WI1421LR Calculus I
Calculus I
WI1421LR
Delft University of Technology
1
, WI1421LR Calculus I
Table of Contents
Vectors……………………………………….………………………………………………………………………………………………………3
(Inverse) Trigonometric Functions……………………………..……………………………………………………………………….3
Implicit Differentiation (Tangent Lines)……………………….……………………………………………………………………..4
Limits…………………….…………………………………………………………………………………………………………..............……4
Linearization……………………………………………………………………………………………………………………………………….5
The Chain Rule……………………………………………………………………………………………………………………………………5
Integration………………………………………………………………………………………………………………………………………….5
- Integration by parts……………………………………………………………………………………………………………….6
- The substitution rule………………………………………………………………………………………………………………6
Infinite Intervals………………………………………………………………………………………………………………………………….7
Differential Equations…………………………………………………………………………………………………………………………7
Tank & Concentrations (Differential Equations)………………………………………………………………………………….8
Complex Numbers………………………………………………………………………………………………………………………………9
2
, WI1421LR Calculus I
Vectors
𝑎 = (𝑎1 , 𝑎2 , 𝑎3 ), 𝑏 = (𝑏1 , 𝑏2 , 𝑏3 ), 𝑐 = (𝑐1 , 𝑐2 , 𝑐3 )
𝑎 × 𝑏 = (𝑎2 𝑏3 − 𝑎3 𝑏2 , 𝑎3 𝑏1 − 𝑎1 𝑏3 , 𝑎1 𝑏2 − 𝑎2 𝑏1 )
|𝑎 × 𝑏| = |𝑎||𝑏| sin 𝜃
𝑎 ∙ 𝑏 = 𝑎1 𝑏1 + 𝑎2 𝑏2 + 𝑎3 𝑏3 = |𝑎||𝑏| cos 𝜃
𝑎 − 𝑏 = (𝑎1 − 𝑏1 , 𝑎2 − 𝑏2 , 𝑎3 − 𝑏3 )
𝑎 + 𝑏 = (𝑎1 + 𝑏1 , 𝑎2 + 𝑏2 , 𝑎3 + 𝑏3 )
1
Area of triangle formed by 𝑎, 𝑏 & 𝑐 = 2 |(𝑏 − 𝑎) × (𝑐 − 𝑎)|
Area of parallelogram spanned by vectors 𝑎, 𝑏 & 𝑐 = |(𝑏 − 𝑎) × (𝑐 − 𝑎)|
Example:
If 𝐴 = (−2, 3, 1), 𝐵 = (1, 0, −2) & 𝐶 = (2, −3, 1). What is the area of triangle ABC?
1
Area of triangle formed by 𝐴, 𝐵 & 𝐶 = 2 |(𝐵 − 𝐴) × (𝐶 − 𝐴)|
𝐵 − 𝐴 = (3, −3, −3) & 𝐶 − 𝐴 = (4, −6, 0)
1
Area of triangle formed by 𝐴, 𝐵 & 𝐶 = 2 |(−18, −12, −6)| = 3|(3, 2, 1)| = 3√14
(Inverse) Trigonometric Functions
Keep in mind the following equations, these will also be provided on
the formula sheet:
sin(2𝑎) = 2 sin(𝑎) cos(𝑎)
cos(2𝑎) = 2 cos2 (𝑎) − 1 = 1 − 2 sin2 (𝑎) = cos2 (𝑎) − sin2 (𝑎)
If we get a difficult expression to rewrite with inverse trigonometric
functions and normal trigonometric functions the easiest thing to do
is to make a small sketch and just take an angle.
Example:
What is the expression cos(2 arctan(𝑥)) equal to?
