14.1 Vector-Valued Functions

A vector-valued function is of the form r(t)=x(t)i+y(t)j

Each value of t returns a vector that points from the origin to (x(t), y(t)). This can also be done in three dimensions (just add z(t)k). For all functional purposes, they work like parametric functions. Don't forget to draw arrows for orientation! (That's the direction where t is increasing).

You can graph parametric functions by eliminating the parameter or making a table of values (or using a computer). For functions in 3-space, drawing a rectangular prism makes graphing easier.

Example: r(t)=<t, -t^2+5>

x=t, y=-t^2+5

y=-x^2+5

r(t)=<2cos(t), 2sin(t)>, 0=<t=<2π

x=2cos(t), y=2sin(t)

x^2=4cos^2(t), y^2=4sin^2(t)

x^2+y^2=4

The domain of a vector-valued function is the intersection of the domain of each component.

Circular Helix: r(t)=<acos(t), asin(t), ct> These are not fun to draw.

13.6 - Planes in 3-Space

No. Not that kind.

This kind:

In order to do anything with planes we need a standard equation for them. For example...
This:
Ax+By+Cz+D=0

But let's face it, although we can use that equation to figure out the x, y and z-intercepts, and use it to find intercepts with other planes etc. we can't really use that form to find an equation for a specific plane.

To find the equation of a plane we need two things, a point on said plane, and a vector that is perpendicular to everything on the plane or, more simply that is normal to it.

So we have point (x_o, y_o, z_o) in the plane and vector <A, B, C> which is normal to the plane. How do we take these and make them into an equation for a plane? Simple! We know that a ∙ b=0 if a and b are perpendicular so if we say that point (x, y, z) is also on the plane we can say that <x-x_o, y-y_o, z-z_o> is also on the plane. Meaning:

<A, B, C> ∙ <x-x_o, y-y_o, z-z_o>=0

We can solve this dot product to get:

A(x-x_o) + B(y-y_o) + C(z-z_o)=0

Which is the point normal form of a plane.

To break it down, A, B and C are the x, y, and z coordinates of a vector normal to a plane.  x_o, y_o, and z_o are the x, y and z coordinates of a point on the plane and, x,y, and z are the variables in the equation.

IMPORTANT: In standard form the plane equation ALWAYS equals zero, if it doesn't it's not in standard form. Always keep track of the constant D value. Ye have been warned.

Armed with the knowledge of plane equations we can do all sorts of stuff with planes like find out if two planes are parallel, perpendicular, their line of intersection, their angle of intersection and even the distance between two planes!

Lets start out simple, determining if two planes are parallel or perpendicular.

If two planes are parallel, their normals are scalar multiples. For example planes with normals <a,b,c> and <ak,bk,ck> are parallel planes because one of the normals is simply the other multiplied by scalar k.

Can't tell at a glance if two vectors are scalar multiples? If a/a₁= b/b₁ = c/c₁ then they're parallel.

Two planes are perpendicular if the dot product of their normals is zero.

Simple enough right?

Now adding lines into the mix doesn't make anything harder but it is very important to remember that the vector given in the line equation is parallel to the line not perpendicular so to find if a line is parallel to a plane the dot product of the vector slope of the line and the normal vector must be zero and if they are perpendicular the vectors will be scalar multiples.

One other little tidbit before moving on, if a point is plugged into a plane equation in standard or point-normal form and it is equal to zero the point is on the plane, if it's positive it's above the plane and if it's negative it's below the plane.

Now we know how to tell if two planes are parallel and where a point is in relation to a plane, what if we want to solve for the line where two planes intersect?

This is actually quite simple to do but the logic and reasoning behind it can be a little hard to grasp at first. Say we have two planes x+2y+3z=0 and 3y+4z=0 and want to find their line of intersection. The first thing we need to do is recall the standard equation for a line in 3-space <x, y, z>=<x_o, y_o, z_o>+t<a,b,c> and that it requires a point (x_o, y_o, z_o) for and intercept and a vector <a, b, c> for a slope. The point (x_o, y_o, z_o) we can find by finding a point that is in both our planes like (0, 0, 0). Now that we have a point we need to find the slope, this can seem difficult at first but the slope of this line is actually the cross product of the normals of the intersecting planes. Each normal is perpendicular to its respective plane so the vector perpendicular to those must be parallel to both the planes, in other words have the same slope of the line of intersection.

