1. Rational Functions and Models
Chapter 3
Section 3.5
Rational Functions and Models
This module continues our exploration of function types with the introduction of rational functions. A number of useful models can be constructed from this type of function.
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2. Rational Functions and Models
Definition
f(x) is a rational function if and only if f(x) =
Examples
1. f(x) =
2. f(x) =
3. f(x) =
p(x)
q(x)
where p(x) and q(x) are polynomial functions with q(x)  0
3x2 + 4x + 1
x3 – 1
Rational Functions: Definition
The most common error students make in identifying rational functions is assuming that any quotient function is a rational function. Care must be taken to identify each of the numerator and denominator as a polynomial. No other type of quotient function is a rational function.
Note that some rational functions can be reduced to simple linear functions, provided the zeros of the denominator are excluded. This is the case in Example 2.
Note that not all quotient type functions are rational functions, as is illustrated in Example 3. The fractional power in the numerator indicates the cube of a square root of x. Since polynomials must have integer powers in all terms, the numerator is not a polynomial and hence the quotient f(x) is not a rational function.
3x2 – 27
x – 3 =
3(x – 3)(x + 3)
x – 3 =
3x + 9
, for x ≠ 3
x3/2 – 8
x2 + 1
Question:
Is this a rational function ? 2
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3. Rational Functions and Models
Domains
Where are rational functions defined?
Examples
1. f(x) =
2. f(x) =
3. f(x) =
3x2 + 4x + 1
x3 – 1
3x2 + 4x + 1
(x – 1)(x2 + x + 1) =
Dom f(x) = { x | x ≠ 1 }
3x2 – 27
x – 3
Dom f(x) = { x | x ≠ 3 }
Domains of Rational Functions
Rational functions are not defined where the denominator polynomial has value zero. Thus, finding the zeros of polynomials is important.
Note that in the first example the second degree factor in the denominator is not zero for any real numbers.
6x2 – x – 2
x2 + x – 6
6x2 – x – 2
(x + 3)(x – 2) =
Dom f(x) = { x | x ≠ – 3, x ≠ 2 }
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4. Rational Functions and Models
Asymptotes
Asymptotes are lines that the graph of a function f(x) approaches closely as x approaches some k or ±
Vertical Asymptote: line x = k such that either f(x)
Horizontal Asymptote: line y = k such that f(x)
as x ∞ ∞
or f(x) as x approaches k ∞ – k ∞
or x ∞ –
Rational Functions: Asymptotes
Because the behavior of the graph of a rational function depends on the behavior of the numerator and denominator polynomials, relative to each other, it is possible that the graph may approach certain lines as closely as desired, possibly never touching them. These lines are called asymptotes.
By the way, it is possible for a graph to cross its horizontal asymptote. We shall take this up in detail a bit later.
Question:
Can the graph of f(x) cross its asymptotes ?
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5. Rational Functions and Models
Asymptote Examples
1. f(x) =
2. f(x) = x y 1 x
y = 0
x = 0 x y 1
x + 1
Asymptote Examples
These simple examples illustrate the behavior of rational functions in general and their differences from simpler types such as polynomial functions.
Asymptotes are a common characteristic of rational functions. Vertical asymptotes marking values of the independent variable where the function is not defined, i.e. values of the independent variable x that are not in the domain of the function. These values of x are typically where the denominator of the function has a zero value. Horizontal asymptotes are lines the graph approaches with the ultimate behavior of the function, i.e. a value the functional values approach as x grows without bound in either direction.
Using this criterion, we see in the first example that the denominator is zero exactly when x = 0. So, the vertical line x = 0 (the y-axis) is a vertical asymptote for the graph of the function. Since f(x) approaches 0 as x grows without bound in either direction, the graph of f(x) has a horizontal asymptote at y = 0, the x-axis.
In the second example, the denominator is zero when x = -1, which is also the equation for the vertical line which is the vertical asymptote for the function. As before, as x grows without bound in either direction, the value of the function approaches zero. This give the graph of the function a horizontal asymptote at y = 0, i.e, the x=axis.
x = –1
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6. Rational Functions and Models
Asymptote Examples
3. f(x) =
4. f(x) = x y
y = 1
x + 1
x – 5
x = 5 x y
Asymptote Examples
In Example 3, both numerator and denominator grow at roughly the same rate as x grows in either direction. For very large x, as x grows large in either positive or negative direction, f(x) is approximately 1 and is positive. Thus the line y = 1 is a horizontal asymptote for the graph of f(x).
