First Order Differential Equations
2.5 Applications of First-Order ODE
A. Introduction
Mathematical modeling is the process of translating real-world problems into mathematical language. This involves formulating, developing, and rigorously testing models to represent and solve complex issues. Differential equations, including both ordinary and partial types, are instrumental in these models. They relate some function with its derivatives, representing rates of change. This makes them particularly suited to modeling dynamic systems where understanding how things evolve is crucial.
In this section, we will explore how first-order differential equations are applied across various domains, including growth and decay processes, substance mixing, Newton’s law of cooling, the dynamics of falling objects, and the analysis of electrical circuits.
B. Population Growth and Decay
One of the most common applications of first-order differential equations is in modeling population growth or decline. The models provide insights into how populations change over time due to births, deaths, immigration, and emigration. The simplest model for population growth is the Exponential Growth Model, which assumes an unlimited resource environment. It is represented by the differential equation:
[asciimath](dP)/(dt)=rP[/asciimath]
where [asciimath]P[/asciimath] is the population size, and [asciimath]r[/asciimath] is the constant of proportionality. The solution to this separable differential equation is
[asciimath]P(t)=P_0e^(rt)[/asciimath]
where [asciimath]P_0[/asciimath] is the initial population at time [asciimath]t=0[/asciimath].
If [asciimath]rlt0[/asciimath] the population decays exponentially and if [asciimath]r>0[/asciimath] the population grows exponentially. This model implies that the population grows continuously and without bounds, which is unrealistic in the long term for any population due to limitations in resources, space, etc. However, it is a good approximation for populations with no significant constraints on resources or for short-term predictions.
When dealing with problems where there are different rates of population entering and exiting a region, the key is to understand that the overall rate of change of the population is the result of the difference between the rate of population entering (immigration or birth) and the rate of population leaving (emigration or death). This can be represented as a differential equation that models the net change in population over time. The general approach is to set up a balance equation reflecting these rates:
[asciimath](dP)/(dt)=R_("in")-R_("out")[/asciimath]
Here [asciimath]R_("in")[/asciimath] is the rate at which the population enters the region, and [asciimath]R_("out")[/asciimath] is the rate at which the population exits the region.
A fish population in a lake grows at a rate proportional to its current size. Without outside factors, the fish population doubles in 10 days. However, each day, 5 fish migrate into the area, 16 are caught by fishermen, and 7 die of natural causes. Determine if the population will survive over time and, if not, when the population will become extinct. The initial population is 200 fish.
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[asciimath](dP)/(dt)=rP(t)-18,[/asciimath] [asciimath]P(0)=200[/asciimath]
Before we solve this IVP, we need to find [asciimath]r[/asciimath] using the information about doubling the population in 10 days without outside factors. If the initial population is 200, then in 10 days it will become 400.
[asciimath](dP)/(dt)=rP(t),[/asciimath] [asciimath]P(10)=400[/asciimath]
The general solution to this separable differential equation is
[asciimath]P(t)=P_0e^(rt)[/asciimath]
Applying the initial condition, we obtain
[asciimath]P(10)=400[/asciimath]
[asciimath]200e^(10r)=400[/asciimath]
[asciimath]e^(10r)=2[/asciimath]
[asciimath]r=ln(2)/10[/asciimath]
Now, we return to the original differential equation.
[asciimath](dP)/(dt)=ln(2)/10 P-18,[/asciimath] [asciimath]P(0)=200[/asciimath]
This is a linear differential equation. we write it in standard form:
[asciimath](dP)/(dt)-ln(2)/10 P=-18[/asciimath]
The integrating factor is
[asciimath]u(x)=e^(-int(ln(2))/10 dt)=e^(-(ln(2))/10 t)[/asciimath]
The general solution is
[asciimath]P(t)=e^((ln(2))/10 t) [int -18 e^(-(ln(2))/10 t) dt+C][/asciimath]
[asciimath]P(t)=e^((ln(2))/10 t) [18(10/(ln(2)) )e^(-(ln(2))/10 t) +C][/asciimath]
[asciimath]P(t)=180/(ln(2)) +Ce^((ln(2))/10 t)[/asciimath]
Applying the initial condition gives
[asciimath]P(0)=200[/asciimath]
[asciimath]180/(ln(2)) +Ce^0=200[/asciimath]
[asciimath]C~~-59.6851[/asciimath]
Thus the specific solution is
[asciimath]P(t)=180/(ln(2)) -59.6851 e^((ln(2))/10 t)[/asciimath]
The exponential term has a positive exponent and thus grows exponentially. However, since the coefficient of the exponential term is negative, the whole population declines and becomes extinct eventually. To determine when the population will become extinct, we set [asciimath]P=0[/asciimath] and solve for [asciimath]t[/asciimath].
