Addition and Subtraction Properties of Equality: Let , , and represent algebraic expressions. Questions on solving linear and quadratic equations, simplifying expressions including expressions with fractions, finding slopes of lines are included.
Compare the slopes, y - intercepts, and their effects on the graphs. They have kindly allowed me to create 3 editable versions of each worksheet, complete with answers. See Notes, p. Linear equations worksheet with answers pdf.
Linear equations worksheet with answers pdf grade 8. Find the intercepts of the graph of the equation. Math workbook 1 is a content-rich downloadable zip file with Math printable exercises and pages of answer sheets attached to each exercise.
Exploring slope formula using an online interactive program. Source: saylordotorg. Order of Operations — Challenge Problems —. Free Printable Easter Pictures.
This is shown in the following examples Example Pin On Algebra. Downloads: x. Two step equation word problems. Solving equations where one is linear and the other quadratic. Graph a linear equation given an equation. Find the value of y. Detailed solutions are included.
Answer key solving equations and inequalities worksheet. This worksheet will give some practice in one variable, then move on to solving equations in two variables, then solving quadratics. If no fractions, combine like terms.
Determine the solution to a system of linear equations in two variables from the graphs of the equations. Steps example step 1. For the equation, complete the table for the given values of x. Divide this sum by e. Answer: 14, 16 or , 16 4. Now that we have x, we use it to find y. Just begin by. In a linear equation, y is called the dependent variable and x is the independent variable. Convert the following statements into equations. Algebra mathematics math solving linear equations.
This free worksheet contains 10 assignments each with 24 questions with answers. These math worksheets should be practiced regularly and are free to download in PDF formats. You must show all your working out. Lesson 6 - Arithmetic Sequences. III and IV. Compare the properties of two functions represented in different ways, including tables, graphs, equations, and written descriptions. Shed the societal and.
Alfredo Tan of Fairleigh Dickinson University. Cafone and to the members of the Priest Community; we also thank Dr. Fredrick Travis, Dr. James Van Oosting, Dr. Molly Smith, and Dr. We also thank Colonel Gary W. Krahn of the United States Military Academy. Barbara Gilson and Ms. Carol Cooper, our contact in the United Kingdom, was equally helpful. Thank you, one and all. Click here for terms of use.
Example 1. With the exceptions of Chapters 31 and 34, the primary focus of this hook mil he ordinary differential equations. The order of a differenlial equation is the order of the highest derivative appearing in the equation.
Equation l. Thus, y, y, and y represent dyldt, d2yldt2, and d3yldt3, respectively. Note that the left side of the differential equation must be nonnegative for every real function y x and any x, since it is the sum of terms raised to the second and fourth powers, while the right side of the equation is negative.
Since no function y x will satisfy this equation, the given differential equation has no solution. We see that some dinerential equations have mnnitely many solutions Example 1. It is also possible that a differential equation has exactly one solution. A particular solution of a differential equation is any one solution.
The general solution of a differential equation is the set of all solutions. The general solution to the differential equation in Example 1. That is, every particular solution of the differential equation has this general form. The general solution of a differential equation cannot always be expressed by a single formula. The subsidiary condi- tions are initial conditions.
If the subsidiary conditions are given at more than one value of the independent variable, the problem is a boundary-value problem and the conditions are boundary conditions. A solution to an initial-value or boundary-value problem is a function y x that both solves the differential equation and satisfies all given subsidiary conditions.
Solved Problems 1. Determine the order, unknown function, and the independent variable in each of the following differential equations: a Third-order, because the highest-order derivative is the third. The unknown function is y; the independent variable is x. The unknown function is y; the independent variable is t. The unknown function is t; the independent variable is s. Raising derivatives to various powers does not alter the number of derivatives involved.
The unknown function is b; the independent variable is p. Determine the order, unknown function, and the independent variable in each of the following differential equations: a Second-order. The unknown function is x; the independent variable is y. The unknown function is x; the independent variable is t. Differentiating y x , it follows that Substituting these values into the differential equation, we obtain Thus, y x is a solution. Since y x is a solution of the differential equation for every value of c1; we seek that value of Cj which will also satisfy the initial condition.
