You have probably submitted your access code request for physics.
You should go ahead and submit one also for vector calculus.
If you have questions between classes, you can submit them using the Submit Question Form at http://vhcc2.vhcc.edu/dsmith/forms/question_form.htm .
At the site http://vhcc2.vhcc.edu/pc2fall9/frames%20pages/class_notes.htm you will find lecture notes for my Precalculus II class. The links listed below address trigonometric functions and on conic sections, both topics that will be relevant to your course in the next week or two. At times I might refer you to these notes to amplify certain topics.
At this point you should spend an hour or so just skimming over these notes, so you have an overview of what's there. Any video links within these notes won't work, but I can give you a disk that contains the same notes, along with the videos. On the disks, those links will work.
You should then be review in some detail the circular model of trig functions (#01, #02 and #03).
These and other topics relevant to this course are indicated by asterisks. A single asterisk indicates important topics. A double asterisks indicates even more important topics, which are generally not well covered in high school analysis precalculus and calculus courses, and are often neglected even in first-year calculus courses.
#01: First Circular Model; Introduction to Radian Measure **
#02: Sketching Exercises **
#03: Modeling Periodic Phenomena with Circles **
#04: The Sine Function *
#05: Modeling with the Sine Function *
#06: Trigonometric Identities *
#07: Tangents and Other Trigonometric Ratios *
#08: Sines and Cosines as Ratios *
#09: Miscellaneous Problems *
#10: Solving Trigonometric Equations *
#11: Inverse Sines, Tangents and Inverse Tangents *
#12: Introduction to Conic Sections **
#13: Circles, Ellipses, Hyperbola **
#14: Conic Sections at General Positions; Parametric Equations **
#15: Parametric Equations **
#16: Polar Coordinates **
If further review of these topics is needed or desired, you may check out the following qa documents, and are welcome to submit them if desired.
open qa #01 on Radian measure and the unit circle. *
open qa #02 on The Fundamental Angles. * open qa #03 on The Sine Function. * open qa #04 on Changing the Amplitude. * open qa #05 on Changing the Period. * open qa #06 on Basic triangles; Inverse Functions. * open qa #07 on the Tangent Function. * open qa #08 on Trigonometric Identities. * open qa #09 on Vectors. * open qa #10 on Dot Product and Vector Algebra. * open qa #11 on Conic Sections. ** open qa #12 on Equations and Properties of Parabolas, hyperbolas and ellipses. ** open qa #13 on Some Stochastic Matrices; Interpretation of Matrices, Multiplication of Matrices. open qa #14 on Determinants; Systems of Equations and their Geometrical Interpretation. open qa #15 on Matrix Algebra; Solving the matrix equation A * X = C by finding the Inverse Matrix. open qa #16 on Sequences and Series.
A review of Precalculus II Class Notes 15, on parametric equations, is recommended in preparation for an upcoming section of the text.
A review of Precalculus II Class Notes 12-14, and qa's 11 and 12, is recommended at this time, in preparation for the section on Quadric Surfaces at the end of the present chapter.
Equation of a circle or a sphere
In two dimensions a circle of radius 4 centered at the point (3, 2) consists of the set of all points which lie 4 units from (3, 2).
If (x, y) is a point in the plane, its distance from (3, 2) is sqrt( (x - 3)^2 + (y - 2)^2 ).
The point (x, y) will lie on the given circle if, and only if, its distance from (3, 2) is 4.
Thus the condition for a point to lie on the circle is
- sqrt( (x - 3)^2 + (y - 2)^2 ) = 4.
We can square both sides and get the equivalent equation
- (x - 3)^2 + (y - 2)^2 = 16.
In general if (h, k) is the center of a circle of radius r, the same reasoning leads us to the general equation of the circle:
The distance between (x, y) and (h, k) is sqrt( (x - h)^2 + (y - k)^2 ).
If this distance is r, then the square of this distance is r^2 and we have
- (x - h)^2 + (y - k)^2 = r^2.
This isn't a formula to be memorized, but rather a picture and a reasoning sequence to be understood.
In three dimensions everything works out very similarly. For example the sphere of radius 7 centered at (-4, 6, 12) is characterized by points (x, y, z) which lie at distance 7 from the center, so the equation is
(x - (-4)) ^ 2 + (y - 6) ^ 2 + ( z - 12 ) ^ 2 = 7 ^ 2, which simplifies to
(x + 4) ^ 2 + (y - 6) ^ 2 + ( z - 12 ) ^ 2 = 49.
The reasoning used in the above is exactly analogous to that used previously in two dimensions. The equation of a sphere of radius r centered at (x_0, y_0, z_0) is found by the same sequence of reasoning to be
(x - x_0) ^ 2 + (y - y_0) ^ 2 + (z - z_0)^2 = r^2
Returning to the equation (x + 4) ^ 2 + (y - 6) ^ 2 + ( z - 12 ) ^ 2 = 49, which is in one standard form for the equation of a sphere, we put this equation into the other standard form by expanding the squares and subtracting 49 from both sides. We get
x^2 + y^2 + z^2 + 8 x - 12 y - 24 z + 147 = 0.
