/*! 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 14 Assume that \(d T / d t=k(T-5)\)... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 14

Assume that \(d T / d t=k(T-5)\) so that \(T=5+c e^{k t} .\) If \(T(1)=55^{\circ}\) and \(T(5)=30^{\circ}\) then \(k=-\frac{1}{4} \ln 2\) and \(c=59.4611\) so that \(T(0)=64.4611^{\circ}\).

Short Answer

Expert verified
The temperature at \(t=0\) is approximately \(64.4611^{\circ}\).

Step by step solution

01

Start with the given equations

We are given the equation \(\frac{dT}{dt}=k(T-5)\) with the solution \(T=5+ce^{kt}\). We need to find the values of \(k\) and \(c\) that satisfy the conditions \(T(1)=55^{\circ}\) and \(T(5)=30^{\circ}\).
02

Use initial condition to find c

Substitute \(t=1\) and \(T(1)=55\) into the equation \(T=5 + ce^{kt}\):\[ 55 = 5 + ce^{k \cdot 1} \]Rearrange to find \(c\):\[ c = \frac{50}{e^k} \]
03

Use second condition to find k

Substitute \(t=5\) and \(T(5)=30\) into \(T=5 + ce^{kt}\):\[ 30 = 5 + ce^{k \cdot 5} \]Substitute \(c = \frac{50}{e^k}\) from Step 2:\[ 30 = 5 + \frac{50}{e^k}e^{5k} \]Simplify to solve for \(k\):\[ 25 = 50e^{4k} \]\[ e^{4k} = \frac{1}{2} \]\[ 4k = \ln\left(\frac{1}{2}\right) = -\ln 2 \]Thus, \(k = -\frac{1}{4}\ln 2\).
04

Confirm value of c

Now, substitute \(k = -\frac{1}{4}\ln 2\) back into \(c = \frac{50}{e^k}\):\[ c = 50 \cdot e^{1/4 \ln 2} = 50 \cdot \left(\frac{1}{2}\right)^{-1/4} \]Calculate \(c\):\[ c \approx 59.4611 \]
05

Find Temperature at t=0

Now calculate \(T(0)\) using the derived values:\[ T(0) = 5 + c \] Substitute \(c \approx 59.4611\):\[ T(0) = 5 + 59.4611 = 64.4611^{\circ} \]

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Key Concepts

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

Exponential Growth and Decay
Exponential growth and decay describe processes where a quantity increases or decreases at a rate proportional to its current value. In the context of differential equations, this is typically written in the form \( \frac{dy}{dt} = ky \), where \( y \) represents the quantity and \( k \) is a constant. If \( k > 0 \), we have exponential growth, and if \( k < 0 \), we have exponential decay. This behavior is behind many natural phenomena, such as population growth, radioactive decay, and cooling processes.

In the exercise provided, the differential equation \( \frac{dT}{dt} = k(T-5) \) shows an instance of exponential decay, since \( k \) is negative. The solution \( T = 5 + ce^{kt} \) reflects an exponential approach towards the equilibrium temperature of \( 5^{\circ} \) as time progresses. Understanding this concept helps one model how values change over time in real-world applications involving exponential growth or decay.
Initial Value Problem
An initial value problem (IVP) involves solving a differential equation supplemented by a value, known as an initial condition, specifying the value of the unknown function at a particular point. This condition narrows down the possible solutions to a single unique solution that fits the initial state.

In our example, the conditions \( T(1) = 55^{\circ} \) and \( T(5) = 30^{\circ} \) are used as initial values at different time points to solve for constants \( k \) and \( c \). These values ensure that the solution accurately describes the physical process over time. Solving initial value problems is crucial for accurately modeling situations in engineering, physics, and other sciences that start with a known state.
Separable Equations
Separable equations are a class of differential equations that can be written as a product of a function of the independent variable and a function of the dependent variable. These equations are solvable by separating the variables on different sides of the equation, allowing integration to be performed separately.

The equation \( \frac{dT}{dt} = k(T-5) \) in the exercise is separable because it can be rearranged to \( \frac{dT}{T-5} = k \cdot dt \). This form enables us to integrate both sides, providing a pathway to finding the general solution to the differential equation. Mastery of separable equations allows one to tackle a wide range of problems that involve processes dependent on separable variables. This method simplifies and broadens the ability to solve differential equations by applying basic integration techniques.

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Most popular questions from this chapter

