Master AP Calculus AB & BC
Part II. AP CALCULUS AB & BC REVIEW
CHAPTER 2. Calculus Prerequisites
VECTORS AND VECTOR EQUATIONS (BC TOPIC ONLY)
Vector curves complete the triumvirate of BC graph representations. Although they have very peculiar and individual characteristics, they are very closely related (by marriage) to parametric equations, as you will see. A vector is, in essence, a line segment with direction. It is typically drawn as an arrow on the coordinate plane. The diagram below is the vector , with initial point A = (-3,-2) and terminal point B = (4,1).
Although you can draw vectors anywhere in the coordinate plane, it’s really handy when they begin at the origin—such a vector is said to be in standard position.
NOTE. You really don’t need a formula to put a vector in standard form—you can simply count the number of units you traveled along the graph.
Example 21: Put vector (as defined above) in standard position.
Solution: To begin, calculate the slope of the line segment: Therefore, to get from point A to point B, you travel up 3 units and to the right 7 units.
Because vector has the same length and direction wherever it is on the coordinate plane, moving it to standard position did not affect it at all. In fact, we can now write in component form: = <7,3>. Graphing <7,3> is almost equivalent to graphing the point (7,3), except that <7,3> will have a vector leading up to that point.
Thanks to Pythagoras (ya gotta love him), finding the length of a vector, denoted ||||, is very simple. For instance, || from Example 21 is √58. To calculate the length, simply draw a right triangle with the vector as its hypotenuse and apply the Pythagorean Theorem.
Perhaps the most common way a vector is represented is in unit vector form. A unit vector is defined as a vector whose length is 1. Clearly, i = < 1,0> and j = <0,1 > are unit vectors, and these vectors are the backbone of unit vector form.
NOTE. Tie length of a vector is also called the norm of the vector.
Let’s put good old vector = <7,3> from previous examples in unit vector form. This vector is created by moving 7 units right and 3 units up from the origin. As demonstrated by the diagram below, this is the same as seven i vectors and three j vectors. Therefore, = 7i + 3j = 7i + 3j.
NOTE. In textbooks, vector v can be written as or v. Snce handwriting in boldface is not possible, you should designate vectors using the little arrow.
Example 22: If vector v has initial point (—2,6) and terminal point (1,-5), complete the following:
(a) Put v in component form.
In order to travel from the beginning to the end of the vector, you proceed right 3 units and down 11. Thus, v = <3,—11>.
(b) Find ||v||.
As pictured in the diagram below,
(c) Write v in unit vector form.
Any vector has unit vector form ai + bj, so v = 3i — 11j.
Graphing vector curves is quite easy. Although the graphs are created by vectors, the graphs are not covered with arrows. Instead, you can graph vector equations exactly the same as parametric equations.
Example 23: Graph the vector curve r(t) = (t + 1)i + t 3j.
Solution: This vector function can be expressed parametrically. Remember that i refers to horizontal distance and j to vertical distance, exactly like x and y, respectively. Therefore, this vector function has exactly the same graph as the parametric equations x = t + 1, y = t3, which you can graph using a calculator or table of values. Solving x = t + 1 for t and substituting t into the y equation gives you the rectangular form of this graph: y = (x — 1)3.
Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book.
YOU MAY USE A GRAPHING CALCULATOR FOR PROBLEM 4 ONLY.
For numbers 1 through 3, v = <2,3>, w has initial point (3,-3) and terminal point (—1,4), and p = 2v — w.
1. Express p in component form.
2. Find ||p||.
3. What is the unit vector form of p?
4. Given vectors r(t) = eti + (2et + 1)j and s(t) — ti + (2t + 1)j, explain why r and s have the same rectangular form but different graphs.
ANSWERS AND EXPLANATIONS
1. First, put w in component form. You can either count units from the start to the end of the vector or simply find the difference of the beginning and ending coordinates: w = <—1 — 3,4 — (—3)> = <—4,7>. Now, p = 2v — w = 2<2,3> — <—4,7> = <4,6> — <—4,7> = <8,—1>.
2. If you like, you can draw a right triangle to justify your calculations, but it is not necessary.
3. Once p is in component form, all of the hard work is done. In standard unit vector form, p = 8i — 1j.
4. You can express r(t) as the parametric equations x = et, y = (2et + 1). By substitution, y = 2x + 1. You will get the very same parametric equations for s(t), but the graph of s(t) is the entire line 2x + 1, whereas the graph of r(t) is 2x + 1, x > 0. The reason for this is the domain of et. Think back to the major graphs to memorize for the chapter. Remember that et has a positive range and can’t output negative numbers or 0. Therefore, after substituting x = et into y to get 2x + 1, that resulting linear function carries with it the restriction x > 0, limiting the graph accordingly.