Three Dimensional Space - Calculus II - Notes | M 408D, Study notes of Mathematics

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Three Dimensional Space

Paul Dawkins

© 2007 Paul Dawkins i http://tutorial.math.lamar.edu/terms.aspx

Table of Contents

• Three Dimensional Space Preface............................................................................................................................................. ii
  • Introduction
  • The 3-D Coordinate System
  • Equations of Lines....................................................................................................................................
  • Equations of Planes
  • Quadric Surfaces
  • Functions of Several Variables.................................................................................................................
  • Vector Functions
  • Calculus with Vector Functions
  • Tangent, Normal and Binormal Vectors
  • Arc Length with Vector Functions
  • Curvature
  • Velocity and Acceleration
  • Cylindrical Coordinates............................................................................................................................
  • Spherical Coordinates

© 2007 Paul Dawkins iii http://tutorial.math.lamar.edu/terms.aspx

don’t have time in class to work all of the problems in the notes and so you will find that some sections contain problems that weren’t worked in class due to time restrictions.

3. Sometimes questions in class will lead down paths that are not covered here. I try to anticipate as many of the questions as possible in writing these up, but the reality is that I can’t anticipate all the questions. Sometimes a very good question gets asked in class that leads to insights that I’ve not included here. You should always talk to someone who was in class on the day you missed and compare these notes to their notes and see what the differences are.
4. This is somewhat related to the previous three items, but is important enough to merit its own item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!! Using these notes as a substitute for class is liable to get you in trouble. As already noted not everything in these notes is covered in class and often material or insights not in these notes is covered in class.

Three Dimensional Space

Introduction In this chapter we will start taking a more detailed look at three dimensional space (3-D space or ° 3 ). This is a very important topic for Calculus III since a good portion of Calculus III is done in three (or higher) dimensional space.

We will be looking at the equations of graphs in 3-D space as well as vector valued functions and how we do calculus with them. We will also be taking a look at a couple of new coordinate systems for 3-D space.

This is the only chapter that exists in two places in my notes. When I originally wrote these notes all of these topics were covered in Calculus II however, we have since moved several of them into Calculus III. So, rather than split the chapter up I have kept it in the Calculus II notes and also put a copy in the Calculus III notes. Many of the sections not covered in Calculus III will be used on occasion there anyway and so they serve as a quick reference for when we need them.

Here is a list of topics in this chapter.

The 3-D Coordinate System – We will introduce the concepts and notation for the three dimensional coordinate system in this section.

Equations of Lines – In this section we will develop the various forms for the equation of lines in three dimensional space.

Equations of Planes – Here we will develop the equation of a plane.

Quadric Surfaces – In this section we will be looking at some examples of quadric surfaces.

Functions of Several Variables – A quick review of some important topics about functions of several variables.

Vector Functions – We introduce the concept of vector functions in this section. We concentrate primarily on curves in three dimensional space. We will however, touch briefly on surfaces as well.

Calculus with Vector Functions – Here we will take a quick look at limits, derivatives, and integrals with vector functions.

Tangent, Normal and Binormal Vectors – We will define the tangent, normal and binormal vectors in this section.

Arc Length with Vector Functions – In this section we will find the arc length of a vector function.

Curvature – We will determine the curvature of a function in this section.

Velocity and Acceleration – In this section we will revisit a standard application of derivatives. We will look at the velocity and acceleration of an object whose position function is given by a vector function.

Cylindrical Coordinates – We will define the cylindrical coordinate system in this section. The cylindrical coordinate system is an alternate coordinate system for the three dimensional coordinate system.

Spherical Coordinates – In this section we will define the spherical coordinate system. The spherical coordinate system is yet another alternate coordinate system for the three dimensional coordinate system.

d ( P P 1 , 2 ) = ( x 2 - x 1 ) 2 + ( y 2 - y 1 )^2

While the distance between any two points in ° 3 is given by,

d ( P P 1 , 2 ) = ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2 + ( z 2 - z 1 )^2

Likewise, the general equation for a circle with center ( h k , )and radius r is given by,

x - h^2 + y - k^2 = r^2

and the general equation for a sphere with center ( h k l , , )and radius r is given by,

x - h^2 + y - k^2 + z - l^2 = r^2

With that said we do need to be careful about just translating everything we know about ° 2 into ° 3 and assuming that it will work the same way. A good example of this is in graphing to some extent. Consider the following example.