As x approaches 5 from the positive side, the denominator approaches 0, so that the fraction representing f(x) values grows without bound in the positive direction. As x approaches 5 from the negative side, that is x < -5, the numerator is positive but the denominator is negative, making values of f(x) very negative, that is f(x) grows without bound in the negative direction as x gets closer to 5 from the left. Thus, the line x = 5 is a vertical asymptote for the graph of f(x).
Example 4 appears to show a rational function with a denominator that grows faster than the numerator. However, noting that the numerator is a difference of squares,
which is true only for x ≠ 2. So, f(x) does not exist at x = 2, but for all other values of x we see that f(x) = x + 2, which is just a linear function, and of course has no asymptotes. The graph is thus a line with a hole in it at x = 2.
x2 – 4
x – 2 ● 2
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7. Rational Functions and Models
Asymptote Examples
5. f(x) =
Sketch asymptotes, intercepts and the graph x y
Function is linear, but undefined at x = 5/2
4x – 10
2x – 5
(0, 2) ● ●
5/2
4x – 10
2x – 5
f(x) =
2(2x – 5)
2x – 5 =
= 2
Domain ?
… provided x ≠ 5/2
Domain = { x │ x ≠ 5/2 }
Asymptote Examples
Example 5 shows another function that is not what it seems at first blush. Noting that the numerator can be factored in such a way that one of the factors is 2x – 5, identical to the denominator, we see that f(x) can be reduced to 2, for the denominator not equal to 0. The denominator 2x – 5 is zero when x = 5/2, so f(5/2) does not exist, but f(x) = 2 for all other values of x. So the graph is a line with a hole in it.
The domain of f is thus the set of all real numbers except x = 5/2. The range of f has only one value: 2.
As the illustration shows, the graph of f(x) is the horizontal line y = 2, but this is not an asymptote. This function has no asymptotes.
Since every function defined over a domain that includes 0 has a vertical intercept at (0, f(0)), we see that f has a vertical intercept at (0, 2). Since the graph is a horizontal line it has no horizontal intercepts.
= ( – , 5/2 ) ( 5/2 , ) ∞ 
No asymptotes !
One intercept:
(0, 2)
Range ?
Range = { 2 } 7
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8. Rational Functions and Models
Asymptote Examples
6. f(x) =
f(1) does not exist since
x3 – 1 = 0 when x = 1 x y
11x – 2
x3 – 1 
f(x)
Line y = 0
as x ∞
Horizontal asymptote
Line x = 1
Vertical asymptote
Asymptote Examples
This rational function dispels the commonly held myth that a graph can never touch an asymptote. Clearly the graph of this function does cross the horizontal asymptote y = 0 when x = 2/11.
If we alter the function slightly by changing the power of x in the denominator from 3 to 2, giving
the graph still has y = 0 as a horizontal asymptote and the graph still cuts this asymptote.
What is radically different is that the graph now comes in three disjoint pieces with two vertical asymptotes. It is instructive to sketch the graph, or display it on a graphing calculator.
Another interesting example is
with horizontal asymptote as the line y = 1. These are easily displayed on a graphing calculator even if the sine function is not well understood at this time. The graph of f(x) in this case crosses the horizontal asymptote y = 1 infinitely many times.
Question:
Does the graph cross an asymptote ?
YES !
, 0 2 11 ) (
f(x) = 0 at x = 2 11
so the graph cuts the asymptote y = 0 at
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9. Rational Functions and Models
Asymptote Review
Vertical Asymptote: line x = k such that either f(x)
Horizontal Asymptote: line y = k such that f(x)
as x ∞
or f(x) as x approaches k ∞ – k ∞
or x ∞ –
Question:
Can the graph of f(x) cross its asymptote ?
Rational Functions: Asymptotes
Because the behavior of the graph of a rational function depends on the behavior of the numerator and denominator polynomials, relative to each other, it is possible that the graph may approach certain lines as closely as desired, possibly never touching them. These lines are called asymptotes.
By the way, it is possible for a graph to cross its asymptote. Consider the graphs of
and
which have horizontal asymptotes of and , respectively. This is
easily seen by setting and solving for x, giving In the
case of g(x) it is clear that evaluating at yields Hence the graph crosses both asymptotes. Sketch this graph to see how this works.
f(x) =
5x2 + 8x – 3
3x2 + 2
and
g(x)
11x + 2
2x3 – 1
Consider
... and their horizontal asymptotes
What about vertical asymptotes ?