Since y x is a solution of the differential equation for all values of Cj and c2 see Example 1. Thus, to satisfy both boundary conditions simultaneously, we must require c2 to equal both 1 and - 2, which is impossible. Therefore, there does not exist a solution to this problem. Which of the following functions are solutions of the differential equation 1.
In this chapter, and in other parts of ihc book sec Chapter 7. Chapter 14 and Chapter This equation is called Ohm's Law. Ohm , a German physicist, Onee Constructed, some models can he used to predict main physical situations. Kor example, weather forecasting, the growth of a tumor, or the outcome of a roulette wheel, ean all be connected with some form of mathematical modeling.
Chapter 34 introduces the idea of difference equations. These are equations in which we consider discrete variables; that is, uiriables which can lake on only certain values, such as whole numbers.
With few modifications, everything presented about modeling w i t h differential equations also holds true viith regard to modeling with difference equations. Research may be able to model this situation in the form of a "very difficult" differential equation. Technology may be used to help us solve the equation computer programs give us an answer. The technological answers are then interpreted or communicaied in light of the real-life situation the amount of radio-active material.
Once they are formulated, models may be virtually impossible to solve analytically. This is a valid approach, provided the simplification does not overly compromise the "real-world" connection, and therefore, its usefulness. This represents a qualitative approach. While we do not possess an exact, analytical solution, we do obtain some information which can shed some light on the model and its application.
Technological tools can be extremely helpful with this approach see Appendix B. Solved Problems Problems 2. Assume the models are valid, even in the cases where some of the variables are discrete.
This model converts temperatures from degrees on the Celsius scale to degrees on the Fahrenheit scale. This models ideal gases and is known as the Perfect Gas Law. What does Boyle's law tell us? Another way of stating this, is that the pressure and volume are inversely proportional.
Discuss the model: This formula is used in electricity; I represents the current amperes , q represents the charge coulombs , t is the time seconds. Problems involving this model will be presented in both Chapter 7 and Chapter Discuss the model: This is a classic model: a forced, mass-spring system.
Variations of this model can be used in problems ranging from shock absorbers on an automobile to answering questions about the human spinal column.
The differential equation uses a number of classical concepts, including Newton's second law and Hooke's law. We will revisit this equation in Chapter Assume M f represents the mass of an element in kgs. Set up a model for this relationship. This equation will be classified as a "separable equation" see Chapter 3.
The solution to this differential equation, which is qualitatively described as "exponential decay", will be explored in Chapter 4. Consider the previous problem. Assume research revealed that the rate of decay is proportional to the square root of the amount present. Model this situation. We note here that the units of k are The solution of this type of differential equation will be explored in Chapter 4.
Model a population P t , if its rate of growth is proportional to the amount present at time t. This is the sister problem to Problem 2. Assume the population described in Problem 2. Discuss this model. We are forced to conclude that this is most probably not a reasonable model, due to the fact that our growth is unlimited. We do add, however, that this model might be helpful over a short period of time.
Consider the assumptions in the two previous problems. Further, suppose the rate of growth of P t is proportional to the product of the amount present and some "maximum population" term, , - P f , where the , represents the carrying capacity. This would closely approximate exponential growth. So, for "small" P t , there would be little difference between this model and the previous model discussed in Problems 2.
So "in the large", P t "levels off" to ,, the carrying capacity of the population. In this problem, we used a qualitative approach: we were able to decipher some information and express it in a descriptive way, even though we did not possess the solution to the differential equation. This type of equation is an example of a logistic population model and is used extensively in sociological studies.
Also see Problem 7. Sometimes differential equations are "coupled" see Chapter 17 and Chapter 25 ; consider the following system: Here, let R represent the number of rabbits in a population, while F represents the number of foxes, and t is time months. Assume this model reflects the relationship between the rabbits and foxes. What does this model tell us? This system of equations 1 mirrors a "predator-prey" relationship. The RF terms in both equations can be interpreted as an "interaction term".
That is, both factors are needed to have an effect on the equations. The -3 coefficient of RF has a negative impact on the rabbit population. Turning our attention to the second equation, we see that F is multiplied by a - 4, indicating that the fox population would decrease if they did not interact with rabbits.
