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INTRODUCTION TO POLYNOMIAL CALCULUS
- Straight Lines
Given two distinct points in the plane, there is exactly one straight line that contains them both. This is one of the important principles of Plane Geometry. If the plane is equipped with a Cartesian coordinate system, it should be possible to write down an equation for such a line in terms of the x and y coordinates. In this section we shall show how to do this. The notion of slope will be very useful in this task. The slope of a line is defined to be the change in the y coordinate (the rise) divided by the change in the x coordinate (the run) as we move from one point on the line to a second point on the line (see Figure 1.1). That is, if (x 1 , y 1 ) and (x 2 , y 2 ) are two points on the line, then the slope of the line is the number m, where
m =
y 2 − y 1 x 2 − x 1
This number does not depend on which two points on the line are chosen. In fact, if two other points (x 3 , y 3 ) and (x 4 , y 4 ) are chosen, then it follows from the similar triangles in Figure 1.2 that y 2 − y 1 x 2 − x 1
y 4 − y 3 x 4 − x 3
and so the points (x 3 , y 3 ) and (x 4 , y 4 ) give the same value for the slope m as do the points (x 1 , y 1 ) and (x 2 , y 2 ). The slope measures whether a line rises or falls as we move to the right and how steeply it does so. Positive slope means the line rises to the right, while negative slope means it falls to the right. Slope zero means the line is horizontal, since it means that the y coordinate does not change at all from point to point on the line. A vertical line has no slope (some would say it has infinite slope). This is because the x coordinates of points on such a line are all the same and so the denominator is zero in the equation defining slope.
Example 1. Find the slope of the line which contains the points (1, 2) and (3, 5). Solution: Here the rise is 5 − 2 = 3 and the run is 3 − 1 = 2 and so the slope is rise run = 3/2.
The point-slope form of the equation of a line. Using slope we can easily write down an equation for any line that is not vertical. Let m be the slope of the line and let (x 0 , y 0 ) be some fixed point on the line. If (x, y) is a variable point on the line, then
m =
y − y 0 x − x 0
and so y − y 0 = m(x − x 0 ) 1
This is called the point-slope form of the equation of a straight line. It gives the equation of the line in terms of the slope m of the line and a point (x 0 , y 0 ) on the line.
Example 2. Find the equation of the line which passes through the point (1, 3) and has slope 2. Solution: Here the slope m is 2 and the point (x 0 , y 0 ) is (1, 3) and so the equation of the line is y − 3 = 2(x − 1) or y = 2x + 1
Example 3. Find the equation of the line which contains the points (− 1 , 2) and (0, 5) Solution: The slope of this line is
m =
and the line contains the point (0, 5); so the equation of the line is
y − 5 = 3(x − 0) or y = 3x + 5
The slope-intercept form of the equation of a line. If a line crosses the y-axis at the point (0, b), then the number b is called the y-intercept of the line. If the line has slope m, then the point-slope form of its equation (using the point (0, b)) is y − b = m(x − 0) or
y = mx + b
This is the point-intercept form of the equation of the line. Note that every line except a vertical line does cross the y-axis and, thus, has a slope-intercept form for its equation.
Example 4. What is the equation of the line with slope −5 and y-intercept 4? Solution: y = − 5 x + 4
The general equation of a line. Every equation of the form
Ax + By + C = 0
is the equation of a line and, conversely, every line has an equation that can be put in this form. A non-vertical line has an equation of the form y = mx + b, which can be written
mx − y + b = 0
(x 1 − x 0 )^2 + y 02 + (x 2 − x 0 )^2 + y^20 = (x 2 − x 1 )^2
which simplifies to 2 x^20 + 2y^20 − 2 x 0 x 1 − 2 x 0 x 2 = − 2 x 1 x 2
On dividing by 2 and rearranging terms this becomes
y^20 = −x^20 + x 0 x 1 + x 0 x 2 − x 1 x 2
or, after factoring, y 02 = −(x 0 − x 2 )(x 0 − x 1 )
or y 0 x 0 − x 2
x 0 − x 1 y 0
which says that
m 2 = −
m 1
where m 1 is the slope of the first line and m 2 is the slope of the second line. Thus, two non horizontal lines are perpendicular if and only if the slope of the second is the negative reciprocal of the slope of the first.