Question:
Are asymptotes always vertical or horizontal ? 9
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10. Rational Functions and Models
Asymptote Examples
7. f(x) =
Since 3x2 + 2 is never zero
for any real x there is no
vertical asymptote
Rewriting
Thus f(x) has a horizontal asymptote of y = x y
5x2 + 8x – 3
3x2 + 2
Horizontal asymptote
y = f(x)
y = 3 5 ● 3 5 19 24 ( ) , 5 x 8 – x2 3 + 2
f(x) = 3 5
Asymptote Examples
In Example 7 we see both numerator and denominator are second degree polynomials, so they grow at about the same rate. Since the lead term in any polynomial dominates the rest of the polynomial as the variable grows without bound in either direction, we expect the fraction to reduce to the ratio of the two lead coefficients, giving a horizontal asymptote. This is exactly what happens and we see f(x) approaching the number 5/3, so that f(x) has a horizontal asymptote of y = 5/3.
Clearly the denominator 3x2 + 2 is never zero, so f(x) has no vertical asymptote.
One question that often arises from misunderstanding of what an asymptote really is concerns whether or not a function’s graph can touch, or even cross, one of its asymptotes. The graph in the illustration should answer this. Setting the expression for f(x) equal to 5/3, we find that the functional value at x = 19/24 ≈ So, indeed the graph can touch (and cross) one of its asymptotes.
Remember: asymptotic behavior deals only with the graph for very large values of the variable, either positive or negative. It’s the ultimate behavior of the graph that counts.
as x ∞ 5 3
Question:
Where does the graph cross its asymptote ?
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11. Rational Functions and Models
Asymptote Examples
8. f(x) = x y
x2 + 2x + 1
x – 1
f(x) = x + 3 + 4
x – 1
By synthetic division ...
y = x + 3 1
Oblique (slant) Asymptote 1 3 1 3 4
Thus
x = 1
f has a vertical asymptote at x = 1
Vertical Asymptote
Asymptote Examples
Example 8 shows a rational function with a numerator that clearly grows faster than the denominator. Also f(x) clearly does not exist at x = 1 and, as x approaches 1 from either side, the functional values increase without bound (in the positive direction for x > 1 and in the negative direction for x < 1). This tells us the graph of f(x) has a vertical asymptote at x = 1.
If we use long division (or synthetic division) to divide x – 1 into x2 + 2x + 1 we get
and since, 4/(x – 1) approaches 0 as x grows in either direction, this forces f(x) to approach x So as x gets very large, f(x) is closer and closer to x That is, the graph of f(x) approaches the graph of y = x + 3, which is a line with slope 1 and y-intercept (0, 3). This line is called an oblique asymptote, a variation of the horizontal asymptote, that describes the ultimate (end) behavior of f(x). Thus long division produces a linear expression that approximates the rational function for large x.
Oblique asymptotes occur when the degree of the numerator is greater than the degree of the denominator by 1.
Note: The converse of the above statement is not true; that is, when deg (numerator) = deg (denominator) + 1 the function might not have either a horizontal or oblique asymptote, as shown in earlier examples of linear functions with a “hole” in the graph. We say this condition is necessary but not sufficient for an oblique asymptote to exist.
and
f(x) y = x + 3 as x ± 
Oblique Asymptote:
line y = ax + b such that f(x) y
as x ∞
Occurs when deg (numerator) = deg (denominator) + 1
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12. Rational Functions and Models
Finding Vertical and Horizontal Asymptotes
Vertical Asymptotes for f(x) =
Find values of x , say x = k, where q(x) = 0
f(x) fails to exist at x = k
… BUT might not have an asymptote there
if (x – k) is also a factor of p(x)
Ensure f(x) is reduced to lowest terms
Check x = 0 for vertical intercept
Horizontal and Oblique Asymptotes for f(x) =
Determine what f(x) approaches as x approaches
Check f(x) = 0 for horizontal intercept
p(x)
q(x)
Finding Vertical and Horizontal Asymptotes
The process for finding the vertical, horizontal, and oblique asymptotes for a rational function is listed here. The procedures are as follows:
For a vertical asymptote for the graph of rational function f(x)
Find the value of x that makes the denominator equal to zero
Ensure the denominator is not a factor of the numerator (else might not have an asymptote)
Ensure the numerator not also zero (else function not in reduced form, i.e. x = k makes both numerator and denominator zero)
For a horizontal or oblique asymptote for the graph of rational function f(x)
Simplify f(x) via long (or synthetic) division if possible
Consider what happens to f(x) as x grows without bound in either direction
Intercepts can also be used to verify the positioning of the graph. Vertical intercepts are easy to find by setting x = 0 and finding f(0).