Find the cross product, which in this case is <-1, -4, 3> and plug the point and vector into the line equation to get  <x, y, z>=<0, 0, 0> +t<-1, -4, 3>. That;s our line on the intersection of the planes  x+2y+3z=0 and 3y+4z=0!

The key step when it comes to finding the angle between planes is recognizing that the angle we want is also the angle between the normals of the two planes, which are simply the coefficients of the equations for the planes.

This means that we can the find the angle between the two normals using the formula from section 13-4, cos(θ)=aŸb/||a||||b||.

Just don’t forget that the angle we want to find is always the acute angle between the planes, so if the angle found from the above equation is obtuse, more than 90˚, then we just need to subtract that value from 180˚ to find it’s supplementary angle.

Distance formula!!!

The distance D between the point P(x0,y0,z0) and the plane ax+by+cz+d=0 is D=|ax0+by0+cz0+d|/√(a2+b2+c2)

Now for the tricky part: If the question asks for the distance between two lines, L1 and L2 for instance, it gets a little more complicated.

First, find a point on L1 by setting t equal to 0. We’ll use this point as P in the distance formula.

Next, find the point normal form of the plane containing L2 by finding the cross product of L1 and L2, which is the normal to the lines, and a point on L2, we’ll call it <x1,y1,z1>, by setting t equal to 0. The point normal form is equal to the normal dotted with the vector from the point we found to the origin, or n1(x-x1)+n2(y-y1)+ n3(z-z1).

Once we have a point on L1 and a plane containing L2, we can solve for the distance between them using the distance formula.

This strategy works for any combination of points, lines, and planes. If we’re given a line and a plane for instance, we just need to find a point on the line by setting t equal to 0, and then we can use the distance formula. As long as we end up with a point and a plane, we’re golden.

By Katelyn and Alice

13.7 Quadric Sections

Hi everyone,
'Tis me, your teacher.

Here are some notes pertaining to quadric sections.

They are partially mine, partially plagiarized (not really, as I've cited them) AND I've also included links to help.

Remember that quadrics are 3-D analogs to conics.
If you're not sure what a "Conic Section" is, go dig out your precalculus text.  If you do, you'll find a picture like this one (from www.dsusd.k12.ca.us)

You'll then find a bunch of equations - if 4 is a bunch - that describe each of these:

Now, I have little doubt that you can brush these right out of the old memory banks...
If you're not sure what an equation of a conic section is or how to identify it, use this simple flowchart:

·             Are both variables squared?

No: It's a parabola.
Yes: Go to the next test....

o   Do the squared terms have opposite signs?

Yes: It's an hyperbola.
No: Go to the next test....

§  Are the squared terms multiplied by the same number?

Yes: It's a circle.

No: It's an ellipse.

Now that that's out of the way... in 3-D, we find it more difficult to plot points like you can in 2-D.  

Rather, we determine what a quadric surface is by looking at the “trace” … that is, setting one variable equal to a constant, and determining what the surface looks like in the plane parallel to the other two axes. (Set a value for Z, look at the trace in the X-Y plane...)When you put all the traces together, you have a picture of the surface.

There are six quadric sections you'll need to be familiar with: Ellipsoids, Hyperbolic Paraboloids, Hyperboloids of 1 sheet, Hyperboloids of 2 sheets, Elliptic Paraboloids, and Elliptic Cones

YOU SHOULD KNOW HOW TO:

1.Recognize the equation of each type of quadric.

2.Describe the trace of each quadric in relevant planes. (That means sketch - not with numbers, just shape and orientation)

Let's look at one example:

The equation

Is a quadric section.  By comparing with the forms of equations in the table below, you should be able to identify it as a hyperbolic paraboloid.  That means that two traces are parabolas, and one is a hyperbola.