Example 6. Find the equation of the line which is parallel to the line with equation 2 x + 3y = 5 and passes through the point (1, 2). Solution: We solve for y in the equation of the first line in order to put its equation in the form
y = −
x +
This shows that the line has slope − 23. The line which has the same slope and passes through the point (1, 2) has the equation
y − 2 = −
(x − 1) or y = −
x +
x or 2 x + 3y = 8
Example 7. Find the equation of the line which is perpendicular to the line 2y + x = 3 and meets it at the point (1, 1). Solution: If we solve the first equation for y it becomes
y = −
x +
which tells us that its slope is − 12. A line perpendicular to this line will, therefore, have slope 2. Since the line we seek must pass through the point (1, 1), its equation is
y − 1 = 2(x − 1) or y = 2x − 1
- Slope of a Curve
We know about the slope of a straight line. It is the change in the y-coordinate divided by the change in the x-coordinate (rise divided by run) as we move from a given point on the line to any other point on the line. The law of similar triangles says that this ratio is independent of the two points on the line that are chosen. What about the slope of a curve that is not a straight line? Does this make sense? If so, how do we calculate it? Let’s look at an example, say the curve
y = x^2
The graph of this curve (Figure 2.1) makes it clear that, if its slope makes sense, it cannot be a fixed number. To the left of x = 0 the curve slopes downward (negative slope) while to the right of x = 0 the curve slopes upward (positive slope) and it rises more steeply the further to the right we go. Thus, if it makes sense at all, the slope must depend on where we are on the curve. In fact, the slope of the curve y = x^2 does make sense at each point (x, y) on the curve (but it changes as the point changes). This is suggested by the fact that if a microscopically tiny piece of the curve is magnified enough to be visible then it looks like a straight line. This is illustrated by the graphs in Figure 2.1 which show the curve near the point (1, 1) magnified by various factors. In other words, the smaller the segment of the curve we look at near (1, 1), the more the curve looks like a straight line. The slope of this straight line should be what we mean by the slope of the curve at the point (1, 1). How can we calculate this slope? We do the same thing we would do if the curve were a straight line. We calculate the change in y divided by the change in x as we move from one point on the curve to another. However, now we choose special points. We let the first one be (1, 1) itself and we choose the second one to be near (1, 1). The nearer to (1, 1) we make it, the more the curve between these two points looks like a straight line and the closer our ratio will be to the slope of this line. Let’s try this for some choices of points on the curve y = x^2 (see Figure 2.2). In each case we will be calculating the rise divided by the run between a first point, (1, 1), and some nearby second point on the curve. In other words we will be calculating the slope of the line joining these two points. With second point (3, 9) we get
slope =
With second point (2, 4) we get
slope =
With second point (1. 5 , 2 .25) we get
slope =
Returning to the curve y = x^2 , we would like a formula for the slope of this curve at any point of the curve, not just at the point (1, 1). We use the same technique. Given a point (x, x^2 ) on the curve, we move to a nearby second point (x + h, (x + h)^2 ), obtained by changing the x coordinate of the first point by an amount h. Then we calculate the slope of the line joining these two points. It is
slope =
(x + h)^2 − x^2 h
2 xh + h^2 h
= 2x + h
As h approaches zero this slope approaches the number 2x. In other words
lim h→ 0
(2x + h) = 2x
We conclude that the slope of the curve y = x^2 at the point (x, x^2 ) is 2x. The preceding discussion leads to the following definition of the slope of a curve which is given as the graph of a function f :
Definition A. The graph of a function f is said to have slope m at a point (x, f (x)) on the graph provided
lim h→ 0
f (x + h) − f (x) h
= m
That is, provided the expression f^ (x+h h)− f^ (x)approaches m as h approaches 0. In this case, we call the number m the derivative of f at x and denote it by f ′(x).
