′ t ෡𝑘
➢ Component form: 𝐯 𝐭 = < x ′ t , y ′ t , z ′ t >
❑ The second derivative 𝐫Ԧ ′′ (𝒕) of vector rԦ is the acceleration
vector function with components
✓ 𝐫Ԧ′′(t) = 𝑣 ′ 𝑡 = 𝒂 𝐭 = x ′′ t 𝑖Ƹ + y ′′ t 𝐽መ + z ′′ t ෡𝑘
➢ Component form: 𝒂 𝐭 = < x ′′ t , y ′′ t , z ′′ t >
10/1/2021
DESTACAMENTO X CALINGIN |
DIFFERENTIATING VECTOR FUNCTIONS
15
Example
Sketch and find the 𝐯 𝐭 , speed, and 𝒂 𝐭 of the position
vector function 𝒓 𝐭 = t 2 − 4 𝒊Ƹ + 𝒕𝒋Ƹ at t=0 and t=2.
Let’s try to graph the function:
𝒙 = 𝑡2 − 4
(1)
𝒚=𝒕
(2)
Substitute the value of t in (1)
from (2)
time
x
y
0
-4
0
1
-3
1
2
0
2
𝒙 = 𝑦2 − 4
[𝑷𝒂𝒓𝒂𝒃𝒐𝒍𝒂 𝒔𝒊𝒅𝒆𝒘𝒂𝒚𝒔 ]
10/1/2021
DESTACAMENTO X CALINGIN |
DIFFERENTIATING VECTOR FUNCTIONS
16
Example
Sketch and find the 𝐯 𝐭 , speed, and 𝒂 𝐭 of the position
vector function 𝒓 𝐭 = 𝐭 𝟐 − 𝟒 𝒊Ƹ + 𝒕𝒋Ƹ at t=0 and t=2.
Solution:
𝒅
𝒅 𝟐
𝒅
𝒓 𝒕 =
𝐭 − 𝟒 𝒊Ƹ +
(𝒕𝒋)Ƹ
𝒅𝒕
𝒅𝒕
𝒅𝒕
Remember:
𝑓 𝑥
𝑐
ⅆ𝒇(𝒙)
= 𝑓′ 𝑥
ⅆ𝑥
0
𝑘𝑥
𝑘
𝑥𝑛
𝑛𝑥 𝑛−1
10/1/2021
𝒓′
𝒅 𝟐
𝐭 −𝟒
𝒅𝒕
𝒅
(𝒕)
𝒅𝒕
= 𝟐𝐭 𝟐−𝟏 − 𝟎
= 𝟏𝐭 𝟏−𝟏
= 𝒗(𝒕)
=𝟏
= 𝟐𝐭
𝒓′ 𝐭 = 𝟐𝐭 𝒊Ƹ + (𝟏)𝒋Ƹ
DESTACAMENTO X CALINGIN |
𝐭
DIFFERENTIATING VECTOR FUNCTIONS
𝒗 𝒕 = 𝟐𝒕𝒊Ƹ + 𝒋Ƹ
17
Example
Sketch and find the 𝐯 𝐭 , speed, and 𝒂 𝐭 of the position
vector function 𝒓 𝐭 = 𝐭 𝟐 − 𝟒 𝒊Ƹ + 𝒕𝒋Ƹ at t=0 and t=2.
