/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 The differential equation for th... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 16

The differential equation for the first container is \(d T_{1} / d t=k_{1}\left(T_{1}-0\right)=k_{1} T_{1},\) whose solution is \(T_{1}(t)=c_{1} e^{k_{1} t}\) since \(T_{1}(0)=100\) (the initial temperature of the metal bar), we have \(100=c_{1}\) and \(T_{1}(t)=100 e^{k_{1} t}\). After 1 minute, \(T_{1}(1)=100 e^{k_{1}}=90^{\circ} \mathrm{C},\) so \(k_{1}=\ln 0.9\) and \(T_{1}(t)=100 e^{t \ln 0.9} .\) After 2 minutes, \(T_{1}(2)=100 e^{2 \ln 0.9}=\) \(100(0.9)^{2}=81^{\circ} \mathrm{C}\). The differential equation for the second container is \(d T_{2} / d t=k_{2}\left(T_{2}-100\right),\) whose solution is \(T_{2}(t)=\) \(100+c_{2} e^{k_{2} t} .\) When the metal bar is immersed in the second container, its initial temperature is \(T_{2}(0)=81,\) so $$T_{2}(0)=100+c_{2} e^{k_{2}(0)}=100+c_{2}=81$$ and \(c_{2}=-19 .\) Thus, \(T_{2}(t)=100-19 e^{k_{2} t} .\) After 1 minute in the second tank, the temperature of the metal bar is \(91^{\circ} \mathrm{C},\) so $$\begin{aligned} T_{2}(1) &=100-19 e^{k_{2}}=91 \\ e^{k_{2}} &=\frac{9}{19} \\ k_{2} &=\ln \frac{9}{19} \end{aligned}$$ and \(T_{2}(t)=100-19 e^{t \ln (9 / 19)} .\) Setting \(T_{2}(t)=99.9\) we have $$\begin{aligned} 100-19 e^{t \ln (9 / 19)} &=99.9 \\ e^{t \ln (9 / 19)} &=\frac{0.1}{19} \\ t &=\frac{\ln (0.1 / 19)}{\ln (9 / 19)} \approx 7.02 \end{aligned}$$ Thus, from the start of the "double dipping" process, the total time until the bar reaches \(99.9^{\circ} \mathrm{C}\) in the second container is approximately 9.02 minutes.

Short Answer

Expert verified
The total time is approximately 9.02 minutes.

Step by step solution

01

Solve the Differential Equation for Container 1

The differential equation for the first container is given by \(\frac{dT_1}{dt}=k_1 T_1\). The solution to this equation is \(T_1(t)=c_1 e^{k_1 t}\) since it's a first-order linear differential equation with constant coefficients.
02

Apply Initial Condition to Container 1

Given the initial condition \(T_1(0)=100\), we substitute in the equation \(T_1(t)=c_1 e^{k_1 t}\) to find \(c_1\). This gives \(100 = c_1 e^{k_1 \cdot 0}\) which simplifies to \(c_1 = 100\). Therefore, \(T_1(t)=100 e^{k_1 t}\).
03

Determine \(k_1\) Using Data After 1 Minute

After 1 minute, the temperature is 90掳C, so \(T_1(1)=100 e^{k_1} = 90\). Solving for \(k_1\), we have \(e^{k_1} = 0.9\), hence \(k_1 = \ln 0.9\). Therefore, \(T_1(t) = 100 e^{t \ln 0.9}\).
04

Calculate Temperature After 2 Minutes in Container 1

Substituting \(t = 2\) into \(T_1(t)\), we have \(T_1(2) = 100 e^{2 \ln 0.9} = 100 (0.9)^2 = 81\,^{\circ}\text{C}\).
05

Solve the Differential Equation for Container 2

The differential equation for the second container is \(\frac{dT_2}{dt}=k_2(T_2-100)\). Its solution is \(T_2(t) = 100 + c_2 e^{k_2 t}\).
06

Apply Initial Condition to Container 2

Initially, \(T_2(0) = 81\). Therefore, from \(T_2(t) = 100 + c_2 e^{k_2 \cdot 0}\), we get \(81 = 100 + c_2\), which implies \(c_2 = -19\). Hence, \(T_2(t) = 100 - 19 e^{k_2 t}\).
07

Determine \(k_2\) Using Data After 1 Minute in Container 2

After 1 minute, \(T_2(1) = 91\), so \(100 - 19 e^{k_2} = 91\). Solving for \(e^{k_2}\), we have \(e^{k_2} = \frac{9}{19}\), hence \(k_2 = \ln \left(\frac{9}{19}\right)\). Thus, \(T_2(t) = 100 - 19 e^{t \ln \left(\frac{9}{19}\right)}\).
08