The positive coefficient for RF indicates a positive impact on the fox population. Predator-prey models are used extensively in many fields ranging from wildlife populations to military strategic planning.
In many of these models qualitative methods are employed. Supplementary Problems 2. Using Problem 2. Discuss Newton's second law of motion: 2. Suppose a room is being cooled according to the model where t hours and T degrees Celsius. Suppose the room in Problem 2.
How long would it take for the room to cool down to its minimum temperature? Consider the model discussed in Problem 2. Describe the motion of displacement, y t. Find a the velocity function; b the acceleration function. Assume a chemical compound, X, is such that its rate of decay is proportional to the cube of its difference from a given amount, M, where both X and M are given in grams and time is measured in hours.
Model this relationship with a differential equation. Suppose A and B are two vats interconnected with a number of pipes and drains. If A t and B t represent the number of gallons of liquid sugar in the respective vats at time t hours , what do A' t and B' t represent? Consider Problem 2. What is happening to the liquid sugar and what are the units of the six constants?
Many, hill not all. The right side of J. Then 3. Bernoulli equations are solved in Chapter 6. Homogeneous equations are solved in Chapter 4. Note: In the general framework of differential equations, the word "homogeneous" has an entirely different meaning see Chapter 8.
Only in the context of first-order differential equations does "homogeneous" have the meaning defined above.
Separable equations are solved in Chapter 4. Solved Problems 3. Solving for y', we obtain or which has form 3. This equation cannot be solved algebraically for y', and cannot be written in standard form. One is which can be written in form 3. We rewrite it as which vhich has the standard form fi or 3. Determine whether any of the differential equations in Problem 3. Exactness is only defined for equations in differential form, not standard form.
The given differential equation has manv differential forms. One such form is given in Problem 3. A second differential form for the same differential equation is given in Eq.
Thus, a given differential equation has many differential forms, some of which may be exact. Prove that a separable equation is always exact.
Does this result violate the theorem? Supplementary Problems In Problems 3. The integrals obtained in Hq.
In such eases. Even if the indicated integrations in 4. In that case, the solution is left in implicit form. Alternatively, the solulion lo hq. Alternatively, the solution to 4. After simplifying, the resulting differential equation will be one with variables this time, u and y separable. Ordinarily, it is immaterial which method of solution is used see Problems 4. Occasionally, however, one of the substitutions 4. In such cases, the better substitution is usually apparent from the form of the differential equation itself.
See Problem 4. Solved Problems 4. Substituting these values into Eq. Solving explicitly for y, we obtain the solution as 4. Solving for y implicitly, we obtain the two solutions and 4. To solve for y explicitly, we first rewrite the solution as In md then take the exponential of both sides.
The differential equation as originally given is also linear. See Problem 6. This equation may be rewritten as which is separable in variables y and t.
Solve This equation may be rewritten in differential form which is separable in the variables x and t. The solution to the differential equation is given by Eq.
Use Eq. Solve This differential equation is not separable, but it is homogeneous as shown in Problem 3. Substituting Eqs. Solve This differential equation is not separable.
Solve the differential equation of Problem 4. Solve Phis differential equation is not separable. Solve This differential equation is homogeneous. Solve The solution to the differential equation is given in Problem 3. Thus, the solution to the initial-value problem is The negative square root is taken, to be consistent with the initial condition.
Solve This differential equation is not separable, but it is homogeneous. Rewriting the differential equation as we have upon using substitutions 4.
Prove that every solution of Eq. Rewrite 4. The result of this substitution is 4. Prove that every solution of system 4. Following the same reasoning as in Problem 4. Show that this definition implies the definition given in Chapter 3. The solution to 5. Equation 5. Occasionally, it is possible to transform 5. A function I x, y is an integrating factor for 5. A solution to 5. If a differential equation does not have one of the forms given above, then a search for an integrating factor likely will not be successful, and other methods of solution are recommended Solved Problems 5.
