DERIVATIVE OF A VECTOR FUNCTION
𝐫(𝑡 + ℎ) − 𝐫(𝑡)
ℎ→𝑜
ℎ
𝐫 ′ (𝑡) = lim
provided this limit exists
Theorem: If 𝐫(𝑡) = ⟨𝑓(𝑡), 𝑔(𝑡), ℎ(𝑡)⟩ then 𝐫 ′ (𝑡) exists if an only if all three component
derivatives exist in which case 𝐫′(𝑡) = ⟨𝑓′(𝑡), 𝑔′(𝑡), ℎ′(𝑡)⟩
𝐫 ′ (𝑡) =
𝐍𝐨𝐭𝐚𝐭𝐢𝐨𝐧:
𝑑
𝑑𝐫
𝐫(𝑡) =
𝑑𝑡
𝑑𝑡
Derivative Rules
1.
𝑑
(𝐜) = 𝟎
𝑑𝑡
2.
𝑑
(𝐮(𝑡) + 𝐯(𝑡)) = 𝐮′ (t) + 𝐯 ′ (𝑡)
𝑑𝑡
3.
𝑑
(𝑓(𝑡)𝐮(𝑡)) = 𝑓′(𝑡)𝐮(t) + 𝑓(𝑡)𝐯 ′ (𝑡)
𝑑𝑡
4.
𝑑
(𝐮(𝑓(𝑡))) = 𝐮′(𝑓(𝑡))𝑓′(𝑡)
𝑑𝑡
5.
𝑑
(𝐮(𝑡) ∙ 𝐯(𝑡)) = 𝐮′ (𝑡) ∙ 𝐯(𝑡) + 𝐮(𝑡) ∙ 𝐯 ′ (𝑡)
𝑑𝑡
6.
𝑑
(𝐮(𝑡) × 𝐯(𝑡)) = 𝐮′ (𝑡) × 𝐯(𝑡) + 𝐮(𝑡) × 𝐯 ′ (𝑡)
𝑑𝑡
sum rule
product rule
chain rule
dot product rule
cross product rule
Integration of vector functions
Suppose 𝑓(𝑡), 𝑔(𝑡), ℎ(𝑡) have respective antiderivatives are 𝐹(𝑡), 𝐺(𝑡), and 𝐻(𝑡).
∫ 𝑓(𝑡)𝐢 + 𝑔(𝑡)𝐣 + ℎ(𝑡)𝐤 𝑑𝑡
=
𝑏
(𝐹(𝑡) + 𝑐1 )𝐢 + (𝐺(𝑡) + 𝑐2 )𝐣 + (𝐻(𝑡) + 𝑐3 )𝐤
(indefinite integral)
∫ 𝑓(𝑡)𝐢 + 𝑔(𝑡)𝐣 + ℎ(𝑡)𝐤 𝑑𝑡
𝑎
𝑏
𝑏
𝑏
= (∫ 𝑓(𝑡)𝑑𝑡 ) 𝐢 + (∫ 𝑔(𝑡)𝑑𝑡 ) 𝐣 + (∫ ℎ(𝑡)𝑑𝑡 ) 𝐤
𝑎
𝑎
𝑎
(definite integral)
Definition: Tangent Vector and Unit Tangent Vector.
zero vector:
Provided 𝐫 ′ (𝑡) exists and is not the
tangent vector: 𝐫′(𝑡)
𝐮𝐧𝐢𝐭 𝐭𝐚𝐧𝐠𝐞𝐧𝐭 𝐯𝐞𝐜𝐭𝐨𝐫: 𝐓(𝑡) =
𝐫′(𝑡)
|𝐫 ′ (𝑡)|
1
Example 1: For 𝐫(𝑡) = ⟨ 5𝑒 5𝑡 , sec 2 𝑡 , 𝑡 ⟩ find 𝐫 ′ (𝑡)
𝐫 ′ (𝑡) =
𝑑 5𝑡
1
⟨𝑒 , tan 𝑡 , ln 𝑡⟩ = ⟨ 5𝑒 5𝑡 , sec 2 𝑡 , ⟩
𝑑𝑡
𝑡
Example 2: For 𝐫(𝑡) = ⟨𝑒 5𝑡 , 6𝑡 2 , cos 2𝑡⟩ find ∫ 𝐫(𝑡) 𝑑𝑡
1
1
∫ 𝐫(𝑡) 𝑑𝑡 = ∫⟨𝑒 5𝑡 , 6𝑡 2 , cos 2𝑡⟩ 𝑑𝑡 = ⟨ 𝑒 5𝑡 , 2𝑡 3 , sin 2𝑡 ⟩ + ⟨𝑐1 , 𝑐2 , 𝑐3 ⟩
5
2
Example 3: Find the unit tangent vector of ⟨𝑡, cos 2𝑡 , 2sin 𝑡 ⟩ at 𝑡 = 0 and 𝑡 = 𝜋/2
Solution:
𝑑
⟨𝑡, cos 2𝑡 , 2sin 𝑡 ⟩ = ⟨1, −2 sin 2𝑡 , 2 cos 𝑡⟩
𝑑𝑡
𝐓(𝑡) =
𝐓(0) =
⟨1,0,2⟩
√5
=⟨
1
√5
⟨1, −2 sin 2𝑡 , 2 cos 𝑡⟩
√12 + (−2 sin 2𝑡)2 + (2 cos 𝑡)2
,0,
2
√5
⟩
𝐓(𝜋/2) =
⟨1,0,0⟩
= ⟨ 1, 0 , 0 ⟩
1
Example 4a:
Let
2
𝐫(𝑡) = ⟨
, 8 − 3𝑡, cos 2(𝜋𝑡) ⟩
2
√5 − 𝑡
Find parametric equations of line that is tangent to the curve corresponding to 𝑡 = 1.
