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Chapter 2: Problem 18

(a) The initial temperature of the bath is \(T_{m}(0)=60^{\circ},\) so in the short term the temperature of the chemical, which starts at \(80^{\circ},\) should decrease or cool. Over time, the temperature of the bath will increase toward \(100^{\circ}\) since \(e^{-0.1 t}\) decreases from 1 toward 0 as \(t\) increases from \(0 .\) Thus, in the long term, the temperature of the chemical should increase or warm toward \(100^{\circ}\). (b) Adapting the model for Newton's law of cooling, we have $$\frac{d T}{d t}=-0.1\left(T-100+40 e^{-0.1 t}\right), \quad T(0)=80$$. Writing the differential equation in the form $$\frac{d T}{d t}+0.1 T=10-4 e^{-0.1 t}$$ we see that it is linear with integrating factor \(e^{\int 0.1 d t}=e^{0.1 t}\) Thus $$\begin{aligned} \frac{d}{d t}\left[e^{0.1 t} T\right] &=10 e^{0.1 t}-4 \\ e^{0.1 t} T &=100 e^{0.1 t}-4 t+c \end{aligned}$$ and $$T(t)=100-4 t e^{-0.1 t}+c e^{-0.1 t}$$ Now \(T(0)=80\) so \(100+c=80, c=-20\) and $$T(t)=100-4 t e^{-0.1 t}-20 e^{-0.1 t}=100-(4 t+20) e^{-0.1 t}$$ The thinner curve verifies the prediction of cooling followed by warming toward \(100^{\circ} .\) The wider curve shows the temperature \(T_{m}\) of the liquid bath.

Short Answer

Expert verified
The chemical cools initially then warms, ultimately reaching \(100^\circ\).

Step by step solution

01

Understand the Problem

The problem involves a chemical with an initial temperature of \(80^\circ\) being placed in a bath that starts at \(60^\circ\) and will rise to \(100^\circ\). In the short term, the chemical cools, but over time, it warms up to the bath's temperature. A differential equation model based on Newton's law of cooling describes this temperature change.
02

Set Up the Differential Equation

The given differential equation model is: \(\frac{d T}{d t} = -0.1\left(T - 100 + 40 e^{-0.1 t}\right)\), with initial condition \(T(0) = 80\). This equation describes the rate of temperature change of the chemical over time.
03

Rewrite Differential Equation in Linear Form

Rewriting the equation in the form: \(\frac{d T}{d t} + 0.1 T = 10 - 4 e^{-0.1 t}\). This reveals the equation as a linear first-order differential equation appropriate for an integrating factor solution.
04

Find the Integrating Factor

The integrating factor, \(e^{\int 0.1 dt} = e^{0.1 t}\), is used to simplify and solve the differential equation. It allows the equation to be manipulated so that differentiation is straightforward.
05

Solve Using Integrating Factor

Multiplying through by the integrating factor, we have: \(\frac{d}{d t}[e^{0.1 t} T] = 10 e^{0.1 t} - 4\). Integrating both sides with respect to \(t\) gives \(e^{0.1 t} T = 100 e^{0.1 t} - 4 t + c \).
06

Solve for Constant of Integration

Substitute the initial condition into the integrated equation: When \(t=0\), \(T=80\), so \(100 + c = 80\). Solving for \(c\), we get \(c = -20\).
07