[asciimath]0=180/(ln(2)) -59.6851 e^((ln(2))/10 t)[/asciimath]
[asciimath]t~~21.2132[/asciimath] days
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C. Mixing Problems
Mixing problems involve combining substances or quantities and observing how they interact over time. This can refer to pollutants in a lake, different chemicals in a reactor, or even sugar dissolving in coffee. The common element in these scenarios is the change in concentration of substances in a mixture over time. Through differential equations, specifically first-order ones, we can model and solve these dynamic situations.
In mixing problems, [asciimath]Q(t)[/asciimath] represents the substance amount dissolved in the fluid, changing over time at a rate ([asciimath](dQ)/(dt)[/asciimath]). The rate is influenced by the inflow and outflow of the substance.
For a typical mixing problem, you might have a tank that contains a certain amount of fluid into which another substance is being mixed. The concentration of the substance in the tank changes as more of the substance is added or removed. The general first-order differential equation for such a scenario is similar to what we discussed for the population change of a region.
[asciimath](dQ)/(dt)=R_("inflow")-R_("outflow")[/asciimath]
Here [asciimath]R_("inflow")[/asciimath] is the rate at which the substance enters the system, and [asciimath]R_("outflow")[/asciimath] is the rate at which the substance leaves the system.
Consider a tank holding 2000 liters of fresh water. Starting at [asciimath]t=0[/asciimath], water containing 0.1 kilograms of salt per liter is poured into the tank at the rate of [asciimath]8 \ "liters/min"[/asciimath]. The mixture is kept uniform by stirring and is drained from the tank at the same rate it is filled. a) Formulate a differential equation for the quantity of salt in the tank ([asciimath]Q(t)[/asciimath]) at any given time, and solve the equation to determine [asciimath]Q(t)[/asciimath]. b) Determine when the concentration of the salt in the tank will reach [asciimath]0.04 \ "kg/L"[/asciimath].
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Given information
- The volume of water in the tank ([asciimath]V[/asciimath]) is constant since water inflow and outflow are equal: [asciimath]V = 2000\ L[/asciimath]
- Water inflow rate [asciimath]= 8 \ "L/min"[/asciimath]
- Water outflow rate [asciimath]= 8 \ "L/min"[/asciimath]
- Concentration of incoming salt: [asciimath]0.1 \ "kg/L"[/asciimath]
a) Our task is to determine the rate at which salt enters the tank ([asciimath]R_("inflow")[/asciimath]) and the rate at which it leaves the system. Remember that the rate at which water enters and leaves the tank is different from the rate at which salt enters and leaves the tank.
[asciimath](dQ)/(dt)=R_("inflow")-R_("outflow")[/asciimath]
The rate at which salt enters the tank is the product of the salt concentration of the incoming water and the water inflow rate:
[asciimath]R_("inflow") = (0.1 "kg/L")(8 "L/min")[/asciimath]
[asciimath]=0.8 \ "kg/min"[/asciimath]
The rate at which salt leaves the tank is the concentration of salt in the tank (ratio of the salt in the tank to the volume of water in the tank), multiplied by the water outflow rate. At any time, the quantity of salt in the tank is [asciimath]Q(t)[/asciimath].