This equation has the form of Eq. Solve the differential equation given in Problem 5. This equation was shown to be exact. We now determine a function g x, y that satisfies Eqs. Integrating both sides of this equation with respect to x, we find or Note that when integrating with respect to x, the constant with respect to x of integration can depend on y.
We now determine h y. Substituting this expression into 1 yields The solution to the differential equation, which is given implicitly by 5. Here which are not equal, so the differential equation as given is not exact. Determine whether the differential equation is exact. We now seek a function g x, y that satisfies 5. Substituting M x, y into 5. Solve Rewriting this equation in differential form, we obtain Here, and and, since the differential equation is exact.
Substituting this h y into 1 , we obtain The solution to the differential equation is given implicitly by 5. This is an equation for the unknown function y t. This equation was shown to be exact, so the solution procedure given by Eqs. Here Integrating both sides with respect to t, we have Differentiating 1 with respect to y, we obtain Hence, where the right side of this last equation is the coefficient of dy in the original differential equation.
The solution to the differential equation is given implicitly by 5. This is an equation for the unknown function x t. In terms of the variables t and x, we find so the differential equation is exact. We seek a function g t, x having the property that dg is the right side of the given differential equation. Here Integrating both sides with respect to t, we have or Differentiating 1 with respect to x, we obtain Hence, where the right side of this last equation is the coefficient of dx in the original differential equation.
Solve The differential equation has the differential form given in Problem 5. Its solution is given in 2 of Problem 5. Solve This differential equation in standard form has the differential form of Problem 5. Solving for y directly, using the quadratic formula, we have where the negative sign in front of the radical was chosen to be consistent with the given initial condition.
This differential equation in standard form has the differential form of Problem 5. Solving for x directly, using the quadratic formula, we have where the positive sign in front of the radical was chosen to be consistent with the given initial condition. It was shown in Problem 5. Multiplying it by —llx2, we obtain Equation 1 has the form of Eq. Now so 1 is exact, which implies that -llx2 is an integrating factor for the original differential equation. Equation 1 can be solved using the steps described in Eqs.
Determine whether —ll xy is also an integrating factor for the differential equation defined in Problem 5.
Now so 1 is exact, which implies that —llxy is also an integrating factor for the original differential equation. Solve Problem 5. Using the results of Problem 5. Note, however, that if terms are strategically regrouped, the differential equation can be rewritten as The group of terms in parentheses has many integrating factors see Table This differential equation is not exact, and no integrating factor is immediately apparent. Note, however, thai the differential equation can be rewritten as The first group of terms has many integrating factors see Table Solve Rewriting this equation in differential form, we have which is not exact.
Furthermore, no integrating factor is immediately apparent. But is a function of x alone. Using Eq. But is a function of y alone.
Convert y — into an exact differential equation. Since are not equal, 1 is not exact. Supplementary Problems In Problems 5.
Problem 5. When both sides of 6. This equation can he solved by ihe method described in Chapter 5. A simpler procedure is to rewrite 6. The substitution transforms 6. The differential equation has the form of Eq. Here so 6. Solve the differential equation in the previous problem. Multiplying the differential equation by the integrating factor defined by 1 of Problem 6.
Multiplying the differential equation by I x , we obtain Integrating both sides of the last equation with respect to x to integrate the right side, we use integration by parts twice , we find 6.
From Problem 6. Note that the differential equation is also separable. Solve This is a linear differential equation for the unknown function z x.
It has the form of Eq. The solution to this differential equation is given in Problem 6. Solve This is a linear differential equation for the unknown function i x. The integrating factor is Multiplying the differential equation by I x , we obtain or Upon integrating both sides of this last equation, we have whereupon 6.
Solve This is a linear differential equation for the unknown function Q t. Enter the email address you signed up with and we'll email you a reset link. Need an account? Click here to sign up.
Download Free PDF. Applications of First-Order Differential Equations. Jays Dejaresco. A short summary of this paper. See Problems 7. For population problems, where N t is actually discrete and integer-valued, this assumption is incorrect.
Nonetheless, 7. See Problem 7. Let T denote ihe temperature of the body and lei T,H denote the temperature of the surrounding medium. Click here for terms of use. Assume that both gravity and mass remain constant and, for convenience, choose the downward direction as the positive direction.