If you set x = 0, you'll see that you're left with the equation -y^2/16 = z.
This is the trace of the surface in the YZ plane.  It's a parabola which opens in the -z direction.

Likewise, if you set y = 0, you're left with x^2/9 = z.  This is the trace of the surface in the XZ plane.  It's  a parabola which opens in the +z direction.

The final trace is found when Z = 0.  Well, when Z = 0, you get a pair of lines which intersect at the origin 
(y = +/- 3/4x).  If is greater than or equal to 1 you start to generate hyperbolas.  When z is a negative number, there is no trace!
  
Here's a picture of the conic section traces for the three planes where x, y and z are each set to zero (top row) and then the fourth, where z = 1.
 

 
So, now that we know what the traces look like, we can imagine what something looks like that has two parabolas and a hyperbola.  It looks like this:

Ta Da!!!  See - easy.

 All this information, as well as several examples of how to do just that are provided in these fabulous videos listed below - which are all on You Tube.  The first is a brief introduction, while the specific ones go through lots of detailed examples.
Woohoo!!!

Introduction to Quadric Surfaces 

Ellipsoid

Hyperboloid of 1 Sheet 

Hyperboloid of 2 Sheets 

Elliptical Paraboloid 

Hyperbolic Paraboloid 

Elliptical Cone

Another helpful resource - the Interactive Gallery of Quadric Sections

http://www.math.umn.edu/~rogness/quadrics/index.shtml

(You do not have to graph them in three dimensions, but you can certainly try!)

Here's a table (from your text) summarizing the parent functions as well as the major characteristics of each(pardon the watermark - I downloaded a "restricted" copy from somewhere... legal for my educational purposes, as we all have the text!
 

 

The following is a link to an Interactive Gallery, where you can play with the parameters of each.  It also gives you great pictures!

http://www.math.umn.edu/~rogness/quadrics/index.shtml 

(Ain't technology grand!)
 

Remember - this section is nothing fancy... just keep basic traces in mind.
 Have fun!
Lisa

  

13.5 Parametric Equations of Lines

13.5 Parametric Equations of Lines

A line is defined by a vector multiplied t times in 2-space.

You can multiply each component by t to find the vector’s parametric equations.

Take the vector <2,3>:

x would equal 2t and

y would equal 3t.

To translate a vector so that it doesn’t pass through the origin, we add another vector.

For example, if the vector must go through the point (1,0), we add the vector <1,0> to the original vector.

L= <1,0>+t<2,3>, where t is assumed to be defined as all real numbers

x=1+2t

y=0+3t

At time t=0, x and y equal 1 and 0.

This method can be easily applied in 3-space by including the third dimension.

The general formula for a 3-space vector that must pass through a specific point for V=<a,b,c> and point (x,y,z) is:

L=<x,y,z>+t<a,b,c>

The equation above is known as the vector equation of a line.

This can be simplified into parametric equations for each component by simply simplifying the equation for each point:

x=at+x

y=bt+y

z=ct+z

For example, a line that has a point at(<3,2,6) and is parallel to vector <1,2,5>

is expressed as

L = <3,2,6>+t<1,2,5>
x=3+t

y=2+2t

z=6+5t

To figure out if two lines are parallel, figure out if they are scalar multiples for each other. For instance, if one line was parallel to <1,2,3> and the other was parallel to <-2,-4,-6>, they would be parallel because all the terms in the first vector can be multiplied by the same coefficient (in this case it’s -2) to become all the terms in the second vector.

