Calculus 3 and calculus 4, Cheat Sheet of Calculus

Vector calculus and parametrizations

Typology: Cheat Sheet

2024/2025

Uploaded on 03/05/2025

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Parametrize the Intersection Example

eq .1 y

2

z

2

= x − 2 ∧ eq .2 y

2

  • z

2

  1. Solve for z:

y

2

  • z

2

= 9 =¿ z =√ 9 − y

2

  1. Sub z into eq 1.

y

2

9 − y

2

= x − 2

, then set y=t and solve for x

x = 2 t

2

, then find components to solve for intersection

x ( t )= 2 t

2

− 7 , y ( t )= t , z ( t )=

9 − t

2

Vector parametrization with sphere

r ⃗ ( t )=¿ 3 , 1 , − 4 >+ t ← 2 , − 2

( x − 1 )

2

+( y + 3 )

2

  • z

2

  1. x ( t )= 3 − 2 t , y ( t )= 1 − 2 t

( 3 − 2 t )− 1

2

( 1 − 2 t ) + 3

( 2 − 2 t )

2

+( 4 − 2 t )

2

+(− 4 + 3 t

Parallel Vectors ⃗

AB ∧
PQ

λ

V

x

W =

V x

λ

W

=¿ λ x

V
W

when A=(1,1) B=(3,4) P=(1,1) Q=(7,10)

  1. Find
AB ∧

PQ vectors

AB =

~ square root trick

27 s + 8 =( 4 + 9 t

2

3

2

( 27 s + 8 )

2

3

= 4 + 9 t

2

Volume of parallelopiped

Volume =| u ⃗ ∙ (

V x

W )|=¿

Points of intersection from graph

and equation in xy plane example

r ( t )=⟨ sin ( t ) , cos ( t ) , sin ( t ) cos ( 2 t )

  1. Components:

x ( t )=sin ( t ) , y ( t )=cos ( t ) , z ( t )=si

  1. Set z=0 and solve for two different values of t

0 =sin ( t )=¿ t = ∧ 0 =cos ( 2 t )=

Simplify arc length

r

'

( t )‖=

t

t

2

( 2 t + 1 )

2

t

2

2 t + 1

t

Simplifying during

arc length

r

'

( t )‖=

( 2 t + 1 )

2

t

2

**Arc length parametrization always 0 to t

Example:

r

t

=⟨ cos

4 t

, sin

4 t

, 3 t

  1. First derivative, then plug into integral

0

t

(− 4 sin ( 4 t ) )

2

+( 4 cos ( 4 t ) )

2

2

dt =¿

0

t

  1. Inverse:

s = 5 t =¿ t =

s

and plug inverse

x = rcosθ = h + rcos

s

r

y = rsinθ = k + rsin

s

r

,

z = z

** if given speed,

A)

4 x

2

  • 4 y

2

  • z

2

B)

Find components from

P=(3,2) and Q=(2,7)

PQ =⟨ 2 − 3 , 7 − 2 ⟩ =⟨− 1 , 5 ⟩ Determin

e Vectors Equivalent

AB

PQ?

A=(1,1) B=(3,7) P=(4,-1) Q=(6,5)

AB =⟨ 3 − 1 , 7 − 1 ⟩ =⟨ 2 , 6 ⟩

PQ =

**

Yes equivalent b/c same

Vector Parametrization Perpendicular to Plane Rules

  1. Perpendicular to xy plane = changes to z only
  2. Perpendicular to xz plane = changes to y only
  3. Perpendicular to yz plane = changes to x only

Show two vectors define same line:

Parametrize the Intersection Example

eq .1 y

2

z

2

= x − 2 ∧ eq .2 y

2

  • z

2

  1. Solve for z:

y

2

  • z

2

= 9 =¿ z =

9 − y

2

  1. Sub z into eq 1.

y

2

−( 9 − y

2

)= x − 2

, then set

y=t and solve for x

Two Different Parametrizations Through Same

Line

Example: (no intersection)

r

1

( t )=⟨ 1 , 1 , 0 ⟩ + t ⟨− 2 , 1 , 3 ⟩

r

2

( t )=⟨− 3 , 3 , 6 ⟩ + t ⟨ 4 , − 2 ,

  1. Set r 0 from r 1 equal to r 2

⟨ 1 , 1 , 0 ⟩ =⟨− 3 , 3 , 6 ⟩ + t ⟨ 4 , −

  1. Parametric equations of components

x ( t )=¿ 1 =− 3 + 4 t =¿ t =

Time for meteor to hit ground in xy from

parametric:

r ⃗ ( t )=¿ 2 , 1 , 4 >+ t < 3 , 2 , − 1 >¿