Example 1 Graph x = 3 in ° , °^2 and °^3.

Solution In ° we have a single coordinate system and so x = 3 is a point in a 1-D coordinate system.

In ° 2 the equation x = 3 tells us to graph all the points that are in the form ( 3, y ). This is a

vertical line in a 2-D coordinate system.

In ° 3 the equation x = 3 tells us to graph all the points that are in the form ( 3, y z , ). If you go

back and look at the coordinate plane points this is very similar to the coordinates for the yz -plane except this time we have x = 3 instead of x = 0. So, in a 3-D coordinate system this is a plane that will be parallel to the yz -plane and pass through the x -axis at x = 3.

Here is the graph of x = 3 in °.

Here is the graph of x = 3 in °^2.

Finally, here is the graph of x = 3 in °^3. Note that we’ve presented this graph in two different styles. On the left we’ve got the traditional axis system and we’re used to seeing and on the right we’ve put the graph in a box. Both views can be convenient on occasion to help with perspective and so we’ll often do this with 3D graphs and sketches.

Note that at this point we can now write down the equations for each of the coordinate planes as well using this idea. 0 plane 0 plane 0 plane

z xy y xz x yz

Let’s take a look at a slightly more general example.

Let’s take a look at one more example of the difference between graphs in the different coordinate systems.

Example 3 Graph x^2 + y^2 = 4 in ° 2 and °^3.

Solution As with the previous example this won’t have a 1-D graph since there are two variables.

In ° 2 this is a circle centered at the origin with radius 2.

In ° 3 however, as with the previous example, this may or may not be a circle. Since we have not specified z in any way we must assume that z can take on any value. In other words, at any value of z this equation must be satisfied and so at any value z we have a circle of radius 2 centered on the z- axis. This means that we have a cylinder of radius 2 centered on the z -axis.

Here are the graphs for this example.

Notice that again, if we look to where the cylinder intersects the xy -plane we will again get the circle from ° 2.

We need to be careful with the last two examples. It would be tempting to take the results of these and say that we can’t graph lines or circles in ° 3 and yet that doesn’t really make sense. There is no reason for there to not be graphs lines or circles in °^3. Let’s think about the example of the circle. To graph a circle in °^3 we would need to do something like x^2 + y^2 = 4 at z = 5. This would be a circle of radius 2 centered on the z -axis at the level of z = 5. So, as long as we specify a z we will get a circle and not a cylinder. We will see an easier way to specify circles in a later section.

We could do the same thing with the line from the second example. However, we will be looking at line in more generality in the next section and so we’ll see a better way to deal with lines in °^3 there.

The point of the examples in this section is to make sure that we are being careful with graphing equations and making sure that we always remember which coordinate system that we are in.

Another quick point to make here is that, as we’ve seen in the above examples, many graphs of equations in °^3 are surfaces. That doesn’t mean that we can’t graph curves in ° 3. We can and will graph curves in °^3 as well as we’ll see later in this chapter.

In this sketch we’ve included the position vector (in gray and dashed) for several evaluations as well as the t (above each point) we used for each evaluation. It looks like, in this case the graph of the vector equation is in fact the line y = 1.

Here’s another quick example. Here is the graph of r^ r^ ( t ) = 6 cos ,3sin t t.

In this case we get an ellipse. It is important to not come away from this section with the idea that vector functions only graph out lines. We’ll be looking at lines in this section, but the graphs of vector function do not have to be lines as the example above shows.

We’ll leave this brief discussion of vector function with another way to think of the graph of a vector function. Imagine that a pencil/pen is attached to the end of the position vector and as we increase the variable the resulting position vector moves and as it moves the pencil/pen on the end sketches out the curve for the vector function.

Okay, we now need to move into the actual topic of this section. We want to write down the equation of a line in ° 3 and as suggested by the work above we will need a vector function to do this. To see how we’re going to do this let’s think about what we need to write down the

equation of a line in ° 2. In two dimensions we need the slope ( m ) and a point that was on the line in order to write down the equation.