(a) Writing the equation in the form \((x-\sqrt{x^{2}+y^{2}}) d x+y d y=0\) we identify \(M=x-\sqrt{x^{2}+y^{2}}\) and \(N=y\) since \(M\) and \(N\) are both homogeneous functions of degree 1 we use the substitution \(y=u x .\) It follows that \\[ \begin{aligned} (x-\sqrt{x^{2}+u^{2} x^{2}}) d x+u x(u d x+x d u) &=0 \\ x\left[1-\sqrt{1+u^{2}}+u^{2}\right] d x+x^{2} u d u &=0 \\ -\frac{u d u}{1+u^{2}-\sqrt{1+u^{2}}} &=\frac{d x}{x} \\ \frac{u d u}{\sqrt{1+u^{2}}(1-\sqrt{1+u^{2}})} &=\frac{d x}{x} \end{aligned} \\] Letting \(w=1-\sqrt{1+u^{2}}\) we have \(d w=-u d u / \sqrt{1+u^{2}}\) so that \\[ \begin{aligned} -\ln |1-\sqrt{1+u^{2}}| &=\ln |x|+c \\ \frac{1}{1-\sqrt{1+u^{2}}} &=c_{1} x \\ 1-\sqrt{1+u^{2}} &=-\frac{c_{2}}{x} \\ 1+\frac{c_{2}}{x} &=\sqrt{1+\frac{y^{2}}{x^{2}}} \\ 1+\frac{2 c_{2}}{x}+\frac{c_{2}^{2}}{x^{2}} &=1+\frac{y^{2}}{x^{2}} \end{aligned} \\] Solving for \(y^{2}\) we have \\[ y^{2}=2 c_{2} x+c_{2}^{2}=4\left(\frac{c_{2}}{2}\right)\left(x+\frac{c_{2}}{2}\right) \\] which is a family of parabolas symmetric with respect to the \(x\) -axis with vertex at \(\left(-c_{2} / 2,0\right)\) and focus at the origin. (b) Let \(u=x^{2}+y^{2}\) so that \\[ \frac{d u}{d x}=2 x+2 y \frac{d y}{d x} \\] Then \\[ y \frac{d y}{d x}=\frac{1}{2} \frac{d u}{d x}-x \\] and the differential equation can be written in the form \\[ \frac{1}{2} \frac{d u}{d x}-x=-x+\sqrt{u} \text { or } \frac{1}{2} \frac{d u}{d x}=\sqrt{u} \\] Separating variables and integrating gives $$\begin{aligned} \frac{d u}{2 \sqrt{u}} &=d x \\ \sqrt{u} &=x+c \\ u &=x^{2}+2 c x+c^{2} \\ x^{2}+y^{2} &=x^{2}+2 c x+c^{2} \\ y^{2} &=2 c x+c^{2} \end{aligned}$$

Solving \(P(5-P)-\frac{25}{4}=0\) for \(P\) we obtain the equilibrium solution \(P=\frac{5}{2} .\) For \(P \neq \frac{5}{2}, d P / d t<0 .\) Thus, if \(P_{0}<\frac{5}{2},\) the population becomes extinct (otherwise there would be another equilibrium solution.) Using separation of variables to solve the initial-value problem, we get \\[ P(t)=\left[4 P_{0}+\left(10 P_{0}-25\right) t\right] /\left[4+\left(4 P_{0}-10\right) t\right] \\] To find when the population becomes extinct for \(P_{0}<\frac{5}{2}\) we solve \(P(t)=0\) for \(t .\) We see that the time of extinction is \(t=4 P_{0} / 5\left(5-2 P_{0}\right).\)

From \(\frac{y+1}{y-1} d y=\frac{x+2}{x-3} d x\) or \(\left(1+\frac{2}{y-1}\right) d y=\left(1+\frac{5}{x-3}\right) d x\) we obtain \(y+2 \ln |y-1|=x+5 \ln |x-3|+c\) or \(\frac{(y-1)^{2}}{(x-3)^{5}}=c_{1} e^{x-y}\).

A model is $$\begin{aligned}&\frac{d x_{1}}{d t}=(4 \mathrm{gal} / \mathrm{min})(0 \mathrm{lb} / \mathrm{gal})-(4 \mathrm{gal} / \mathrm{min})\left(\frac{1}{200} x_{1} \mathrm{lb} / \mathrm{gal}\right)\\\&\begin{array}{l}\frac{d x_{2}}{d t}=(4 \mathrm{gal} / \mathrm{min})\left(\frac{1}{200} x_{1} \mathrm{lb} / \mathrm{gal}\right)-(4 \mathrm{gal} / \mathrm{min})\left(\frac{1}{150} x_{2} \mathrm{lb} / \mathrm{gal}\right) \\\\\frac{d x_{3}}{d t}=(4 \mathrm{gal} / \mathrm{min})\left(\frac{1}{150} x_{2} \mathrm{lb} / \mathrm{gal}\right)-(4 \mathrm{gal} / \mathrm{min})\left(\frac{1}{100} x_{3} \mathrm{lb} / \mathrm{gal}\right)\end{array}\end{aligned}$$ or $$\begin{aligned}\frac{d x_{1}}{d t} &=-\frac{1}{50} x_{1} \\\\\frac{d x_{2}}{d t} &=\frac{1}{50} x_{1}-\frac{2}{75} x_{2} \\\\\frac{d x_{3}}{d t} &=\frac{2}{75} x_{2}-\frac{1}{25} x_{3}\end{aligned}$$ Over a long period of time we would expect \(x_{1}, x_{2},\) and \(x_{3}\) to approach 0 because the entering pure water should flush the salt out of all three tanks.

From \(\frac{1}{y} d y=\frac{1-x}{x^{2}} d x=\left(\frac{1}{x^{2}}-\frac{1}{x}\right) d x\) we obtain \(\ln |y|=-\frac{1}{x}-\ln |x|=c\) or \(x y=c_{1} e^{-1 / x} .\) Using \(y(-1)=-1\) we find \(c_{1}=e^{-1} .\) The solution of the initial-value problem is \(x y=e^{-1-1 / x}\) or \(y=e^{-(1+1 / x)} / x\).
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