Thus, the derivative of a function f is another function f ′^ of the same variable x. Its value at x is the slope of the graph of f at the point (x, f (x)). In other words, it is the instantaneous rate of change of y with respect to x as we pass through the point (x, f (x)) while moving along the graph of f.
For now we will take the statement “ f^ (x+h h)− f^ (x) approaches m as h approaches 0” as intuitively understood but later in the course we will need to make this idea more precise. We will do this when we study limits. As we shall see in the next section, when
f is a polynomial it is quite easy to see what happens to the expression f^ (x+h h)− f^ (x)as h approaches 0 and so to study derivatives of polynomials we do not need a sophisticated study of limits. The discussion preceding the above definition shows that the derivative of x^2 is 2x. We give two other examples of the calculation of a derivative.
Example 2. Find the derivative of a constant function f (x) = c. Solution: f ′(x) = lim h→ 0
f (x + h) − f (x) h
= lim h→ 0
c − c h
= lim h→ 0
Thus, the derivative of a constant function is 0. This just reflects the fact that the graph of a constant function is a horizontal line and, thus, has slope 0.
Example 3. Find the derivative of the function f (x) = 3x − 2. Solution:
f ′(x) = lim h→ 0
f (x + h) − f (x) h
= lim h→ 0
3(x + h) − 2 − (3x − 2) h
= lim h→ 0
Thus, the derivative of the function f (x) = 3x − 2 is the constant 3. This is not surprising, since the graph of the function f (x) = 3x − 2 is a straight line with slope 3.
- Derivative of a Polynomial
In the last section we defined the derivative f ′(x) of a function f (x) to be the slope of the curve y = f (x) at the point (x, f (x)). This, in turn, is defined to be the number that
the expression f^ (x+h h)− f^ (x)approaches as h approaches 0. In other words
f ′(x) = lim h→ 0
f (x + h) − f (x) h
We will now use this definition to calculate the derivative of any polynomial. We begin by calculating the derivative of the monomial xn^ for each value of n. For n = 0, 1 , 2 , 3 this was done in the previous section and its problem set. However, we will repeat these calculations here in order to demonstrate the pattern that emerges.
If f (x) = x^0 = 1 then f ′(x) = lim h→ 0
h
= lim h→ 0
If f (x) = x^1 = x then
f ′(x) = lim h→ 0
x + h − x h
= lim h→ 0
If f (x) = x^2 then
f ′(x) = lim h→ 0
(x + h)^2 − x^2 h
= lim h→ 0
(2x + h) = 2x
If f (x) = x^3 then
f ′(x) = lim h→ 0
(x + h)^3 − x^3 h
= lim h→ 0
(3x^2 + 3xh + h^2 ) = 3x^2
The pattern here suggests that, for any natural number n, the derivative of xn^ should be nxn−^1. This is, in fact, true. The proof is not difficult. From the definition of derivative, we have that
(xn)′^ = lim h→ 0
(x + h)n^ − xn h
Example 1. Find the derivative of 3x^2 + 2x + 5. Solution:
(3x^2 + 2x + 5)′^ = (3x^2 )′^ + (2x)′^ + (5)′^ Th B(2) = 3(x^2 )′^ + 2(x)′^ + (5)′^ Th B(1) = 3 · 2 x + 2 · 1 + 0 Th A = 6x + 2
Example 2. Find the derivative of x^5 − 11 x^3 + 9x + 2. Solution:
(x^5 − 11 x^3 + 9x + 2)′^ = (x^5 )′^ + (− 11 x^3 )′^ + (9x)′^ + (2)′ = 5x^4 − 11 · 3 x^2 + 9 · 1 + 0 = 5x^4 − 33 x^2 + 9
Example 3. Find the slope of the curve y = x^4 − 3 x^2 + 2 at the point (1, 0). Solution: We first find the derivative of x^4 − 3 x^2 + 2:
(x^4 − 3 x^2 + 2)′^ = 4x^3 − 3 · 2 x + 0 = 4x^3 − 6 x
