The magnitude of the range R of exhaust temperatures (in...

The magnitude of the range $R$ of exhaust temperatures (in degrees Fahrenheit) in an experimental diesel engine is approximated by the model $R=\frac{5000(4-3 x)}{(11-7 x)(7-4 x)}, \quad 0<x \leq 1$ where $x$ is the relative load (in foot-pounds). (a) Write the partial fraction decomposition of the equation. (b) The decomposition in part (a) is the difference of two fractions. The absolute values of the terms give the expected maximum and minimum temperatures of the exhaust gases for different loads. $\mathrm{Ymax}=| 1$ st term $|\quad \mathrm{Ymin}=| 2 \mathrm{nd}$ term $|$ Write the equations for Ymax and Ymin. (c) Use a graphing utility to graph each equation from part (b) in the same viewing window. (d) Determine the expected maximum and minimum temperatures for a relative load of 0.5.

The magnitude of the range $R$ of exhaust temperatures (in degrees Fahrenheit) in an experimental diesel engine is approximated by the model
$R=\frac{5000(4-3 x)}{(11-7 x)(7-4 x)}, \quad 0<x \leq 1$
where $x$ is the relative load (in foot-pounds).
(a) Write the partial fraction decomposition of the equation.
(b) The decomposition in part (a) is the difference of two fractions. The absolute values of the
terms give the expected maximum and minimum temperatures of the exhaust gases for different loads.
$\mathrm{Ymax}=| 1$ st term $|\quad \mathrm{Ymin}=| 2 \mathrm{nd}$ term $|$
Write the equations for Ymax and Ymin.
(c) Use a graphing utility to graph each equation from part (b) in the same viewing window.
(d) Determine the expected maximum and minimum temperatures for a relative load of 0.5.

Key Concepts

Graphing Functions

Graphing involves plotting a function to visually examine its behavior, such as identifying maximum and minimum values, asymptotes, and intervals of increase or decrease. It is an important tool for interpreting the results of a function in various applications and aids in verifying analytical solutions.

Partial Fraction Decomposition

This is an algebraic technique used to express a rational function as a sum or difference of simpler fractions. It simplifies the handling of complex fractions by breaking them into parts that are easier to integrate, differentiate, or analyze, especially when the denominator can be factorized into linear or irreducible quadratic factors.

Rational Functions

These are functions defined by the ratio of two polynomials. Understanding their structure, including how the degrees of the numerator and denominator affect the function's behavior, is crucial for simplifying expressions, determining asymptotic behavior, and solving equations.

Application of Mathematical Models

Mathematical models, such as those in engineering or the sciences, use equations to represent real-world phenomena. They allow for the estimation of outcomes based on variable inputs and often require techniques like partial fraction decomposition and graphing to analyze and interpret the behavior of the system under study.

Transcript

00:01 In this cution, rational expression, r equal to 5 ,000 times 4 negative 3 times x upon 11 negative 7 times x time 7 negative 4 times x where x greater than to 0 less than an equal to 1.

00:32 The magnitude of the range are of exhaust temperature in the an experimental diesel engine is approximated by the model which is given and here is limit x greater than to zero x less than an equal to 1, where x is a relative load.

00:58 First of all, we have to write partial fraction decomposition of the equation.

01:03 So 5 ,000 times 4 negative 3 times x upon 11 negative 7 times x times 7 negative 4 times x equal to now you can see that here the degree of denominator is greater than to the degree of numinator.

01:32 So we can write form of decomposition a upon 11 negative 7 time x plus b upon 7 negative 4 time x now we have to take least common denominator and we get 11 negative 7 time x time 7 negative 4 time x in the denom numerator we get a time 7 negative 4 time x plus b time 11 negative 7 time x now we have to multiply both sides by 11 negative 7 time x time 7 times x time 7 negative 4 time x then we get 5 ,000 time 4 negative 3 times x equal to a time 7 negative 4 times x plus b time 11 negative 7 time x now we put x equal to 7 upon 4 so we get negative 6 ,250 equal to negative 5 upon 4 time b so b equal to positive 5, 5 ,000 now we put x equal to 11 .11 upon 7, then we get negative 25 ,000 upon 7 equal to 5 upon 7 time so a equal to negative 5 ,000.

03:49 So now we put that value in this equation and we get 5 ,000.

03:56 Time 4 negative 3 times x upon 11 negative 7 time x time 7 negative 4 time x equal to negative 5 ,000 upon 11 negative 7 time x plus 5 ,000 upon 7 negative 4 times x plus 5 ,000 upon 7 negative 4 time x so now you can see that it is our partial fraction decomposition of the equation...

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