Calc 3 Unit 13 - 15 Notations

Question 1

Explain how to go about solving for the Two (𝓾 & 𝓿) Vectors and their components:

4𝓾 + 6𝓿 = 𝓲 , 𝓾 - 𝓿 = 𝓳

Answer 1

Start by knowing that 𝓲 = <1,0> and 𝓳 = <0,1>, we can the set up a matrix using the two equations:

4𝓾 + 6𝓿 = < 1,0>
𝓾 - 𝓿 = <0,1>

Then solve for the value of (𝓾 & 𝓿)’s vectors using linear reduced row echelon.

Question 2

An ant walks due east at a constant speed of 2mi/hr on a sheet of paper that rests on a table. Suddenly the sheet of paper starts moving southeast at √(2)mi/hr. Describe the motion of the insect relative to the table.

How should this equation be set up ?

Answer 2

We know that the vector for the ant’s movement relative to the stationary table is a sum of all movement in that system. so we can take the equation:

𝓥(table) = 𝓥(antpaper) + 𝓥(P)

since the paper moves SouthEast:
𝓥(P) = √(2) ⋅ < √(2)/2 , -√(2)/2 >mi/hr = < 1, -1 >mi/hr
𝓥(antpaper) = < 2,0 >mi/hr

𝓥(table) = < 2,0 >mi/hr + < 1, -1 >mi/hr
𝓥(table) = <3 , -1>mi/hr

Question 3

How to find the projection of Vector 𝓤 onto 𝓥 ?

projᵥ𝓤

Answer 3

projᵥ𝓤 = |𝓤|cosθ[𝓥 / |𝓥|]

or

(𝓤⋅𝓥 / 𝓥⋅𝓥)𝓥

Question 4

How to find Scale of Vector 𝓤 onto 𝓥 ?

scalᵥ𝓤

Answer 4

scalᵥ𝓤 = |𝓤|cosθ = 𝓤⋅𝓥 / |𝓥|

Question 5

How to find Parallelogram using Cross Product ?

Answer 5

Find the Cross Product of the two vectors which form the parallelogram then solve for its magnitude

Question 6

What is the Vector Equation and Parametric Equations of a line passing through (0,0,5) in the direction 𝓥 = <6,-5,0>

Answer 6

Note that the Vector Equation is

𝓻 = 𝓻₀ + t𝓥

Vector Equation: 𝓻 = <0,0,5> + t<6,-5,0>

Parametric Equations:
𝔁 = 6t
𝔂 = -5t
𝔃 = 5

Question 7

How do we solve for the Normal and Parallel Forces created

Where:
θ = 30º
F = 10N

Answer 7

We start by identifying the Parallel force as being in the same direction as gravitational force projected onto the slope.
F = <0,-10> given its direction
𝓥 = <cos(-30),sin(-30)> = <√3/2 , -1/2>

We can find out projᵥF using:
projᵥF = (F⋅𝓥 / 𝓥⋅𝓥)𝓥

projᵥF = (<0,-10> ⋅ <√3/2, -1/2>)/1 * <√3/2, -1/2>

= 5 * <√3/2, -1/2> =

projᵥF = <5√3/2 , -5/2>

Now we can find the Normal component using the equation: F = N + projᵥF

N = F - projᵥF → N = <0,-10> - <5√3/2 , -5/2>
N = <-5√3/2, -15/2>

Question 8

How is Torque of an object found:

Answer 8

Take the cross product of the Location the Force is applied (wrt the screw) by the Force applied

Շ = 𝓻 X 𝙁

When applying Right Hand Orientation:
𝙁 = y component
𝓻 = x component
Շ = Resultant component

Question 9

Explain how to find parametric equations for the line segment joining the first point to the second point.

<-4,-9,5> and <-7,5,-3>

Answer 9

We start by finding 𝓥 the distance between our first point and our second point

𝓥 = <-3,14,-8>

Now our Vector Equation will be:
<x,y,z> = <-4,-9,5> + t<-3,14,-8>

We can now define our Parametric Equations using basic vector algebra:

x = -4 - 3t
y = -9 +14t
z = 5 - 8t

Question 10

How to find the point of intersection of two parametric equations defined as:

𝓻 = <2,4,1> + t<6,-6,1>

𝓡 = <18,-4,13> + s<4,-2,3>

Answer 10

Start by converting our Vector Forms into parametric form by performing basic vector addition

𝓻 = {2 + 6t, 4 - 6t, 1 + t}
𝓡 = {18 + 4s, -4 -2s, 13 +3s}

All we need to do now is solve for t and s by setting each of these equations equal to each-other:

x; 2 + 6t = 18 + 4s
y; 4 - 6t = -4 -2s
z; 1 + t = 13 +3s

We can then plug the found values into each equation to solve for the interception for <x,y,z>

Question 11

What is the Equation of a Plane ?