We then evaluate the derivative at x = 1 to get the slope of the curve at the point (1, 0). Thus, the answer is 4 − 6 = − 2
Example 4. Find the rate of change of the function f (x) = x^3 − 4 x with respect to x when x = 2. Solution: We first find the derivative of f (x):
f ′(x) = (x^3 − 4 x)′^ = 3x^2 − 4
We then evaluate the derivative at x = 2 to find the rate of change of f (x) with respect to x at x = 2. Thus, the answer is
Velocity and Acceleration
Much of Physics and Engineering is concerned with the mathematics of moving bodies. Suppose a body moves along a straight line. If we fix an origin for this line and units for measuring distance on the line, then the position of the body at any time t is described by its coordinate on the line, often denoted by s(t). The velocity v(t) of the body is then defined to be the rate of change of s(t) with respect to t - that is, the derivative s′(t) of s(t). Similarly, the acceleration a(t) of the body is defined to be the rate of change of v(t) with respect to t - that is, the derivative v′(t) of v(t). In summary,
v(t) = s′(t)
a(t) = v′(t)
Example 5. If a ball if dropped off the top of a 64 foot high building, then its height s(t) above the ground t seconds later is described by the formula
s(t) = − 16 t^2 + 64
What is its velocity when it hits the ground? What is its acceleration at any time? Solution: The ball hits the ground when
− 16 t^2 + 64 = 0.
This happens when t^2 = 4; that is, when t = 2. Thus, we need to know the velocity of the ball when t = 2. But v(t) = s′(t) = − 16 · 2 t + 0 = − 32 t
At time t = 2 this is −64. Thus, the ball hits the ground with velocity −64 ft/sec. The acceleration at any time t is
a(t) = v′(t) = (− 32 t)′^ = − 32
Thus, the acceleration is a constant -32 ft/sec/sec.
Theorem A. If n is a non-negative integer, then
∫ xn^ dx =
xn+ n + 1
where c ranges over all constants.
Remember, this theorem simply says that the set of antiderivatives for xn^ is the set of
functions of the form x
n+ n+1 +^ c. If we apply this for^ n^ = 0,^1 ,^ 2 we get:
∫ 1 dx =
x^0 dx = x + c
∫ x dx =
x^1 dx =
x^2 2
∫ x^2 dx =
x^3 3
If h is an antiderivative for f and a is a constant, then
(ah)′^ = ah′^ = af
and so ah is an antiderivative for af. Similarly, if h is an antiderivative for f and k is an antiderivative for g, then (h + k)′^ = h′^ + k′^ = f + g
and so h + k is an antiderivative for f + g. Thus, we have proved the following theorem
Theorem B. If f and g are functions and a is a constant, then
∫ af (x) dx = a
(1) f (x) dx ∫ (f (x) + g(x)) dx =
f (x) dx +
(2) g(x) dx
In other words, the integral of a constant times a function is the constant times the integral of the function and the integral of the sum of two functions is the sum of the integrals of the two functions.
Theorems A and B combined allow us to integrate (find the indefinite integral of) any polynomial.
Example 1. Integrate x^3 + 3x − 6.
Solution: ∫ (x^3 + 3x − 6) dx =
x^3 dx +
3 x dx +
− 6 dx Th B(2)
x^3 dx + 3
x dx − 6
1 dx Th B(1)
x^4 4
3 x^2 2
− 6 x + c Th A
Note that we can always check our answer to an integration problem by differentiating it to see if we get back the original function. For example, let’s differentiate the answer in the preceding example:
( x^4 4
3 x^2 2
− 6 x + c
4 x^3 4
3 · 2 x 2
− 6 + 0 = x^3 + 3x − 6
Indeed, we do get back our original function and this verifies that our answer was correct.