Find Time to Reach 99.9掳C in Container 2

We set \(T_2(t) = 99.9\) and solve: \(100 - 19 e^{t \ln \left(\frac{9}{19}\right)} = 99.9\). This yields \(e^{t \ln \left(\frac{9}{19}\right)} = \frac{0.1}{19}\). Solving for \(t\), we have \(t = \frac{\ln \left(\frac{0.1}{19}\right)}{\ln \left(\frac{9}{19}\right)} \approx 7.02\).
09

Calculate Total Time for Both Containers

The total time consists of 2 minutes in the first container and 7.02 minutes in the second, resulting in a total time of approximately 9.02 minutes.

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Key Concepts

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

First-order Linear Differential Equation
A first-order linear differential equation is a differential equation of the form \( \frac{dy}{dt} + P(t) y = Q(t) \). These equations only involve the first derivative of the unknown function. As a primary feature, the name "linear" indicates the absence of products or powers of \( y \) and its derivatives. In a rigorous sense, these equations are linear in the unknown function and its derivative.In the context of our problem, the equation for Container 1 \( \frac{dT_1}{dt} = k_1 T_1 \) is a classic example of a first-order linear differential equation. Here, *"first-order"* implies that the derivative appearing is the first derivative (\( \frac{dT_1}{dt} \)), and *"linear"* suggests that the function \( T_1 \) and its derivative have no exponents other than one. Being earlier solved through initial conditions transforms it into a useful predictive tool.
Constant Coefficients
Differential equations can greatly simplify when they have constant coefficients. "Constant coefficients" implies that terms multiplying the unknown function and its derivatives are constants, not dependent on the independent variable, typically time \( t \). In our initial problem, \( \frac{dT_1}{dt} = k_1 T_1 \) and \( \frac{dT_2}{dt} = k_2 (T_2 - 100) \), the \( k_1 \) and \( k_2 \) terms are constant coefficients.The simplification comes from the fact that solutions to such equations often involve exponential functions. The significance of constant coefficients is that it allows us to use well-understood mathematical techniques to find solutions, often involving interpreting or calculating exponentials based on empirical data, such as the temperature changes observed in the metal bar example.
Exponential Solutions
An exponential solution refers to a solution of a differential equation that involves exponential functions. These solutions are common for first-order linear differential equations with constant coefficients. The general solution format for a differential equation \( \frac{dy}{dt} = ky \) is \( y(t) = ce^{kt} \), where \( c \) is a constant determined by initial conditions, and \( k \) is a constant coefficient. In our exercise, the temperature \( T_1 \) decreases exponentially according to the function \( T_1(t) = 100 e^{t \ln 0.9} \), which results from solving \( \frac{dT_1}{dt} = k_1 T_1 \).Such exponential models are particularly powerful in describing processes like radioactive decay, cooling of objects, or even population growth, where the rate of change at any point is proportional to the current state.

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Most popular questions from this chapter

Write the differential equation as \(d P / d t-a P=-b P^{2}\) and let \(u=P^{-1}\) or \(P=u^{-1}\). Then $$\frac{d p}{d t}=-u^{-2} \frac{d u}{d t},$$ and substituting into the differential equation, we have $$-u^{-2} \frac{d u}{d t}-a u^{-1}=-b u^{-2} \quad \text { or } \quad \frac{d u}{d t}+a u=b.$$ The latter differential equation is linear with integrating factor \(e^{\int a d t}=e^{a t},\) so $$\frac{d}{d t}\left[e^{a t} u\right]=b e^{a t}$$ and $$\begin{aligned} e^{a t} u &=\frac{b}{a} e^{a t}+c \\ e^{a t} P^{-1} &=\frac{b}{a} e^{a t}+c \\ P^{-1} &=\frac{b}{a}+c e^{-a t} \\ P &=\frac{1}{b / a+c e^{-a t}}=\frac{a}{b+c_{1} e^{-a t}}. \end{aligned}$$