To find where two lines intersect (if they do), take the two equations for the x component of the parametric, set them equal to each other, and solve for t(1) in terms of t(2), or vice versa. If the lines intersect, these values of t will set the y and z variables also equal. If they do not come out as equal with the initially found t values, the lines do not intersect. We’ll use the lines

x=3+2t(1)

y=t(1)-1

z=3t(1)

and

x=2t(2)

y=5

z=t(2)+4

First we set them equal to each other:

2t(2)=3+2t(1)

5=t(1)-1

t(2)+4=3t(1)

The easiest t to find is t(1):

5=t(1)-1

t(1)=6

Plugging this into

t(2)+4=3t(1),

we find t(2)

t(2)+4= 18

so t(2)= 14

Finally, plugging those values into the third equation

2t(2)=3+2t(1),

we test for intersection

2(14)=3+2(6)

28 is not equal to 15

So the lines do not intersect.

If two lines are not parallel and do not intersect, they are skew.

Using time as a part of the equations of lines and vectors might be weird at first, but once you get used to it, it’s actually really easy and somewhat fun!

-Lance, Ben, and Jack <3 <3 <3

13.4 - Cross Product

You can express vectors as matrices

<1,3,6>
<2,1,4>     Is the same as  --------------------->
<1,1,1>


Determinant: A function that assigns a numerical value to a square array of numbers. (2x2, 3x3, etc)

In 2-space the determinant is the area of the parallelogram formed by two vectors.

In 3-space the determinant is related to the volume of the parallel pipet formed by three vectors.

How to calculate the determinant

---

Step 1:

Take the top of the left column, cross out the rest of the row and column

Then take the 2x2 vector remaining, multiply the upper left with the lower right, and subtract the upper right multiplied by the lower left. Multiply the result by the circled number. In this case you would do:

((4*2)-(-4*3))(3)

Step 2:

Repeat the same step, using the center of the top row:

((1*2)-(-4*0))(-2)

Step 3:

Repeat the same step, now using the upper right corner:

((1*3)-(4*0))(-5)

Finally, add the results from the last three steps.

Note, alternating steps are subtracted, resulting in:

(3)(8 + 12) - (-2)(2 - 0) + (-5)(3 - 0) = 49

Note:

• Lines are used to denote determinants, while [brackets] denote matrices.
• If two rows of a matrix are the same the determinant is zero.
• If two rows of a matrix are switched the determinant is multiplied by -1.

To check your work, you can also enter the matrices into the calculator in:
2nd -> MATRX -> EDIT
Once entered, run 2nd -> MATRX -> MATH -> det( [A] ) to calculate the determinant!


Cross Products

---

If U = <u1, u2, u3> and V = <v1, v2, v3> then the cross product of U and V (UxV) is defined as:

Properties of determinants and cross products

---

The cross product of two vectors is a third vector in a perpendicular direction, and it has a length equal to that of the parallelogram formed by the two original vectors.

Example:

Find the area of the parallelogram formed by A = <1,2,3> and B = <-1, 1, -2>

A x B = <-7, -1, -3>

|| <-7, -1, -3> || = √( 59 )

To find the area of the triangle between the vectors, just cut the parallelogram in half!

Note:  You must take the absolute value of the vector length, because areas and volumes are always positive.

Scalar Triple Product

---

The same concept can be applied in 3-Space, when you have a parallel pipet formed by three vectors.

Given vectors A, B, and C, the volume of the pipet is: (A x B) • C

Example: Use a scalar triple product to show that the vectors A <1,4,-7> B <2,-1,4> and C <0,-9,18> are coplanar.

Note: If three vectors are coplanar then the volume will be zero, because the shape will be flat.

<18, -36, -18> • <1, 4, -7> = 0

Relations with cross and dot products:

U • V = ||U|| ||V|| cos ø

|| U x V || = ||U|| ||V|| sin ø

Calculator Note:
Below is the source code to a calculator program which does cross products for you! Use it for good:

Disp "VCTR A"
Input A
Input B
Input C
Disp "VCTR B"
Input D
Input E
Input F
Disp ((B*F)-(C*E))
Disp (-1)((A*F)-(C*D))
Disp (A*E)-(B*D)

XOXO, 
Milo and Allison 

13.3 - Dot Products and Projections

Hey guys! This is Ilana and Coulter with the notes on 13.3 from Thursday and Friday.

Lisa started off class with the song "Matchmaker" form the movie Fiddler on the Roof.