Solution:
𝒗 𝐭 = 2𝑡𝒊Ƹ + 𝒋Ƹ
𝒗 𝐭 =< 2,1,0 >
𝒔𝒑𝒆𝒆𝒅 = | 𝒗 𝐭 |
|𝒗 𝐭 |=
|𝒗 𝐭 |=
10/1/2021
(2𝑡)2 + 1
2
4𝑡 2 + 1
DESTACAMENTO X CALINGIN |
𝒅
(𝒗 𝐭) = 𝒂 (𝒕)
𝒅𝒕
𝑑
𝒂 𝐭 = [2𝑡𝒊Ƹ + 𝒋]Ƹ
𝑑𝑡
𝑑
𝑑
𝒂 𝐭 =
2𝑡𝒊Ƹ + [1 𝒋]Ƹ
𝑑𝑡
𝑑𝑡
𝒂 𝐭 =< 2,0,0 >
𝒂 𝐭 = 2𝒊Ƹ
DIFFERENTIATING VECTOR FUNCTIONS
18
Example
Sketch and find the 𝐯 𝐭 , speed, and 𝒂 𝐭 of the position
vector function 𝒓 𝐭 = 𝐭 𝟐 − 𝟒 𝒊Ƹ + 𝒕𝒋Ƹ at t=0 and t=2.
Solution:
𝒗 𝐭 = 2𝑡𝒊Ƹ + 𝒋Ƹ
𝒗 𝐭 =< 2,1,0 >
At t=0
𝒗 𝐭 = 2(0)𝒊Ƹ + 𝒋Ƹ = 𝒋Ƹ
𝒗 𝐭 = < 0,1,0 >
|𝒗 𝐭 |=
(2𝑡)2 + 1
|𝒗 𝐭 |=
4𝑡 2 + 1
𝒂 𝐭 = 2𝒊Ƹ
𝒂 𝐭 =< 2,0,0 >
10/1/2021
2
𝒗 𝐭
=1
At t=2
𝒗 𝐭 = 2(2)𝒊Ƹ + 𝒋Ƹ
= 4𝒊Ƹ + 𝒋Ƹ
𝒗 𝐭 =< 4,1,0 >
|𝒗 𝐭 |= 5
𝒂 𝐭 = 2𝒊Ƹ = 2𝒊Ƹ
𝒂 𝐭 = 2𝒊Ƹ = 2𝒊Ƹ
𝒂 𝐭 =< 2,0,0 >
𝒂 𝐭 =< 2,0,0 >
DESTACAMENTO X CALINGIN |
DIFFERENTIATING VECTOR FUNCTIONS
19
𝒓 𝐭 = 𝐭 𝟐 − 𝟒 𝒊Ƹ + 𝒕𝒋Ƹ
Click to see the graph: https://www.geogebra.org/m/ubratnur
10/1/2021
DESTACAMENTO X CALINGIN |
DIFFERENTIATING VECTOR FUNCTIONS
20
Remember:
Find the 𝐯 𝐭 , speed, and 𝒂 𝐭 of the position vector
෡
function 𝒓 𝐭 = 𝟒𝐜𝐨 𝐬 𝒕 𝒊Ƹ + 𝟒𝒔𝒊𝒏 𝒕 𝒋Ƹ + 𝒕𝒌.
𝑓 𝑥
𝑓′ 𝑥
cos 𝑥
− sin 𝑥
sin x
cos 𝑥
෡
𝒗 𝐭 = 𝟒(−𝒔𝒊𝒏 𝒕 )𝒊Ƹ + 𝟒(𝒄𝒐𝒔 𝒕 )𝒋Ƹ + 𝒌
෡
𝒗 𝐭 = −𝟒𝒔𝒊𝒏 𝒕 𝒊Ƹ + 𝟒(𝒄𝒐𝒔 𝒕 )𝒋Ƹ + 𝒌
𝒗 𝐭 =< −𝟒𝒔𝒊𝒏 𝒕 , 𝟒 𝒄𝒐𝒔 𝒕 , 𝟏 >
𝟐
+ (𝟒𝒄𝒐𝒔 𝒕
𝟐
+ 𝟏𝟐
|𝒗 𝐭 |=
(−𝟒𝒔𝒊𝒏 𝒕
|𝒗 𝐭 |=
𝟏𝟔𝒔𝒊𝒏𝟐 (𝒕) + 𝟏𝟔𝒄𝒐𝒔𝟐 (𝒕) + 𝟏
𝒂 𝐭 = −𝟒𝒄𝒐𝒔 𝒕 )𝒊Ƹ −𝟒𝒔𝒊𝒏 𝒕 𝒋Ƹ
𝒂 𝐭 =< −𝟒𝒄𝒐𝒔 𝒕 , −𝟒𝒔𝒊𝒏 𝒕 , 𝟎 >
10/1/2021
DESTACAMENTO X CALINGIN |
DIFFERENTIATING VECTOR FUNCTIONS
21
Remember:
Find the 𝐯 𝐭 and 𝒂 𝐭 of the position vector function
𝜋
𝒓 𝐭 = 𝟒𝐜𝐨 𝐬 𝒕 𝒊Ƹ + 𝟑𝒔𝒊𝒏 𝒕 𝒋Ƹ at t = 4 .