[asciimath]R_"outflow" = "Concentration"xx"Water outflow rate"[/asciimath]
[asciimath]R_"outflow" = ((Q(t))/2000 \ "kg/L")*(8 \ "L/min")[/asciimath]
[asciimath]=(Q(t))/250[/asciimath]
The tank initially has pure fresh water without any salt, so [asciimath]Q(0)=0[/asciimath]. Therefore, the differential equation with an initial condition becomes
[asciimath](dQ)/(dt)=R_("inflow")-R_("outflow")[/asciimath]
[asciimath](dQ)/(dt)=0.8-(Q(t))/250[/asciimath] , [asciimath]Q(0)=0[/asciimath]
The differential equation is separable (and linear) and can be solved easily. The solution to the IVP is
[asciimath]Q(t)=200(1-e^(-t/250))[/asciimath]
This equation gives us the amount of salt in the tank [asciimath]Q(t)[/asciimath] in kilograms at any time t after the process starts.
b) To determine when the concentration of salt in the tank reaches [asciimath]0.04 \ "kg/L"[/asciimath], we first need to find an equation for the concentration in terms of time. Concentration is the ratio of the salt quantity and the volume of the water. The volume remains constant at 2000 liters. Therefore, the concentration [asciimath]C(t)[/asciimath] at time t is the amount of salt divided by the total volume [asciimath]V[/asciimath]:
[asciimath]C(t)=(Q(t))/V \ "kg/L"[/asciimath]
[asciimath]C(t)=200/2000(1-e^(-t/250))[/asciimath]
[asciimath]=0.1(1-e^(-t/250))[/asciimath]
Now, we need to solve for t when [asciimath]C(t)=0.04 \ "kg/L"[/asciimath].
[asciimath]C(t)=0.04[/asciimath]
[asciimath]0.1(1-e^(-t/250))=0.04[/asciimath]
[asciimath]e^(-t/250)=1.4[/asciimath]
[asciimath]t = -250ln(0.6)[/asciimath]
[asciimath]~~127.71 min[/asciimath]
The concentration of salt in the tank will reach 0.04 kg/L approximately at [asciimath]t~~127.71[/asciimath] minutes after the process starts.
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D. Newton’s Law of Cooling
Newton’s Law of Cooling describes the rate at which an object’s temperature changes when it is exposed to a surrounding environment with a different, constant temperature. The fundamental principle is that the rate of change of temperature ([asciimath](dT)/(dt)[/asciimath]) is proportional to the difference between the object’s temperature ([asciimath]T[/asciimath]) and the surrounding temperature ([asciimath]T_s[/asciimath]). Therefore, the differential equation representing Newton’s Law of Cooling is
[asciimath](dT)/(dt)=-k(T-T_s)[/asciimath]
Here, [asciimath]T[/asciimath] represents the object’s temperature at any time [asciimath]t[/asciimath], [asciimath]T_s[/asciimath]is the constant surrounding temperature, [asciimath]k[/asciimath] is a positive constant dependent on the characteristics of the object and its environment, and [asciimath](dT)/(dt)[/asciimath] is the rate of change of temperature. When the initial temperature is denoted by [asciimath]T_0[/asciimath], the initial value problem is
[asciimath](dT)/(dt)=-k(T-T_s), \ \ T(0)=T_0[/asciimath]
This differential equation is separable (and linear), which has the solution
[asciimath]T(t) = T_s +(T_0-T_s)e^(-kt)[/asciimath](2.5.1)
The negative sign in the exponent indicates that the temperature difference between the object and its surroundings decreases exponentially over time. This formula applies whether the object is initially hotter or cooler than the surroundings, depicting both cooling and warming processes under the law’s assumptions.
Consider a microprocessor that operates in an environment where the room temperature is constant at 25 ◦C. After a long period of operation, the microprocessor’s temperature is at 75 ◦C. Once the device is turned off, the microprocessor begins to cool down to room temperature. Suppose the characteristic cooling constant [asciimath]k[/asciimath] for this scenario, which depends on the heat transfer properties of the microprocessor and its cooling system, is 0.07/min. a) Find the equation of the microprocessor’s temperature. b) What will be the temperature of the microprocessor 10 minutes after the device is turned off? c) How long will it take for the microprocessor to cool down to 35 ◦C?