Example 2. Find the antiderivative of x^3 − 3 x^2 that has value 0 when x = 1. Solution: All antiderivatives of x^3 − 3 x^2 are given by ∫ (x^3 − 3 x^2 ) dx =
x^4 4
− x^3 + c
When x = 1 this becomes
1 / 4 − 1 + c = − 3 /4 + c
Thus, if we want the antiderivative that has value 0 when x = 1, we should choose c = 3/4. The answer is then x^4 4
− x^3 +
Example 3. The acceleration experienced by an object due to the force of gravity is −32ft/sec^2. A projectile is fired straight up with an initial velocity of 128 ft/sec, after which the only force acting on it is gravity. What is its velocity t seconds later? When does it reach its maximum height? Solution: Acceleration is the rate of change of velocity with respect to time - that is, it is the derivative of velocity as a function of time. We know the acceleration is −32ft/sec^2. Thus, the velocity v(t) is an antiderivative of the constant function −32. So
v(t) =
− 32 dt = − 32 t + c
- Area is well-defined; regions we shall consider in this course have area. Again, calculation of areas has been an ongoing problem since the beginning of recorded history, and the Calculus provides a most effective way of doing this, if we know that it makes sense. As we shall see below, area of a curvilinear figure is found by approximating the figure by collections of rectangles. That the approximations lead to a good concept of area is a subtle problem which for the time being, we shall ignore.
From time to time in this course we will touch on the difficulties with these assumptions, but for the present, we shall assume them as true statements.
Suppose that y = f (x) is a nonnegative function on the interval [a, b]. The area under the curve y = f (x) from a to b is denoted
∫ (^) b
a
f (x)dx.
Theorem. The area under y = f (x) is equal to F (b) − F (a), for any indefinite integral F (x) of f (x).
Demonstration. For x any point between a and b , let A(x) be the area under the curve from a to x. Let’s calculate A′(x). For a small increment h the area, A(x + h) up to x + h is approximated by A(x) + hf (x), where hf (x) is the area of the rectangle of base h and height f (x). Thus
(1) [A(x + h) − A(x)] = f (x)h + error,
where “error” is bounded by [f (x + h) − f (x)]h, if, for example, f is an increasing function. Thus A(x + h) − A(x) h
− f (x)
is bounded by f (x + h) − f (x) which will go to zero if f is at all a reasonable function (we say, f is continuous). We conclude
lim x−> 0
A(x + h) − A(x) h
= f (x).
Thus A(x) is an indefinite integral of f. If F is any other indefinite integral,
(2) A(x) − F (x) = C,
where C is some constant which we can easily find since A(a) = 0. Substituting in (2) we obtain C = −F (a). Thus, the area under the curve from a to b is A(b) = F (b) − F (a).
Examples
- Find the area under y = xn^ from 0 to 1.
An indefinite integral is F (x) = (^) n^1 +1 xn+1, so the area is F (1) − F (0) = (^) n+1^1.
- Find the area under the curve y = x^2 + 2x^3 from 2 to 4. ∫ (^4)
2
(x^2 + 2x^3 )dx =
x^3 + 2
x^4 |^42 =
The definite integral will be defined later in this course as the limit of a sum, as follows. Suppose that y = f (x) is defined over the interval [a, b]. Let us subdivide the interval by a sequence of partition points {xi}:
x 0 = a < x 1 < x 2 < · · · < xn− 1 < xn = b
Consider the sum
f (x 1 )(x 1 − x 0 ) + f (x 1 )(x 2 − x 1 ) + · · · + f (xn)(xn − xn− 1 )
which can be written, using the summation notation
Σn 1 f (xi)(xi − xi− 1 ).
If, for example, we were calculating the area under the graph of a positive function, this would be the area of a polygonal figure approximating the curvilinear figure, and would thus give an approximate value for the area. If we divide the figure more finely, using more points {xi}, we would expect to get a better approximation to the area. Under suitable conditions (the continuity of the function f ), these sums do approach a limit which is the definite integral of the function f from a to b.
The Fundamental Theorem of the Calculus will tell us that the definite integral can be computed as F (b) − F (a), where F is any indefinite integral of f.
Example. Suppose a particle moves in a straight horizontal line so that its velocity directed to the right at time t is v(t) = t^2 − t^3 meters per minute. How far to the left or right of the initial position of the particle will it be after 2 minutes? Answer: ∫ (^2)
0
v(t))dt =
0
(t^2 − t^3 )dt =
t^3 −
t^4 |^20 =
the particle is 4/3 of a meter to the left of its initial position.