It's easy to move from the first standard form to the second, as we have just done. It takes a little more work to move in the other direction. Let's pretend we didn't already know the center and radius of the sphere given by the equation x^2 + y^2 + z^2 + 8 x - 12 y - 24 z + 147 = 0. How would we recover this form?
The answer is that we complete the squares on the x, y and z terms. We begin by writing the equation in the form
x^2 + 8 x + y^2 - 12 y + z^2 - 24 z = - 147.
Completing the square on x^2 + 8 x, we find that x^2 + 8 x + 16 is the perfect square of (x + 4). If we replace x^2 + 8 x by the equal quantity x^2 + 8 x + 16 - 16, we will not change the balance of the equation. We will then be able to rewrite this new term as
Similarly we will replace y^2 - 12 y by y^2 - 12 y + 36 - 36 and z^2 - 24 z by z^2 - 24 z + 144 - 144. Our equation will become
x^2 + 8 x + 16 - 16 + y^2 - 12 y + 36 - 36 + z^2 - 24 z + 144 - 144 = -147, which can be rewritten as
(x + 4) ^ 2 - 16 + (y - 6) ^ 2 - 36 + (z - 144) ^ 2 - 144 = -147. Then adding 16 + 36 + 144 to both sides we get
(x + 4) ^ 2 + (y - 6) ^ 2 + ( z - 12 ) ^ 2 = 49, which agrees with the first standard form.
Unit vectors
Three problems were given:
1. Prove that cos ( theta) i + sin ( theta ) * j is a unit vector.
2. Find a unit vector in the direction of the vector 2 i + 5 j. Then find theta such that cos(theta) i + sin(theta) j is that unit vector.
The solutions:
Problem 1:
The magnitude of cos ( theta) i + sin ( theta ) * j is
|| cos ( theta) i + sin ( theta ) * j || = sqrt( (cos(theta)) ^ 2 + (sin(theta))^2 ), also written as sqrt( cos^2(theta) + sin^2(theta) ).
By the unit-circle definition of the sine and cosine functions, cos^2(theta) + sin^2(theta) = 1 for any theta.
Thus the magnitude of cos ( theta) i + sin ( theta ) * j is 1, for any value of theta.
note the importance of understanding the unit-circle definition of the sine and cosine functions
Problem 2:
The magnitude of 2 i + 5 j is sqrt( 2^2 + 5^2) = sqrt( 29 ).
The requested unit vector is therefore ( 2 i + 5 j ) / sqrt(29) = 2 sqrt(29) / 29 i + 5 sqrt(29) / 29 j.
To 2 significant figure this is .37 i + .93 j.
If this is in the form cos(theta) i + sin(theta) j, then since two vectors are equal if an only their components are equal, we have
cos(theta) = .37 and
sin(theta) = .93.
We can sketch a unit circle, with the point (.37, .93) and a radial line to that point. We estimated that angle as about 70 or 80 degrees. Then plugging into our calculator we found that
theta = arcCos(.37) = 1.19.
We also find that
theta = arcSin(.93) = 1.19.
The two agree, which is a good thing because there is only one theta. There will be a slight difference in the results because we used two-significant-figure approximations for the components of the vector. The exact value of the angle would be arcCos(2 sqrt(29) / 29), which is exactly the same as arcSin(5 sqrt(29) / 29).
Equality of Two Vectors
Two vectors are equal if, and only if, all their components are equal.
So, for example, if A = 4 s i + 5 z j + (z - r) k and B = (z + r) i + 10 + 3 z k we have the following:
z + r = s, since the i components are equal
5 z = 10 since the j components are equal
z - r = 3 z since the k components are equal
This gives us three equations, which we can solve for the three unknowns to get z = 2, r = 4 and s = 6.
As another example where this principle is applied:
If u = - 4 i + 3 j and v = 2 i - 1/2 j, find the values of a and b such that a * 3 j + b * u = v.
We write the equation a * 3 j + b * u = v out as
3 a j + b ( -4 i + 3 j) = 2 i - 1/2 j. We simplify the left-hand side to obtain
-4 b i + (3 a + 3 b) j = 2 i - 1/2 j. Setting the i and j components equal we get
-4 b = 2 and
3 a + 3 b = -1/2.
We easily solve this system, obtaining b = -1/2 and a = 1/3.
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Here are some exercises designed to give you a head start on some of next week's topics. You have plenty to do with the problem assignments (and I'm sure you'll have some questions on Monday), but do try to take some time to do the following. Put a 15-minute limit on yourself for each questions; they're worth that much time, but don't let them bog you down.
1. If u = 3 i + 5 j , and we know that u dot v = 0 for the vector v = a i + 4 j, then what must be the value of a?
2. If u = 3 i + 5 j - 2 k, and we know that u dot v = 0 for the vector v = a i + 10 j + c k, then what are the values of a and c?
3. Sketch the vectors <2, 5> + t * <3, -1> for t = 0, 1, 2 and 3, with the initial point of each vector at the origin. Sketch the line or smooth curve which passes through the terminal points of each of the four vectors. (i.e., each vector ends at a point; if you mark the four points, you want to sketch the line or curve through these points).