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Given information:
- Surrounding temperature: [asciimath]T_s=25^@C[/asciimath]
- Initial temperature of the microprocessor: [asciimath]T_0=75^@C[/asciimath]
- Cooling constant: [asciimath]k=0.07 \ "min"^-1[/asciimath]
a) Plugging the given values into the solution to Newton’s Law of Cooling equation, Equation 2.5.1, gives the formula for [asciimath]T(t)[/asciimath].
[asciimath]T(t) = T_s +(T_0-T_s)e^(-kt)[/asciimath]
[asciimath]T(t) = 25 +(75-25)e^(-0.07t)[/asciimath]
b) To find the temperature of the microprocessor 10 minutes after the device is turned off, plug in [asciimath]t=10[/asciimath] minutes into [asciimath]T(t)[/asciimath].
[asciimath]T(10) = 25 +(75-25)e^(-0.07(10))[/asciimath]
[asciimath]~~49.83^@C[/asciimath]
c) To find the time when the temperature is 35 ◦C, rearrange the formula when [asciimath]T(t)=35^@C[/asciimath].
[asciimath]25 +(75-25)e^(-0.07t)=35[/asciimath]
[asciimath]e^(-0.07t)=1/5[/asciimath]
[asciimath]t=ln(5)/0.07[/asciimath]
[asciimath]t~~23[/asciimath] minutes
It takes 23 minutes for the microprocessor to cool down to 35 ◦C.
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E. Dynamics of Falling Objects
The dynamics of falling objects represent a classic example of how differential equations model real-world situations. This phenomenon is directly connected to Newton’s Second Law of Motion, which states that the force acting on an object is equal to the mass of the object times its acceleration.
[asciimath]F=ma[/asciimath]
In this equation, the force may depend on time ([asciimath]t[/asciimath]), displacement ([asciimath]y[/asciimath]), and velocity ([asciimath]v[/asciimath]). To focus on first-order differential equations, we typically consider problems where [asciimath]F[/asciimath] doesn’t depend on [asciimath]y[/asciimath], as inclusion often leads to higher-order equations. Given that the object’s acceleration ([asciimath]a[/asciimath]) is [asciimath]dv//dt[/asciimath], the equation for Newton’s Second Law of Motion becomes
[asciimath]mv'=F(t,v)[/asciimath] .
Solving this equation yields [asciimath]v[/asciimath] as a function of time.
Basic Model
The simplest model of a falling object applies Newton’s Second Law by considering gravity as the only force acting on the object. Here, the force due to gravity is, leading to the differential equation
where is the acceleration due to gravity, and the mass is assumed to be constant. This model assumes no air resistance and that the gravitational field is uniform. The approximate value of [asciimath]g[/asciimath] are [asciimath]g=9.8 \ "m"//"s"^2[/asciimath] (metric unit) or [asciimath]g=32 \ "ft"//"s"^2[/asciimath] (British unit). Depending on the direction convention you set for a problem, the sign of [asciimath]F_g[/asciimath] changes. For example, if you decide that the upward direction is positive, then since the force due to gravity is downward the equation is simplified to
Including Air resistance
In reality, as an object falls, it encounters air resistance, which opposes the motion of the object. The net force on the object then becomes a combination of gravity and air resistance, modifying the equation to
[asciimath]mv'=F_g+F_A[/asciimath](2.5.2)
where [asciimath]F_A[/asciimath] is the force of air resistance.
The force of air resistance is often proportional to the velocity of the object and thus [asciimath]F_A=-kv[/asciimath], where [asciimath]k[/asciimath] is a constant of proportionality (a positive value) that represents the coefficient of air resistance. When solving problems involving forces and motions, it is important to ensure consistent conventions for positive and negative directions.
As the object falls, air resistance increases with velocity until it balances the gravitational force. At this equilibrium point, the net force is zero, and the object no longer accelerates, reaching a constant velocity, known as terminal velocity.