Separating variables, we have \\[\frac{d y}{y^{2}-y}=\frac{d x}{x} \quad \text { or } \quad \int \frac{d y}{y(y-1)}=\ln |x|+c.\\] Using partial fractions, we obtain $$\begin{aligned} \int\left(\frac{1}{y-1}-\frac{1}{y}\right) d y &=\ln |x|+c \\ \ln |y-1|-\ln |y| &=\ln |x|+c \\ \ln \left|\frac{y-1}{x y}\right| &=c \\ \frac{y-1}{x y} &=e^{c}=c_{1}.\end{aligned}$$ Solving for \(y\) we get \(y=1 /\left(1-c_{1} x\right) .\) We note by inspection that \(y=0\) is a singular solution of the differential equation. (a) Setting \(x=0\) and \(y=1\) we have \(1=1 /(1-0),\) which is true for all values of \(c_{1} .\) Thus, solutions passing through (0,1) are \(y=1 /\left(1-c_{1} x\right)\). (b) Setting \(x=0\) and \(y=0\) in \(y=1 /\left(1-c_{1} x\right)\) we get \(0=1 .\) Thus, the only solution passing through (0,0) is \(y=0\). (c) Setting \(x=\frac{1}{2}\) and \(y=\frac{1}{2}\) we have \(\frac{1}{2}=1 /\left(1-\frac{1}{2} c_{1}\right),\) so \(c_{1}=-2\) and \(y=1 /(1+2 x)\). (d) \(\operatorname{Setting} x=2\) and \(y=\frac{1}{4}\) we have \(\frac{1}{4}=1 /\left(1-2 c_{1}\right),\) so \(c_{1}=-\frac{3}{2}\) and \(y=1 /\left(1+\frac{3}{2} x\right)=2 /(2+3 x)\).

(a) The second derivative of \(y\) is \\[\frac{d^{2} y}{d x^{2}}=-\frac{d y / d x}{(y-1)^{2}}=-\frac{1 /(y-3)}{(y-3)^{2}}=-\frac{1}{(y-3)^{3}}\\] The solution curve is concave down when \(d^{2} y / d x^{2}<0\) or \(y>3\), and concave up when \(d^{2} y / d x^{2}>0\) or \(y<3 .\) From the phase portrait we see that the solution curve is decreasing when \(y<3\) and increasing when \(y>3\). (b) Separating variables and integrating we obtain \\[ \begin{aligned} (y-3) d y &=d x \\ \frac{1}{2} y^{2}-3 y &=x+c \\ y^{2}-6 y+9 &=2 x+c_{1} \\ (y-3)^{2} &=2 x+c_{1} \\ y &=3 \pm \sqrt{2 x+c_{1}}.\end{aligned}\\] The initial condition dictates whether to use the plus or minus sign. When \(y_{1}(0)=4\) we have \(c_{1}=1\) and \(y_{1}(x)=3+\sqrt{2 x+1}\). When \(y_{2}(0)=2\) we have \(c_{1}=1\) and \(y_{2}(x)=3-\sqrt{2 x+1}.\) When \(y_{3}(1)=2\) we have \(c_{1}=-1\) and \(y_{3}(x)=3-\sqrt{2 x-1}.\) When \(y_{4}(-1)=4\) we have \(c_{1}=3\) and \(y_{4}(x)=3+\sqrt{2 x+3}\).

(a) \(\operatorname{From} 2 W^{2}-W^{3}=W^{2}(2-W)=0\) we see that \(W=0\) and \(W=2\) are constant solutions. (b) Separating variables and using a CAS to integrate we get \\[ \frac{d W}{W \sqrt{4-2 W}}=d x \quad \text { and } \quad-\tanh ^{-1}\left(\frac{1}{2} \sqrt{4-2 W}\right)=x+c \\] Using the facts that the hyperbolic tangent is an odd function and \(1-\tanh ^{2} x=\operatorname{sech}^{2} x\) we have \\[ \begin{aligned} \frac{1}{2} \sqrt{4-2 W} &=\tanh (-x-c)=-\tanh (x+c) \\ \frac{1}{4}(4-2 W) &=\tanh ^{2}(x+c) \\ 1-\frac{1}{2} W &=\tanh ^{2}(x+c) \\ \frac{1}{2} W &=1-\tanh ^{2}(x+c)=\operatorname{sech}^{2}(x+c) \end{aligned} \\] Thus, \(W(x)=2 \operatorname{sech}^{2}(x+c)\) (c) Letting \(x=0\) and \(W=2\) we find that \(\operatorname{sech}^{2}(c)=1\) and \(c=0.\)

The left-hand derivative of the function at \(x=1\) is \(1 / e\) and the right- hand derivative at \(x=1\) is \(1-1 / e\). Thus, \(y\) is not differentiable at \(x=1\).
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