Lisa explained this by having us figure out who we would best fit with according to certain personal likes and dislikes, such as our opinion of body piercing, sushi and classical music.

We chose a rank between -5 and 5 to represent our views, with -5 as "repulsive" and 5 as "amazing". We used our number for each item as a specified coordinate for a vector: i, j and k. We then milled around the room. For each person we met, we multiplied our i values together, then our j values and ended with the k values. Once we had those three numbers, we added them together to get a single number.

Sounds pretty easy, right?

That is when Lisa told us that that was actually something called DOT PRODUCTS, a scary name for something simple.

We started off class with a bit of review of vector addition and scaler multiplication. We then moved onto Dot Products, also known as vector multiplication, named for the dot used as a multiplication symbol.

It is important to remember that TWO VECTORS ARE PERPENDICULAR IFF THEIR DOT PRODUCT IS ZERO. THEY FORM AN ACUTE ANGLE IFF THEIR DOT PRODUCT IS POSITIVE. THEY FORM AN OBTUSE ANGLE IFF THEIR DOT PRODUCT IS NEGATIVE.

We then went over some properties to keep in mind while dealing with Dot Products. Afterward, we found a useful equation to use to find an angle from the Law of Cosines.

(Keep the projection stuff in mind for later.)

Then, came more practice. After all, practice makes perfect. :)

Now comes the PROJECTION STUFF! Make sure to notice the two equations and how the vecter projection is the scaler projection (component) multiplied by a unit vector.

And, that is where Lisa reminded us that she is first and foremost a physics teacher. :) The last example can be found in the textbook.

Then, came one last (physics related) example before Lisa gave us homework and told us about our QUIZ ON THURSDAY!!!!!!! and that our HOMEWORK IS DUE ON TUESDAY!!!!!!! 

Lisa sent us off with a hint for the last homework problem:

THAT'S ALL FOLKS.

UNTIL NEXT TIME. :)

-Ilana and Coulter 

Section 13.2, Vectors!

Vectors!

What is a vector? 

Well, it's magnitude and direction represented by an arrow!

i, j and k are all unit vectors with length 1

i is in the x direction

j is in the y direction

k is in the z direction

Vector Notation:

In 3-Space, vector V starts at the origin and terminates at the point (Vx, Vy, Vz), therefore it has components <Vx, Vy, Vz>. Vx indicated the magnituse of the x-component of the vector, Vy the magnitude of the y-component and Vz the z-component. 

<Vx, Vy, Vz>  can be translated anywhere along the graph. In other words, two vectors are equal if and only if their corresponding components are equal.

Properties of Vectors:

When adding or subtracting vectors, simply add/subtract the components

Figure 13.2.6: Subtracting vectors is the same as adding the negative vector

Some other properties of Vectors. They essentially follow the same principles of  everyday Math!

The resultant vector is always the sum of two vectors!

Here's an example of manipulating vectors

More Properties and Terms:

Finding a vector when given the initial point, P1, and the terminal point, P2:

In both 2 and 3 space,  the components of vector P1P2 are found by subtracting P1 from P2

Using Laws of Sines and Cosines to Find the Magnitude and Angle of a Resultant Vector:

Throwback to Trig!! #TBTT :

Red indicates usage of the Law of Cosines. Blue indicates usage of the Law of Sines

Here's an example!!

Normalizing Vectors:

(Watch Out! 

Finding the norm and normalizing 

are NOT the same!!)

How to find the Norm of a vector (which is the same as the magnitude!)

Normalizing: Creating a unit vector with the same angle as vector  V

 Heres an example!

We normalized the vector,  creating a unit vector with the components 2/3i + 2/3j - 1/3k

Here's Some More Helpful Links:

Graphing Vectors in 3D: http://www.youtube.com/watch?v=51vgIfdBlAk

Properties of Vectors:http://www.youtube.com/watch?v=bGRrZvzNzqY

Properties of vectors (plus a little more we haven't learned yet):http://www.wyzant.com/help/math/calculus/multivariable_vectors/properties_of_vectors