Consider an object that has a mass of 25 kg and is initially moving downward with a velocity of -29 m/s. The object is falling through the atmosphere, which exerts a resistive force against its motion. This resistive force is proportional to the object’s velocity. Specifically, when the object’s velocity is 2 m/s, the resistive force is known to be 20 N. a) Write the differential equation that describes the motion of the object in terms of its velocity and time. b) Solve the differential equation to find the velocity of the object as a function of time, [asciimath]v(t)[/asciimath]. c) Determine the terminal velocity of the object
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Given information
- mass of the object: [asciimath]m=25 \ "kg"[/asciimath]
- Initial velocity: [asciimath]v_0=-29 \ "m/s"[/asciimath]
- The acceleration due to gravity: [asciimath]g=9.8 \ "m"//"s"^2[/asciimath]
a) Downward velocity is expressed as a negative value. Therefore, the upward direction is positive and the downward direction is negative.
Two primary forces acting on the object are gravity and air resistance. The force of gravity always acts downward, which we consider negative in our coordinate system, and is given by [asciimath]-mg[/asciimath].
On the other hand, air resistance acts in the opposite direction of the object’s motion, providing an upward force when the object is falling downward. This force is represented as [asciimath]-kv[/asciimath]. The negative sign in [asciimath]-kv[/asciimath] ensures that the air resistance force always opposes the motion: it is positive (upward) when the object is falling ( [asciimath]v[/asciimath] is negative), and negative (downward) when the object is moving upward ([asciimath]v[/asciimath] is positive).
Combining these forces, the equation of motion is
[asciimath]mv'=-F_g+F_A[/asciimath]
[asciimath]mv'=-mg-kv[/asciimath]
we can use the information about the magnitude of air resistance to be [asciimath]20 \ "N"[/asciimath] when velocity is [asciimath]2 \ "m/s"[/asciimath] to find [asciimath]k[/asciimath]:
[asciimath]F_A=k|v|[/asciimath]
[asciimath]k=F_A/|v|[/asciimath]
[asciimath]k=20/2=10\ "kg"//"s"[/asciimath]
Plugging in the values with initial condition [asciimath]v(0)=-29\ "m/s"[/asciimath] , we obtain the IVP
[asciimath]25v'=-245-10v, v(0)=-29[/asciimath]
b) This is a separable (and linear) differential equation. The general solution of the equation is
[asciimath]v(t)=1/2(Ce^(-2/5t)-49)[/asciimath]
Applying the initial condition yields
[asciimath]v(t)=1/2(-9e^(-2/5t)-49)[/asciimath]
c) The terminal velocity is
[asciimath]lim_(x->oo)v(t)=-49/2 \ "m/s"[/asciimath]
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F. Electrical Circuits: RL and RC
Electrical circuits are integral to technological advancements, functioning based on the interplay of components such as resistors, inductors, and capacitors. In this section, we specifically discuss the application of first-order differential equations to analyze electrical circuits composed of a voltage source with either a resistor and inductor (RL) or a resistor and capacitor (RC), as illustrated in Fig. 2.5.1 Circuits containing both an inductor and a capacitor, known as RLC circuits, are governed by second-order differential equations, a topic we will revisit in the following chapter.
(a)
(b) 
Figure 2.5.1 (a) RL Series circuit and (b) RC series circuit
Kirchhoff’s laws—current law and voltage law—form the foundational principles governing electrical circuits. Kirchhoff’s current law states that the total current entering a junction must equal the total current leaving, implying that the algebraic sum of currents in a node is zero. Kirchhoff’s voltage law asserts that the algebraic sum of all voltages around any closed loop in a circuit must equal zero.
Kirchhoff’s current law implies that the same current passes through all elements in circuits in Figure 2.5.1. To apply Kirchhoff’s voltage law, understanding the voltage drop across each component is crucial:
a) Ohm’s law dictates that the voltage drop [asciimath]E_R[/asciimath] across a resistor is proportional to the current I flowing through it, expressed as [asciimath]E_R=RI[/asciimath], where [asciimath]R[/asciimath] is the resistance.
b) Faraday’s law, complemented by Lenz’s law, describes that the voltage drop [asciimath]E_L[/asciimath] across an inductor is proportional to the rate of change of current, given as [asciimath]E_L=L(dI)/(dt)[/asciimath], where [asciimath]L[/asciimath] is the inductance.