Final Form of Solution

Substitute \(c\) back into the equation: \(T(t) = 100 - 4 t e^{-0.1 t} - 20 e^{-0.1 t} = 100 - (4t + 20)e^{-0.1t}\). This equation describes how the temperature of the chemical changes over time.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
Differential equations are mathematical equations that involve derivatives, which represent rates of change. They are essential in modeling various physical phenomena, such as the cooling of a chemical in a bath as seen in the given problem. These equations express the relationship between a function and its derivatives, providing a powerful tool to describe dynamic processes.
In everyday terms, a differential equation lets us predict how quantities change continuously over time or space. In our example, the differential equation describes how the temperature of the chemical changes over time as it interacts with the temperature of the surrounding bath. This allows us to model the cooling and eventual warming behavior of the chemical.
Differential equations can be complex, but understanding their purpose is straightforward: they're essentially rules describing how something changes.
Initial Value Problem
An initial value problem (IVP) is a type of differential equation problem that requires finding a solution that fits particular starting conditions. In the case of our exercise, the starting condition is the initial temperature of the chemical, which is \(T(0) = 80^{\circ}\). This gives us a specific point to begin solving the differential equation, making it possible to find a unique solution.
Initial values act like guideposts, ensuring that the solution to a differential equation precisely fits the starting circumstances of a problem. Without these values, it is often difficult to find a single, exact solution as many different curves could satisfy the equation at different points.
By specifying an initial condition, we tighten down one particular solution path amidst the multitude of possible trajectories, guiding us to the correct outcome for our specific problem.
Integrating Factor
The integrating factor is a clever technique used to solve linear first-order differential equations. It's a mathematical tool that simplifies the process significantly. The primary goal of an integrating factor is to convert the differential equation into a form where integration becomes direct and more manageable.
In our example, the integrating factor is found by computing \(e^{\int 0.1 \, dt} = e^{0.1 t}\). Once this factor is determined, it is multiplied throughout the differential equation. This manipulation results in simplification, creating an equation that can be easily integrated.
The beauty of the integrating factor lies in how it transforms a complex equation into a straightforward one, letting us find the solution with greater ease. It's like finding a clever shortcut through a maze.
Linear First-Order Differential Equation
A linear first-order differential equation is one of the simplest forms of differential equations. It is defined by having the term with the highest derivative鈥攁 first derivative鈥攊n a linear form. This doesn't mean there are only straight lines involved but rather that the mathematical operations are straightforward and predictable.
In the given exercise, the equation \(\frac{d T}{d t} + 0.1 T = 10 - 4 e^{-0.1 t}\) is a typical form of a linear first-order differential equation. You can spot these by noticing the presence of the derivative of the unknown function plus some other function multiplied by the unknown function itself.
The approach used for solving these equations generally involves finding an integrating factor, as we've discussed. Once the equation is in this linear form, solving it becomes methodical, following a predictable set of steps. This reliability makes them often one of the first differential equations students learn to solve.

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Most popular questions from this chapter

(a) Separating variables and integrating, we have \\[\left(-2 y+y^{2}\right) d y=\left(x-x^{2}\right) d x\\] and \\[-y^{2}+\frac{1}{3} y^{3}=\frac{1}{2} x^{2}-\frac{1}{3} x^{3}+c\\]. Using a CAS we show some contours of \\[f(x, y)=2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}\\]. The plots shown on \([-7,7] \times[-5,5]\) correspond to \(c\) -values of -450,-300,-200,-120,-60,-20,-10,-8.1,-5 \(-0.8,20,60,\) and 120. (b) The value of \(c\) corresponding to \(y(0)=\frac{3}{2}\) is \(f\left(0, \frac{3}{2}\right)=-\frac{27}{4}\) The portion of the graph between the dots corresponds to the solution curve satisfying the intial condition. To determine the interval of definition we find \(d y / d x\) for \\[2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-\frac{27}{4}\\]. Using implicit differentiation we get \(y^{\prime}=\left(x-x^{2}\right) /\left(y^{2}-2 y\right)\) which is infinite when \(y=0\) and \(y=2 .\) Letting \(y=0\) in \(2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-\frac{27}{4}\) and using a CAS to solve for \(x\) we get \(x=-1.13232 .\) Similarly, letting \(y=2,\) we find \(x=1.71299 .\) The largest interval of definition is approximately (-1.13232,1.71299). (c) The value of \(c\) corresponding to \(y(0)=-2\) is \(f(0,-2)=-40\) The portion of the graph to the right of the dot corresponds to the solution curve satisfying the initial condition. To determine the interval of definition we find \(d y / d x\) for \(2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-40\). Using implicit differentiation we get \(y^{\prime}=\left(x-x^{2}\right) /\left(y^{2}-2 y\right)\) which is infinite when \(y=0\) and \(y=2 .\) Letting \(y=0\) in \(2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-40\) and using a CAS to solve for \(x\) we get \(x=-2.29551 .\) The largest interval of definition is approximately \((-2.29551, \infty)\).