p(x)
q(x) ∞
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13. Rational Functions and Models
Find Horizontal / Oblique Asymptotes
9. f(x) =
10. f(x) =
11. f(x) =
4x3 – 2
x + 2
6x2 – x – 2
2x2 – 3x + 4
Horizontal / Oblique Asymptote Examples
Example 9 shows a rational function with a numerator that clearly grows faster than the denominator. Also f(x) clearly does not exist at x = –2, indicating a vertical asymptote at x = –2. Long division gives
Clearly, as x grows “large” the trailing fraction approaches zero, leaving the parabolic form 4x2 – 8x + 16 as the “ultimate” graph; that is, the graph of f(x) approaches this parabola as x grows large. Thus, there is no horizontal or oblique asymptote. On a graphing calculator, entering f(x) as one function and the parabola y = 4x2 – 8x + 16 as a second function and using the TRACE facility, it is clear that the graph of f(x) approaches the graph of the parabola asymptotically.
In Example 10, long division yields
Clearly, in this case the trailing fraction approaches zero as x grows large, leaving the constant 3. Thus, the line y = 3 is the horizontal asymptote.
Example 11 shows a function with a vertical asymptote at x = 1. It is a repeat of Example 8 and is included just for practice.
x2 + 2x + 1
x – 1
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14. Rational Functions and Models
Rational Functions for “Large” x
For the function
R(x) =
3x + 1
x – 2
fill in the table x
3x + 1
x – 2
R(x)
, , , ,000,000 10 31
301
3,001
30,001
300,001
3,000,001 1 8 98
998
9,998
99,998
999,998 10
3.9
3.07
3.007
3.0007
Thus as x
Rational Functions with “Large” x
The purpose of this exercise is to show the student how the various parts of a rational function grow as the value of the input variable grows very large. This will give meaning to such expressions as “x → ” and “R(x) → 3”.
While much larger values of x can be chosen, the ones selected here should give a good sampling of “large” values for x. By examining the corresponding values of 3x + 1, x – 2 , and R(x) one can see something of the behavior of these expressions when x grows quite large.
By seeing that R(x) really does approach 3 as x →  it should be easier to understand that the line y = 3 is a horizontal asymptote for the graph of the function R(x).
Another approach to finding horizontal asymptotes for rational functions is to apply the technique of dividing both numerator and denominator by the highest power of x appearing in either. In this case we would divide numerator and denominator by x:
This causes all fractions to have x in the denominator which forces those fractions to 0 as x grows without bound. What is left are the pieces that have no x and make up the value of the horizontal asymptote.
1,000,000 we see that
3x ,000,001 ≈ 3,000,000 = 3x
x – ,998 ≈ 1,000,000 = x
R(x) ≈ 3 = (3x)/x
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15. Solving Rational Equations
Rational Functions and Models
Rational Functions and Models
h(x) =
g(x)
f(x)
How do we solve equations of form:
Method 1: Clear Fractions
Examples:
1. Solve:
= 15 23
x + 2
= 15 23
x + 2
· (x + 2)
23 = 15x + 30
Solving Rational Equations: Method 1 – Clear Fractions
For equations with rational functions, a reasonable first step might be to simplify the equation by clearing the fractions. In the first example, this can be done by multiplying both sides of the equation by the binomial (x + 2), giving the simpler linear equation 23 = 15x + 30, which is easily solved as shown.
–7 = 15x = –7 15 x
Solution Set: { } 7 15 – 15
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16. Solving Rational Equations
Rational Functions and Models
Rational Functions and Models
Method 1: Clear Fractions
2. Solve: =
x + 1
x + 2
x + 5
x + 7
· (x + 2)(x + 7) =
x + 1
x + 2
x + 5
x + 7
(x + 5)(x + 2) =
(x + 1)(x + 7)
x2 + 7x + 10 =
x2 + 8x + 7
7x + 10 =
8x + 7
Solving Rational Equations: Method 1 – Clear Fractions
In this example the numerators and denominators of both sides are 1st-degree polynomials. By multiplying both sides by the common denominator of the two rational functions we can clear the fractions on both sides of the equation at the same time. Of course this could be done in two separate operations, first multiplying by (x + 2) on both sides and then by (x + 7) on both sides.