c) The voltage drop [asciimath]E_C[/asciimath] across a capacitor is proportional to the electric charge q stored on it, represented as [asciimath]E_C=1/Cq[/asciimath], with [asciimath]C[/asciimath] being the capacitance
RL Circuit Model
In this section, we derive the mathematical model for an RL circuit as shown in Figure 2.5.1, while the model derivation for an RC circuit is left as an exercise. Consider [asciimath]E(t)[/asciimath] to be the voltage source for the RL circuit. By applying Kirchhoff’s voltage law, we have
[asciimath]V_L+V_R=E(t)[/asciimath]
[asciimath]E_L+E_R=E(t)[/asciimath]
where [asciimath]E_L=L (dI)/dt[/asciimath] is the voltage across the inductor and [asciimath]E_R=RI[/asciimath] is the voltage across the resistor. Substituting these into the equation yields a first-order linear differential equation
[asciimath]L (dI)/dt+RI=E(t)[/asciimath]
or in the standard form
[asciimath](dI)/dt+R/L I=(E(t))/L[/asciimath]
To solve this linear differential equation. we use an integrating factor
[asciimath]u(x)=e^(int R/L dt)=e^(Rt//L)[/asciimath]
The general solution for the current [asciimath]I(t)[/asciimath] is then:
[asciimath]I(t)=e^(-Rt//L)[int e^(Rt//L) \ (E(t))/L dt +C][/asciimath](2.5.3)
With specific [asciimath]E(t)[/asciimath] and an initial condition, such as [asciimath]I(0)[/asciimath], one can determine the current I(t) using the above equation. Once [asciimath]I(t)[/asciimath] is known, the voltage across the resistor and inductor can be determined.
Consider an RL circuit with a resistor of [asciimath]3\ Omega[/asciimath] and an inductor of [asciimath]0.01\ H[/asciimath], powered by a voltage [asciimath]E(t)=sin(10t)\ V[/asciimath] voltage source. Initially, the current through the resistor, [asciimath]I(0)[/asciimath], is 0 A. Calculate the following: a) The current [asciimath]I(t)[/asciimath] in the circuit as a function of time. b) The voltage across the inductor as a function of time. c) The voltage across the resistor as a function of time.
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Given information:
- Resistor: [asciimath]R=3\ Omega[/asciimath]
- Inductor: [asciimath]L=0.01 \ H[/asciimath]
- Voltage source: [asciimath]E(t)=sin(10t)\ V[/asciimath]
- Initial condition: [asciimath]I(0)=0 \ A[/asciimath]
a) Finding the current [asciimath]I(t)[/asciimath]
The differential equation for an RL series circuit using Kirchhoff’s voltage law is
[asciimath]V_L+V_R=E(t)[/asciimath]
[asciimath]L (dI)/dt+RI=E(t)[/asciimath]
Plugging in the given values, we obtain
[asciimath]0.01 (dI)/dt+3I=sin(10t)[/asciimath]
This is a first-order linear non-homogeneous differential equation.
[asciimath]u(t)=e^(int 3/0.01 dt)=e^300t[/asciimath]
Equation 2.5.3 gives the solution to this differential equation.
[asciimath]I(t)=e^(-300t)[100int e^(300t) \ sin(10t) dt +C][/asciimath]
The right-hand side involves an integral with the exponential and sinusoidal terms that is typically solved using integration by parts. We only provide the final solution of the integral, leaving the detailed integration steps as an exercise for further exploration.
[asciimath]I(t)=e^(-300t)[-10/901e^(300t)cos(10t)+300/901e^(300t)sin(10t)+C][/asciimath]
Which further simplifies to
[asciimath]I(t)=-10/901cos(10t)+300/901sin(10t)+Ce^(-300t)[/asciimath]
Applying the initial condition yields
[asciimath]I(0)=0[/asciimath]
[asciimath]-10/901cos(0)+300/901sin(0)+Ce^(0) =0[/asciimath]
[asciimath]-10/901+C=0[/asciimath]
[asciimath]C=10/901[/asciimath]
Therefore, the current is
[asciimath]I(t)=-10/901cos(10t)+300/901sin(10t)+10/901e^(-300t)[/asciimath]
b) Finding the voltage across the inductor [asciimath]V_L(t)[/asciimath]
To find the voltage across the inductor, we first need to differentiate [asciimath]I(t)[/asciimath].