𝑓 𝑥
𝑓′ 𝑥
cos 𝑥
− sin 𝑥
sin x
cos 𝑥
𝒗 𝐭 = 𝟒(−𝒔𝒊𝒏 𝒕 )𝒊Ƹ + 𝟑(𝒄𝒐𝒔 𝒕 )𝒋Ƹ
𝒗 𝐭 = −𝟒𝒔𝒊𝒏 𝒕 𝒊Ƹ + 𝟑(𝒄𝒐𝒔 𝒕 )𝒋Ƹ
𝒗 𝐭 =< −𝟒𝒔𝒊𝒏 𝒕 , 𝟑 𝒄𝒐𝒔 𝒕 , 𝟎 >
𝒂 𝐭 = −𝟒𝒄𝒐𝒔 𝒕 )𝒊Ƹ −𝟑𝒔𝒊𝒏 𝒕 𝒋Ƹ
𝒂 𝐭 =< −𝟒𝒄𝒐𝒔 𝒕 , −𝟑𝒔𝒊𝒏 𝒕 , 𝟎 >
10/1/2021
DESTACAMENTO X CALINGIN |
DIFFERENTIATING VECTOR FUNCTIONS
22
Find the 𝐯 𝐭 , speed, and 𝒂 𝐭 of the position vector
𝜋
function 𝒓 𝐭 = 𝟒𝐜𝐨 𝐬 𝒕 𝒊Ƹ + 𝟑𝒔𝒊𝒏 𝒕 𝒋Ƹ at t = 4 .
𝜋
𝒂𝒕 𝐭 =
4
Remember:
cos
sin
𝜋
𝟒
𝜋
𝟒
=
=
2
2
2
2
𝒗 𝐭 = −𝟒𝒔𝒊𝒏 𝒕 𝒊Ƹ + 𝟑(𝒄𝒐𝒔 𝒕 )𝒋Ƹ
𝜋
𝜋
𝒗 𝐭 = −𝟒𝒔𝒊𝒏
+ 𝟑 𝒄𝒐𝒔
4
4
2
2
𝒗 𝐭 = −𝟒
𝒊Ƹ + 𝟑
𝒋Ƹ
2
2
𝟑 2
𝒗 𝐭 = −𝟐 𝟐 𝒊Ƹ +
𝒋Ƹ
2
10/1/2021
DESTACAMENTO X CALINGIN |
DIFFERENTIATING VECTOR FUNCTIONS
𝒂 𝐭 = −𝟒𝒄𝒐𝒔 𝒕 𝒊Ƹ − 𝟑𝒔𝒊𝒏 𝒕 𝒋Ƹ
𝒂 𝐭 = −𝟒𝒄𝒐𝒔
𝒂 𝐭 = −𝟒
2
2
𝜋
4
𝒊Ƹ − 𝟑𝒔𝒊𝒏
𝒊Ƹ − 𝟑
𝒂 𝐭 = −𝟐 𝟐 𝒊Ƹ −
2
2
𝒋Ƹ
𝟑 𝟐
𝒋
𝟐
23
𝜋
4
𝒋Ƹ