EN: Calculus variant 2 (2WBB0) is a basis course of the Bachelor College at Eindhoven University of Technology. This means that all Bachelor TUe students should have completed one of the variants of this course. It is given in the first quartile of the first year. Calculus variant 2 is the average ...
Calculus Cheat sheet
Disclaimer: this is not actually meant for cheating. Research shows that making a “cheat
sheet” right before a test with all hard topics and then not actually using it during the test, is
an effective way to practice. This was the “cheat sheet” I made for Calculus variant 2 (2WBB0)
and I hope it can prove to be useful to you.
a d
() ()
b and e are perpendicular if a × d+ b ×e +c ×f =0.
c f
x x
Line l through a
⃗ and b⃗ with both of the form
() ()
y is: y =⃗a + λ ×( ⃗b−⃗a ).
z z
x x
⃗ , e⃗ ,
A plane through 3 vectors d
z () ()
⃗f of the form y is: y = ⃗d+ λ × ( d⃗ −⃗e )+ μ ×( ⃗d−⃗f ).
z
k
⃗ ⃗ ⃗
The normal vector of this is n=( d−⃗e ) × ( d− f )= l .
m ()
So the equation that follows from this is k × x +l × y +m × z=d . d is acquired after substituting d
⃗ into
the equation.
'
(a): 1) Use f ( b )=a to calculate b. 2) Calculate f ' (x).
Calculating ( f −1 )
−1 ' 1
3) Use ( f ) ( a )= to solve the equation.
f ' (b)
For the domain of f-1, determine the range of f by calculating lim f (x ) and lim f (x) .
x →e x→ ∞
A Taylor polynomial is defined as follows:
f ' ( a) f '' (a) 2
f n( a) n
Pn ( x )=f ( a )+ ( x−a )+ ( x −a ) +…+ ( x−a ) .
1! 2! n!
A point is a global maximum of f in x = a if f is decreasing when x ≥ a (for f ’(x) ≤ 0) and f is increasing
when x ≤ a (for f ‘(x) ≥ 0).
The linearization of f is f ( x )=f ( a ) + f ' ( a ) ( x −a ) at x = a.
B
When you have ∫ A dx =B+ c ∫ A dx , do: ( 1−c )∫ A dx=B and then ∫ A dx = 1−c .
dy g( x) dy g( x)
f ( t ) dt=f ( g ( x ) ) × g '( x ) and f ( t ) dt=f ( g ( x ) ) × g' ( x ) −f ( h ( x ) ) ×h' ( x ) .
dx a dx h(x)
f ( x )= A for x< c (with A and B both consisting of x’s and c’s) is
Determining all c for which { B for x ≥ c
continuous can be done as follows:
1) Determine c. 2) Use f(x) which nears x = c from left and right: lim f (x ) and lim f (x ).
x ↓c x ↑c
3) Then use lim f (x )=lim f ( x ) to find c (since that is the condition to being continuous.
x ↓c x ↑c
To be differentiable means to be continuous.
dy
The initial value-problem with =ky and y ( 0 )= y 0 has the unique solution y= y 0 × e kt.
dt
An integral equal to ∞ is divergent. An integral that is a number, e.g. 15 is convergent.
By Isabel Rutten
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