[asciimath](dI)/(dt)=100/901sin(10t)+3000/901cos(10t)-3000/901e^(-300t)[/asciimath]
Therefore, the voltage across the inductor is
[asciimath]V_L=L (dI)/dt[/asciimath]
[asciimath]V_L(t)=0.01[100/901sin(10t)+3000/901cos(10t)-3000/901e^(-300t) ][/asciimath]
[asciimath]V_L(t)=1/901sin(10t)+30/901cos(10t)-30/901e^(-300t)[/asciimath]
c) Finding the voltage across the resistor [asciimath]V_R(t)[/asciimath]
Similarly, the voltage across the resistor is found by
[asciimath]V_R=RI[/asciimath]
[asciimath]V_R(t)=3[-10/901cos(10t)+300/901sin(10t)+10/901e^(-300t) ][/asciimath]
[asciimath]V_R(t)=-30/901cos(10t)+900/901sin(10t)+30/901e^(-300t)[/asciimath]
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Section 2.5 Exercises
- A tank initially contains a solution of 11 kilograms of salt in 2400 liters of water. Water with 0.2 kilograms of salt per liter is added to the tank at 11 L/min, and the resulting solution leaves at the same rate. Let [asciimath]Q(t)[/asciimath] denote the quantity (kg) of salt at time [asciimath]t[/asciimath] (min). a) Write a differential equation for [asciimath]Q(t)[/asciimath]. b) Find the quantity [asciimath]Q(t)[/asciimath] of salt in the tank at time [asciimath]t[/asciimath]. c) Determine when the concentration of the salt in the tank will reach 0.1 kg/L. Round to the nearest minute.
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a) [asciimath]Q'(t)=2.2-11/2400Q(t)[/asciimath]
b) [asciimath]Q(t)=480 -469 e^(-11/2400 t)[/asciimath]
c) 146 min
- A fluid initially at 135 ◦C is placed outside on a day when the temperature is -30 ◦C, and the temperature of the fluid drops 30 ◦C in one minute. Let [asciimath]T(t)[/asciimath] denote the temperature, in Celsius, at time [asciimath]t[/asciimath], in minutes. (a) Find the temperature [asciimath]T(t)[/asciimath] of the fluid for [asciimath]t > 0[/asciimath]. (b) Find the temperature of the fluid 15 minutes after it is placed outside. Round your answer to two decimal places.
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a) [asciimath]T(t)=-30+165 (9/11)^t[/asciimath]
b) [asciimath]T(15)=-21.87^@C[/asciimath]
- An object with mass [asciimath]32 \ "kg"[/asciimath] has an initial downward velocity of [asciimath]-69 \ "m/s"[/asciimath]. Assume that the atmosphere exerts a resistive force with a magnitude proportional to the speed. The resistance is [asciimath]20 \ "N"[/asciimath] when the velocity is [asciimath]4 \ "m/s"[/asciimath]. Use [asciimath]g = 10 \ "m/"s"^2[/asciimath]. a) Write a differential equation in terms of the velocity [asciimath]v[/asciimath], and acceleration [asciimath]v'[/asciimath]. b) Find the velocity [asciimath]v(t)[/asciimath] of the object.
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a) [asciimath]32v'=-320-5v[/asciimath]
b) [asciimath]v(t) = -64-5e^(-5/32*t)[/asciimath]
- Suppose an RL circuit with a [asciimath]3\Omega[/asciimath] resistor and a [asciimath]1 H[/asciimath] inductor is driven by the voltage [asciimath]E(t)=e^(4 t) \ V[/asciimath]. If the initial resistor current is [asciimath]I(0)=0 A[/asciimath], find the current [asciimath]I[/asciimath], the voltages across the inductor [asciimath]E_L[/asciimath] and the resistor [asciimath]E_R[/asciimath] in terms of time [asciimath]t[/asciimath]. Find the current [asciimath]I(t)[/asciimath] .
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[asciimath]I(t)=1/(7)(e^(4t)-e^(-3t))[/asciimath]