Newton's second law of motion: The net force acting on a body is equal to the time rate of change of the momentum of the body; or, for constant mass, where F is the net force on the body and v is the velocity of the body, both at time t.
The minus sign is required because this force opposes the velocity; that is, it acts in the upward, or negative, direction see Fig. Substituting this result into 7. These equa- tions are not valid if, for example, air resistance is not proportional to velocity but to the velocity squared, or if the upward direction is taken to be the positive direction. The problem is to find the amount of salt in the tank at any time t.
The time rate of change of Q, dQIdt, equals the rate at which salt enters the tank minus the rate at which salt leaves the tank. To determine the rate at which salt leaves the tank, we first calculate the volume of brine in the tank at any time t, which is the initial volume V0 plus the volume of brine added et minus the volume of brine removed f t.
For more complex circuits see Chapter The problem is to find another one-parameter family of curves, called the orthogonal trajectories of the family 7. We first implicitly differentiate 7. For many families of curves, one cannot explicitly solve for dyldx and obtain a differential equation of the form 7. We do not consider such curves in this book. Solved Problems 7. Find a the amount in the account after three years, and b the time required for the account to double in value, presuming no withdrawals and no additional deposits.
Let N t denote the balance in the account at any time t. The balance in the account grows by the accumulated interest payments, which are proportional to the amount of money in the account. The constant of proportionality is the interest rate. Substituting these values into 2 and solving for t, we obtain 7. Assuming no additional deposits or withdrawals, how much will be in the account after seven years if the interest rate is a constant 8.
Over the last three years, the interest rate is 9. What constant interest rate is required if an initial deposit placed into an account that accrues interest compounded continuously is to double its value in six years? The balance N t in the account at any time t is governed by 7. Substituting these values into 2 and solving for k, we find An interest rate of A bacteria culture is known to grow at a rate proportional to the amount present.
After one hour, strands of the bacteria are observed in the culture; and after four hours, strands. Find a an expres- sion for the approximate number of strands of the bacteria present in the culture at any time t and b the approximate number of strands of the bacteria originally in the culture.
From 6. The population of a certain country is known to increase at a rate proportional to the number of people presently living in the country. If after two years the population has doubled, and after three years the population is 20,, estimate the number of people initially living in the country. Let N denote the number of people living in the country at any time t, and let NQ denote the number of people initially living in the country.
Then, from 7. Substituting 3 , we obtain 7. A certain radioactive material is known to decay at a rate proportional to the amount present. Each chapter begins with a clear statement of pertinent definitions, principles andtheorems together with illustrative and other descriptive material.
This is followed bygraded sets of solved and supplementary problems. The solved problems serve to illustrateand amplify the theory, bring into sharp focus those fine points without which the studentcontinually feels himself on unsafe ground, and provide the repetition of basic principlesso vital to effective teaching.
Numerous proofs of theorems and derivations of formulasare included among the solved problems. The large number of supplementary problemswith answers serve as a complete review of the material of each chapter. Topics covered include the algebra and the differential and integral calculus of vec-tors, Stokes' theorem, the divergence theorem and other integral theorems together withmany applications drawn from various fields. Added features are the chapters on curvilin-ear coordinates and tensor analysis which should prove extremely useful in the study ofadvanced engineering, physics and mathematics.
Considerably more material has been included here than can be covered in most firstcourses. This has been done to make the book more flexible, to provide a more useful bookof reference, and to stimulate further interest in the topics.
The author gratefully acknowledges his indebtedness to Mr. Henry Hayden for typo-graphical layout and art work for the figures.
The realism of these figures adds greatly tothe effectiveness of presentation in a subject where spatial visualizations play such an im-portant role. Vector algebra. Laws of vector algebra. Unit vectors.
Rectangular unitvectors. Components of a vector. Scalar fields. Vector fields. Cross or vector products. Triple products.
Reciprocal sets ofvectors. Space curves. Continuity and differentiability. Differen-tiation formulas. Partial derivatives of vectors Differentials of vectors. Formulas involving del. Line integrals. Surface integrals.
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