State 𝑦 = arctan(𝑥), thus 𝑥 = tan(𝑦)
This gives: cos(2 arctan(𝑥)) = cos 2(arctan(𝑥)) − sin2 (arctan(𝑥))
2
If we look at the triangle on the right, this is the situation described in the √𝑥 + 1
question. 𝑥
1
1 2 1
This gives: cos2(arctan(𝑥)) = ( ) = 𝑥 2 +1
√𝑥 2 +1
𝑥 2 𝑥2
This also gives: sin2 (arctan(𝑥)) = ( ) = 𝑥 2 +1
√𝑥 2 +1
1 𝑥2 1−𝑥 2
Therefore: cos(2 arctan(𝑥)) = cos2 (arctan(𝑥)) − sin2 (arctan(𝑥)) = 𝑥 2 +1 − 𝑥 2 +1 = 𝑥 2 +1
3
Calculus I
WI1421LR
Delft University of Technology
1
, WI1421LR Calculus I
Table of Contents
Vectors……………………………………….………………………………………………………………………………………………………3
(Inverse) Trigonometric Functions……………………………..……………………………………………………………………….3
Implicit Differentiation (Tangent Lines)……………………….……………………………………………………………………..4
Limits…………………….…………………………………………………………………………………………………………..............……4
Linearization……………………………………………………………………………………………………………………………………….5
The Chain Rule……………………………………………………………………………………………………………………………………5
Integration………………………………………………………………………………………………………………………………………….5
- Integration by parts……………………………………………………………………………………………………………….6
- The substitution rule………………………………………………………………………………………………………………6
Infinite Intervals………………………………………………………………………………………………………………………………….7
Differential Equations…………………………………………………………………………………………………………………………7
Tank & Concentrations (Differential Equations)………………………………………………………………………………….8
Complex Numbers………………………………………………………………………………………………………………………………9
2
, WI1421LR Calculus I
Vectors
𝑎 = (𝑎1 , 𝑎2 , 𝑎3 ), 𝑏 = (𝑏1 , 𝑏2 , 𝑏3 ), 𝑐 = (𝑐1 , 𝑐2 , 𝑐3 )
𝑎 × 𝑏 = (𝑎2 𝑏3 − 𝑎3 𝑏2 , 𝑎3 𝑏1 − 𝑎1 𝑏3 , 𝑎1 𝑏2 − 𝑎2 𝑏1 )
|𝑎 × 𝑏| = |𝑎||𝑏| sin 𝜃
𝑎 ∙ 𝑏 = 𝑎1 𝑏1 + 𝑎2 𝑏2 + 𝑎3 𝑏3 = |𝑎||𝑏| cos 𝜃
𝑎 − 𝑏 = (𝑎1 − 𝑏1 , 𝑎2 − 𝑏2 , 𝑎3 − 𝑏3 )
𝑎 + 𝑏 = (𝑎1 + 𝑏1 , 𝑎2 + 𝑏2 , 𝑎3 + 𝑏3 )
1
Area of triangle formed by 𝑎, 𝑏 & 𝑐 = 2 |(𝑏 − 𝑎) × (𝑐 − 𝑎)|
Area of parallelogram spanned by vectors 𝑎, 𝑏 & 𝑐 = |(𝑏 − 𝑎) × (𝑐 − 𝑎)|
Example:
If 𝐴 = (−2, 3, 1), 𝐵 = (1, 0, −2) & 𝐶 = (2, −3, 1). What is the area of triangle ABC?
1
Area of triangle formed by 𝐴, 𝐵 & 𝐶 = 2 |(𝐵 − 𝐴) × (𝐶 − 𝐴)|
𝐵 − 𝐴 = (3, −3, −3) & 𝐶 − 𝐴 = (4, −6, 0)
1
Area of triangle formed by 𝐴, 𝐵 & 𝐶 = 2 |(−18, −12, −6)| = 3|(3, 2, 1)| = 3√14
(Inverse) Trigonometric Functions
Keep in mind the following equations, these will also be provided on
the formula sheet:
sin(2𝑎) = 2 sin(𝑎) cos(𝑎)
cos(2𝑎) = 2 cos2 (𝑎) − 1 = 1 − 2 sin2 (𝑎) = cos2 (𝑎) − sin2 (𝑎)
If we get a difficult expression to rewrite with inverse trigonometric
functions and normal trigonometric functions the easiest thing to do
is to make a small sketch and just take an angle.
Example:
What is the expression cos(2 arctan(𝑥)) equal to?
State 𝑦 = arctan(𝑥), thus 𝑥 = tan(𝑦)
This gives: cos(2 arctan(𝑥)) = cos 2(arctan(𝑥)) − sin2 (arctan(𝑥))
2
If we look at the triangle on the right, this is the situation described in the √𝑥 + 1
question. 𝑥
1
1 2 1
This gives: cos2(arctan(𝑥)) = ( ) = 𝑥 2 +1
√𝑥 2 +1
𝑥 2 𝑥2
This also gives: sin2 (arctan(𝑥)) = ( ) = 𝑥 2 +1
√𝑥 2 +1
1 𝑥2 1−𝑥 2
Therefore: cos(2 arctan(𝑥)) = cos2 (arctan(𝑥)) − sin2 (arctan(𝑥)) = 𝑥 2 +1 − 𝑥 2 +1 = 𝑥 2 +1
3