Answer 11

a(x - x₀) + b(y - y₀) + c(z - z₀) = 0

or ax+ by + cz = d

Question 12

How to find to the Plane passing through P₀<2,-3,4> with a Normal Vector of
𝓷 = <-1,2,3>

Answer 12

We start with are Equation to the Plane: a(x - x₀) + b(y - y₀) + c(z - z₀) = 0

Then we use our Normal Vector to describe our a,b,c values and P₀ for our x₀,y₀,z₀ values

in Equation form: -1(x - 2) + 2(y + 3) + 3(z - 4) = 0 then we must convert to Standard Form:

-x + 2 + 2y + 6 + 3z - 12 = 0

-x + 2y + 3z = 4

Question 13

How to find Equation of plane that passes through the points:

P<2,-1,3>
Q<1,4,0>
R<0,-1,5>

Answer 13

We start by finding the two vectors which have P as the origin:

PQ = <-1,5,-3> & PR = <-2,0,2>

Then we will take the cross product of the two vectors to find the Normal Vector

PQ X PR = <10,8,10>

We can now create the Equation of the Plane using P as our P₀ and our new Normal Vector as our 𝓷

10(x - 2) + 8(y + 1) + 10(z - 3) = 0 → 10x - 20 + 8y + 8 + 10z - 30 = 0

Equation of Plane: 10x + 8y + 10z = 42

Question 14

How to find Vector Normal to plane: 2x - 3y - z = 6

Answer 14

𝓷 = <2,-3,-1>

Question 15

How to find the points where Q intersects with each coordinate Axis ?

Lets do for 2x - 3y - z = 6

Answer 15

For each axis, set the other two values equal to 0

x-axis: y=z=0, x = 3
y-axis x=z=0, y = -2
z-axis x=y=0, z = -6

Question 16

How do the equation that 2x - 3y - z = 6 intercepts each plane ?

Answer 16

We set the missing value of each plane equal to 0:

x-y plane: z = 0, 2x - 3y = 6
x-z plane: y = 0, 2x - z = 6
y-z plane: x = 0, -3y - z = 6

Question 17

How to find the Unit Tangent Vector T(t) given a Vector 𝖗(t)

Answer 17

T(t) = 𝖗’(t)/|𝖗’(t)|

Take the derivative of 𝖗(t), then the magnitude of 𝖗’(t). then divide them by eachother such as a standard Unit Vector.

Question 18

Take the derivative of 𝓥(t²)

if 𝓥(t) = <sint, 2cost, cost>

Answer 18

This follows the standard chain rule.

d/dt[𝓥(t²)] = 𝓥’(t²) ⋅ 2t

= < cos(t²), -2sin(t²), -sin(t²) > ⋅ 2t

d/dt[𝓥(t²)] = **< 2tcos(t²), -4tsin(t²), -2tsin(t²) > **

Question 19

How does integration work for multivariable’d functions ?

Answer 19

Take the integral of each component then solve for C at each given point:

𝖗(0) = <solved at p₀> + C = < values of p₀ >

Question 20

How to verify if a given “circular” Position and Velocity are truly Circular

Answer 20

Take the derivative of 𝖗(t) for 𝓥(t) then take the dot product of the two functions.

if 𝖗 ⋅ 𝓥 = 0; they’re circular.

Question 21

What is the vector function of acceleration, velocity and position in 2D projectile motion

Answer 21

a(t) = < 0 , -g >; where (g) is gravitational acceleration
v(t) = < u₀ , v₀ - gt >
r(t) = < u₀t + x₀ , v₀ - ½gt + y₀ >

Question 22

What is the equation to Arc-Length of a vector valued function

Answer 22

L = ∫ √( ⨍’(t) + 𝑔’(t) + 𝒽’(t) ) dt or √r’(t)dt

Question 23

What is the equation to the Curvature ᴋ
and how to solve for it

Answer 23

ᴋ = |[T’(t) /|𝓥(t)|] |

or ᴋ [𝓥 × a] / |𝓥|³

the absolute value of the Unit Tangent Vector |T’(t)| divided by the Magnitude of the Velocity Vector |𝓥(t)|

Question 24

How to find Principal Unit Normal Vector to a curve N given r(t)

Answer 24

N = T’(t) / |T’(t)|

Start by finding T(t) = r’(t) / |r’(t)|, then take the derivative of T(t).

Question 25

How to find the Normal Acceleration Vector (a_N) ?

Answer 25

a_N = [𝓥 × a] / |𝓥|

Question 26

How to find the Tangent Acceleration Vector (a_T) ?

Answer 26

a_T = a ⋅T

where T is the Tangent Unit Vector T(t)

Question 27

How to find The Unit Binormal Vector 𝐵 ?

Answer 27

𝐵 = T × N

or 𝐵 = [𝓥 × a] / |𝓥 × a|

Question 28

How to find the Torsion τ of an object ?

Answer 28

τ = 𝐵’ ⋅ N

Where:

𝐵’ is the derivative of the Binormal Vector:
d/dt[𝐵] = d/dt [ [𝓥 × a] / |𝓥 × a|]

and N is the Unit Normal Vector
N = |[T’(t) / |T’(t)|]|

Question 29

How to find the domain of 𝑔(x,y) = √(4 - x² - y²)

What shape does this domain create ?

Answer 29

Since a value in the radical cannot be negative we must set the inner function greater than 0, then solve for (x² + y²)

4 - x² - y² ≥ 0 → x² + y² ≤ 4

This domain is (on the xy-plane) a circle with radius of 2

Question 30

How to find the

limit (x,y) → (2,8): 3x²y + √(xy)

Answer 30

For this type of limit all we need to do is plug in the values.