Solution:
𝐫′(𝑡) = ⟨
𝐫 ′ (1)
=
⟨
2𝑡
, −3, −2𝜋 cos(𝜋𝑡) sin(𝜋𝑡) ⟩
(5 − 𝑡 2 )3/2
2
, −3, −2𝜋 cos(𝜋) sin(𝜋) ⟩
43/2
𝐫(1) = ⟨
The tangent line is
2
√5 −
12
1
⟨ , −3,0⟩
4
=
, 8 − 3(1), cos 2 (𝜋) ⟩
=
=
tangent vector
⟨ 1,5,1⟩
1
𝐫(1) + 𝑡𝐫 ′ (1) = ⟨ 1,5,1⟩ + 𝑡 ⟨ 4 , −3,0⟩
Parametric equations:
1
𝑥 =1+ 𝑡
4
Example 4b:
𝑦 = 5 − 3𝑡
𝑧=1
Let
2
𝐫(𝑡) = ⟨
, 8 − 3𝑡, cos 2(𝜋𝑡) ⟩
2
√5 − 𝑡
Find parametric equations of line that is tangent to the curve at the point (2,14,1)
Solution: We first need to find the value of 𝑡 that gives the point (2,14,0).
Using the 𝑦
coordinate of 𝐫(𝑡) makes it easy: 8 − 3𝑡 = 14 requires 𝑡 = −2. Next, find derivative:
𝐫′(𝑡) = ⟨
2𝑡
, −3, −2𝜋 cos(𝜋𝑡) sin(𝜋𝑡) ⟩
(5 − 𝑡 2 )3/2
so that
𝐫 ′ (−2) = ⟨
2(−2)
, −3, −2𝜋 cos(−2𝜋) sin(−2𝜋) ⟩ = ⟨−4, −3,0⟩ =
13/2
𝐫(−2) = ⟨ 2,14,1⟩ = vector terminating at point on line
tangent vector
The tangent line is
𝐫(−2) + 𝑡𝐫 ′ (−2) = ⟨ 2,14,1⟩ + 𝑡⟨−4, −3,0⟩
Parametric equations:
𝑥 = 2 − 4𝑡
Example 5:
𝑦 = 14 − 3𝑡
𝑧=1
Suppose
𝐫 ′ (𝑡) = derivative = 5𝐢 +
and 𝐫(2) = ⟨1,2,3⟩.
Compute 𝐫(𝑡)
1
1 2
2
⟨5,
𝐣
+
𝑡
𝐤
=
,𝑡 ⟩
𝑡2
𝑡2
Solution:
1
1
𝐫(𝑡) = (5𝑡 + 𝑐1 )𝐢 + (− + 𝑐2 ) 𝐣 + ( 𝑡 3 + 𝑐3 ) 𝐤
𝑡
3
To get 𝑐1 , 𝑐2 , 𝑐3 use the condition 𝐫(2) = ⟨1,2,3⟩.
1
8
𝐫(1) = ( 10
+ 𝑐2 ) 𝐣 + ( + 𝑐3 ) 𝐤
⏟ + 𝑐1 )𝐢 + (−
⏟2
3⏟
1
5
1
2
3
𝑐1 = −9, 𝑐2 = 2 , 𝑐3 = 3
1 5
1
1
𝐫(𝑡) = (5𝑡 − 9)𝐢 + (− + ) 𝐣 + ( 𝑡 3 + ) 𝐤
𝑡 2
3
3