In ° 3 that is still all that we need except in this case the “slope” won’t be a simple number as it was in two dimensions. In this case we will need to acknowledge that a line can have a three dimensional slope. So, we need something that will allow us to describe a direction that is potentially in three dimensions. We already have a quantity that will do this for us. Vectors give directions and can be three dimensional objects.

So, let’s start with the following information. Suppose that we know a point that is on the line,

P 0 = ( x 0 , y 0 , z 0 ), and that v = a b c , ,

r (^) is some vector that is parallel to the line. Note, in all

likelihood, v^ r^ will not be on the line itself. We only need v^ r^ to be parallel to the line. Finally, let

P = ( x y z , , )be any point on the line.

Now, since our “slope” is a vector let’s also represent the two points on the line as vectors. We’ll do this with position vectors. So, let r 0

ur and r^ r^ be the position vectors for P 0 and P respectively.

Also, for no apparent reason, let’s define a^ r^ to be the vector with representation P P 0

uuur .

We now have the following sketch with all these points and vectors on it.

Now, we’ve shown the parallel vector, v^ r^ , as a position vector but it doesn’t need to be a position vector. It can be anywhere, a position vector, on the line or off the line, it just needs to be parallel to the line.

Next, notice that we can write r^ r^ as follows, r = r 0 + a r ur r

If you’re not sure about this go back and check out the sketch for vector addition in the vector arithmetic section. Now, notice that the vectors a^ r^ and v^ r^ are parallel. Therefore there is a number, t , such that

To get a point on the line all we do is pick a t and plug into either form of the line. In the vector form of the line we get a position vector for the point and in the parametric form we get the actual coordinates of the point.

There is one more form of the line that we want to look at. If we assume that a , b , and c are all non-zero numbers we can solve each of the equations in the parametric form of the line for t. We can then set all of them equal to each other since t will be the same number in each. Doing this gives the following,

x x 0 (^) y y 0 (^) z z 0 a b c

This is called the symmetric equations of the line.

If one of a , b , or c does happen to be zero we can still write down the symmetric equations. To see this let’s suppose that b = 0. In this case t will not exist in the parametric equation for y and so we will only solve the parametric equations for x and z for t. We then set those equal and acknowledge the parametric equation for y as follows, x x 0 (^) z z 0 (^) y y 0 a c

Let’s take a look at an example.

Example 1 Write down the equation of the line that passes through the points ( 2, - 1,3 )and

( 1, 4,^ -^3 ). Write down all three forms of the equation of the line.

Solution To do this we need the vector v^ r^ that will be parallel to the line. This can be any vector as long as it’s parallel to the line. In general, v^ r^ won’t lie on the line itself. However, in this case it will. All we need to do is let v^ r^ be the vector that starts at the second point and ends at the first point. Since these two points are on the line the vector between them will also lie on the line and will hence be parallel to the line. So, v^ r = 1, - 5, 6

Note that the order of the points was chosen to reduce the number of minus signs in the vector. We could just have easily gone the other way.

Once we’ve got v^ r^ there really isn’t anything else to do. To use the vector form we’ll need a point on the line. We’ve got two and so we can use either one. We’ll use the first point. Here is the vector form of the line. r^ r = 2, - 1,3 + t 1, - 5, 6 = 2 + t , - - 1 5 ,3 t + 6 t

Once we have this equation the other two forms follow. Here are the parametric equations of the line.

x t y t z t

Here is the symmetric form. 2 1 3 1 5 6

x - (^) = y + (^) = z -

Example 2 Determine if the line that passes through the point ( 0, - 3,8 )and is parallel to the

line given by x = 10 + 3 t , y = 12 t and z = - 3 - t passes through the xz -plane. If it does give the coordinates of that point.

Solution To answer this we will first need to write down the equation of the line. We know a point on the line and just need a parallel vector. We know that the new line must be parallel to the line given by the parametric equations in the problem statement. That means that any vector that is parallel to the given line must also be parallel to the new line.

Now recall that in the parametric form of the line the numbers multiplied by t are the components of the vector that is parallel to the line. Therefore, the vector, v^ r = 3,12, - 1 is parallel to the given line and so must also be parallel to the new line.