This yields a product of binomials on each side of the equation with no common factor. Performing the indicated multiplication and simplifying, we obtain a simple linear equation in x, which is easily solvable.
In case the square term does not cancel, then we have to solve a quadratic equation, which is also solvable. 3 = x
Solution Set: { 3 }
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17. Solving Rational Equations
Rational Functions and Models
Rational Functions and Models
Method 2: Cross Multiplication
Basic Principle:
Examples:
1. Solve: = a b c d
if and only if ad = bc =
x + 1
x + 2
x + 5
x + 7
(x + 2)(x + 5) =
(x + 1)(x + 7)
Solving Equations: Method 2 – Cross Multiplication – Example 1
The basic principle can be established simply by clearing the fractions, as in the previous examples. Multiplying both sides of the initial equation by bd yields the equivalent equation: ad = bc .
In the first example, cross multiplication produces the very same result as in the second step of Method 1 applied to this equation. Since no factors cancel, we carry out the multiplication. We can cancel the square terms, leaving just a simple linear equation to solve. This solution is shown in the illustration.
x2 + 7x + 10 =
x2 + 8x + 7
7x + 10 =
8x + 7 3 = x
Solution Set: { 3 }
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18. Solving Rational Equations
Rational Functions and Models
Rational Functions and Models
Method 2: Cross Multiplication
2. Solve:
Cross multiplying =
x – 3 7 1
x + 3 7 =
(x – 3)(x + 3) 7 =
x2 – 9
Factoring and Zero Product Property
Square Root Property =
x2 – 16 16 = x2
Solving Equations: Method 2 – Cross Multiplication – Example 2
Again the basic principle of clearing the fractions can be used, as in the previous examples, or the immediate consequence of that method – cross multiplication can used.
Multiplying the numerator of the first rational function by the denominator of the second and the denominator of the first by the numerator of the second gives a factored polynomial equation. As shown, this new equation can be solved in either of two ways: by factoring into conjugate binomials and using the zero product property, or by forming a perfect square on both sides and then finding the square roots of each side using the square root property.
Both methods are shown. =
(x + 4)(x – 4) = x2 √ 16 =
x + 4 OR =
x – 4 = x 4
– 4 = x 4 = x
Solution Set: { – 4, 4 }
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19. Solving Rational Equations
Rational Functions and Models
Rational Functions and Models
Method 2: Cross Multiplication
3. Solve:
Cross multiplying =
x + 1
5x – 3 2 3
2(5x – 3) =
3(x + 1)
10x – 6 =
3x + 3 7x = 9 = 9 7 x
Solving Equations: Method 2 – Cross Multiplication – Example 3
Again the basic principle of clearing the fractions can be used, as in the previous examples, or the immediate consequence of that method – cross multiplication can used.
Multiplying the numerator of the first rational function by the denominator of the second and the denominator of the first by the numerator of the second gives a factored linear equation. The usual method of isolating the variable is used to solve the linear equation.
This form of the equation is especially suited to an approximate graphical solution. Graphical verification of the algebraic method is probably the most accurate.
Solution Set: { } 9 7
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20. Solving Rational Equations
Rational Functions and Models
Rational Functions and Models
Method 3: Graphical Approach
1. Solve:
Let y1 = =
x + 1
x – 5 2 x y 3 6 9 11 –2 –3
x + 1
x – 5 y1
Horizontal Asymptote y = 1 y2
and y2 = 2
(11, 2) 
So y1 = y2
For what x is this true ?  
Solving Equations: Method 3 – Graphical Approach
Here we use a technique that we have used in dealing with inequalities. We divide the equation into two expressions and use these expressions to define two distinct functions, y1 and y2. We can then graph these two functions to find the intersection of the graphs – that is, the values of x and y where the two expressions in the original equation are equal.
As shown, we define y1 as
and y2 as y2 = 2. By graphing each of these functions and locating the intersection of their graphs we can find the point (x,y) that they have in common.
The value of x for this point is thus the solution of the original equation. As we see, the graphs intersect at the point (11, 2) and we conclude that, since this point is on both graphs, this ordered pair it satisfies the equation
so that 11 is the value of x that we seek. Note that 11 must be in the domain of both functions. Why?