While the object is in the air its velocity is modelled by the linear differential equation \(m d v / d t=m g-k v .\) Using \(m=160, k=\frac{1}{4},\) and \(g=32,\) the differential equation becomes \(d v / d t+(1 / 640) v=32 .\) The integrating factor is \(e^{\int d t / 640}=e^{t / 640}\) and the solution of the differential equation is \(e^{t / 640} v=\int 32 e^{t / 640} d t=20,480 e^{t / 640}+c\) Using \(v(0)=0\) we see that \(c=-20,480\) and \(v(t)=20,480-20,480 e^{-t / 640} .\) Integrating we get \(s(t)=20,480 t+\) \(13,107,200 e^{-t / 640}+c .\) since \(s(0)=0, c=-13,107,200\) and \(s(t)=-13,107,200+20,480 t+13,107,200 e^{-t / 640}\) To find when the object hits the liquid we solve \(s(t)=500-75=425,\) obtaining \(t_{a}=5.16018 .\) The velocity at the time of impact with the liquid is \(v_{a}=v\left(t_{a}\right)=164.482 .\) When the object is in the liquid its velocity is modeled by the nonlinear differential equation \(m d v / d t=m g-k v^{2} .\) Using \(m=160, g=32,\) and \(k=0.1\) this becomes \(d v / d t=\left(51,200-v^{2}\right) / 1600 .\) Separating variables and integrating we have \\[ \frac{d v}{51,200-v^{2}}=\frac{d t}{1600} \quad \text { and } \quad \frac{\sqrt{2}}{640} \ln \left|\frac{v-160 \sqrt{2}}{v+160 \sqrt{2}}\right|=\frac{1}{1600} t+c \\] Solving \(v(0)=v_{a}=164.482\) we obtain \(c=-0.00407537 .\) Then, for \(v<160 \sqrt{2}=226.274\) \\[ \left|\frac{v-160 \sqrt{2}}{v+160 \sqrt{2}}\right|=e^{\sqrt{2} t / 5-1.8443} \quad \text { or } \quad-\frac{v-160 \sqrt{2}}{v+160 \sqrt{2}}=e^{\sqrt{2} t / 5-1.8443} \\] Solving for \(v\) we get \\[ v(t)=\frac{13964.6-2208.29 e^{\sqrt{2} t / 5}}{61.7153+9.75937 e^{\sqrt{2} t / 5}} \\] Integrating we find \\[ s(t)=226.275 t-1600 \ln \left(6.3237+e^{\sqrt{2} t / 5}\right)+c \\] Solving \(s(0)=0\) we see that \(c=3185.78,\) so \\[ s(t)=3185.78+226.275 t-1600 \ln \left(6.3237+e^{\sqrt{2} t / 5}\right) \\] To find when the object hits the bottom of the tank we solve \(s(t)=75,\) obtaining \(t_{b}=0.466273 .\) The time from when the object is dropped from the helicopter to when it hits the bottom of the tank is \(t_{a}+t_{b}=\) 5.62708 seconds.

$$h-0.1$$ $$h=0.05$$

(a) \(\operatorname{From} d P / d t=\left(k_{1}-k_{2}\right) P\) we obtain \(P=P_{0} e^{\left(k_{1}-k_{2}\right) t}\) where \(P_{0}=P(0)\). (b) If \(k_{1} >k_{2}\) then \(P \rightarrow \infty\) as \(t \rightarrow \infty\). If \(k_{1}=k_{2}\) then \(P=P_{0}\) for every \(t .\) If \(k_{1}< k_{2}\) then \(P \rightarrow 0\) as \(t \rightarrow \infty\).

(a) The differential equation is \(d P / d t=P(5-P)-4 .\) Solving \(P(5-P)-4=0\) for \(P\) we obtain equilibrium solutions \(P=1\) and \(P=4 .\) The phase portrait is shown on the right and solution curves are shown in part (b). We see that for \(P_{0}>4\) and \(1
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