The equation of new line is then, r^ r = 0, - 3,8 + t 3,12, - 1 = 3 , t - 3 + 12 ,8 t - t

If this line passes through the xz -plane then we know that the y -coordinate of that point must be zero. So, let’s set the y component of the equation equal to zero and see if we can solve for t. If we can, this will give the value of t for which the point will pass through the xz -plane.

3 12 0 1 4

• • t = fi t =

So, the line does pass through the xz -plane. To get the complete coordinates of the point all we

need to do is plug 1 4 t = into any of the equations. We’ll use the vector form.

3 1 , 3 12 1 ,8 1 3 , 0,^31 4 4 4 4 4 r = ÊÁ^ ˆ˜^ - + ÊÁ^ ˆ˜ - = Ë ¯ Ë ¯

r

Recall that this vector is the position vector for the point on the line and so the coordinates of the

point where the line will pass through the xz -plane are 3 , 0,^31 4 4

Ê ˆ

Á Ë ˜¯.

A slightly more useful form of the equations is as follows. Start with the first form of the vector equation and write down a vector for the difference.

0 0 0

a b c x y z x y z a b c x x y y z z

g g

Now, actually compute the dot product to get,

a x ( - x 0 ) + b ( y - y 0 ) + c z ( - z 0 )= 0

This is called the scalar equation of plane. Often this will be written as, ax + by + cz = d where d = ax 0 (^) + by 0 (^) + cz 0.

This second form is often how we are given equations of planes. Notice that if we are given the equation of a plane in this form we can quickly get a normal vector for the plane. A normal vector is, n^ r = a b c , ,

Let’s work a couple of examples.

Example 1 Determine the equation of the plane that contains the points P = ( 1, - 2, 0 ),

Q = ( 3,1, 4)and R = ( 0, - 1, 2 ).

Solution In order to write down the equation of plane we need a point (we’ve got three so we’re cool there) and a normal vector. We need to find a normal vector. Recall however, that we saw how to do this in the Cross Product section.

We can form the following two vectors from the given points. PQ = 2,3, 4 PR = - 1,1, 2

uuur uuur

These two vectors will lie completely in the plane since we formed them from points that were in the plane. Notice as well that there are many possible vectors to use here, we just chose two of the possibilities.

Now, we know that the cross product of two vectors will be orthogonal to both of these vectors. Since both of these are in the plane any vector that is orthogonal to both of these will also be orthogonal to the plane. Therefore, we can use the cross product as the normal vector.

2 3 4 2 3 2 8 5 1 1 2 1 1

i j k i j n = PQ ¥ PR = = i - j + k

r r r r r r^ uuur^ uuur^ r^ r r

The equation of the plane is then,

x y z x y z

We used P for the point, but could have used any of the three points.

Example 2 Determine if the plane given by - x + 2 z = 10 and the line given by r^ r^ = 5, 2 - t ,10 + 4 t are orthogonal, parallel or neither.

Solution This is not as difficult a problem as it may at first appear to be. We can pick off a vector that is normal to the plane. This is n^ r^ = - 1, 0, 2. We can also get a vector that is parallel to the line.

This is v = 0, - 1, 4.

Now, if these two vectors are parallel then the line and the plane will be orthogonal. If you think about it this makes some sense. If n^ r^ and v^ r^ are parallel, then v^ r^ is orthogonal to the plane, but v^ r is also parallel to the line. So, if the two vectors are parallel the line and plane will be orthogonal.

Let’s check this.

1 0 2 1 0 2 4 0 0 1 4 0 1

i j k i j n ¥ v = - - = i + j + k π

r r r r r r r r^ r^ r^ r

So, the vectors aren’t parallel and so the plane and the line are not orthogonal.

Now, let’s check to see if the plane and line are parallel. If the line is parallel to the plane then any vector parallel to the line will be orthogonal to the normal vector of the plane. In other words, if n^ r^ and v^ r^ are orthogonal then the line and the plane will be parallel.

Let’s check this. n v^ r r g = 0 + 0 + 8 = 8 π 0

The two vectors aren’t orthogonal and so the line and plane aren’t parallel.

So, the line and the plane are neither orthogonal nor parallel.