The technique illustrated here is one well adapted to solution with a graphing calculator. A demonstration could be done in class to illustrate the method.
Vertical Asymptote x = 5
Intercepts for y1 :
Horizontal : ( –1, 0 )
Intersection at (11, 2)
Vertical : ( 0, –1/5 )
Hence: x = 11
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21. Direct and Inverse Variation
Rational Functions and Models
Rational Functions and Models
Direct Variation
Output varies directly with input
Example: y = kx
Inverse Variation
Output varies inversely with input
Example: y = kx–1
Inverse Variation Functions
Output varies inversely with xn
Example: y = kx–n y x k = OR
k is the constant of variation
yx = k OR
Direct and Inverse Variation
The discussion of rational functions leads directly to the notion of direct versus inverse variation.
We say an output (or dependent variable, or functional value) y varies directly with an input (independent variable) x if changes in y are proportional to changes in x. That is, the ratio of y to x is a constant, called the constant of variation or constant of proportionality.
Notice in each case there is an expression in terms of the variables that is constant:
yx = k inverse variation
yx–1 = k direct variation
yxn = k function of variation
yxn = k OR
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22. Rational Functions and Models
Variation Examples
Rational Functions and Models
Rational Functions and Models
Direct Variation
The resultant force acting on an object of mass m is directly proportional to the acceleration of the object
F = ma
F varies directly with a Newton’s Second Law
Constant of variation is m OR
= m F a
Variation Examples: Direct Variation
In the case of direct variation, the example shows that acceleration increases in direct proportion to the force applied. This is just a statement of Newton’s second law of mechanics. If the force is doubled, so is the acceleration. It is useful to note that this means that the ratio of F to a is constant, and that the constant is simply the mass m of the object. 22
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23. Rational Functions and Models
Variation Examples
Rational Functions and Models
Rational Functions and Models
Direct Variation Example
Weight and Mass
Excluding other external forces, the only force acting on an object of mass m is the force of gravity mg, where g is the acceleration due to gravity
To lift the object, the force of gravity must be overcome
This force is called weight and given by F = W = mg
Clearly weight varies directly with mass
; that is
F = mg
Direct Variation Example
In this case acceleration due to gravity is constant, while the mass m can be varied. Thus the gravitational force, or weight, exerted on the mass by gravity is proportional to the mass m and varies directly with m. If the mass is doubled then the weight is doubled under the constant acceleration of gravity. It is useful to note that this means that the ratio of F to m is constant, and that the constant is simply the gravitational acceleration g. 23
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24. Rational Functions and Models
Variation Examples
Rational Functions and Models
Rational Functions and Models
Inverse Variation
At constant temperature the volume of n moles of gas is inversely proportional to the pressure of the gas
PV = nRT OR P = (nRT)V–1
P varies inversely with V Ideal Gas Law
Constant of variation is nRT
Variation Examples: Inverse Variation
In the case of inverse variation, the example shows that the pressure and volume of gas in a closed system vary inversely with each other. We stress the importance of the closed system because in this situation the amount of gas (measured in moles) does not change. It’s not like blowing up a balloon where we blow more air (more moles of gas) into the balloon; instead we are comparing more to a soft air-filled rubber ball that can be squeezed to reduce the volume of air inside but only by applying more pressure than the ball already has. We note that the when we release the ball it pops back to its normal size and shape, since the pressure is reduced.
In this example, notice that the product of P and V is constant. 24
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25. Rational Functions and Models
Variation Examples
Rational Functions and Models
Rational Functions and Models
Function of Variation
The earth’s gravitational force acting on an object of mass m is inversely proportional to the square of the distance between the mass and the center of the earth F
F varies inversely with r Law of Gravity
Constant of variation is GMm, where G is the earth’s gravitational constant, M is the mass of the earth, and m is the mass of the object
This is just one of many inverse square laws =
GMm r2
OR Fr2 = GMm
Variation Examples: Function of Variation
The example shown here is typical of inverse square laws. The earth, with mass M, attracts other objects of mass m via a gravitational force inversely proportional to the distance of the earth from the objects. This is generally true for any two masses m1 and m2 with a constant G dependent on the nature of the objects.
As the object m moves away from the earth the gravitational attraction drops as the square of the distance. For example, if the distance is doubled, to 2r, then the force is diminished by a factor of four. This looks like
so that the reduced force is only one fourth of the original force. Similarly, if the distance is cut in half, the force is increased to four times the original force. 25
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Think about it ! 26
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