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Chapter 2: Problem 34

Newton's Law of Cooling. According to Newton's law of cooling, if an object at temperature \( T \) is immersed in a medium having the constant temperature \( M \), then the rate of change of \( T \) is proportional to the difference of temperature \( M-T \). This gives the differential equation $$ d T / d t=k(M-T) $$ (a) Solve the differential equation for \( T \) (b) A thermometer reading \( 100^{\circ} \mathrm{F} \) is placed in a medium having a constant temperature of \( 70^{\circ} \mathrm{F} \). After 6 min, the thermometer reads \( 80^{\circ} \mathrm{F} \). What is the reading after 20 min? (Further applications of Newton's law of cooling appear in Section 3.3.)

Short Answer

Expert verified
The thermometer will read approximately 70.48 degrees Fahrenheit after 20 minutes.

Step by step solution

01

Solve The Differential Equation

This is a first order linear differential equation. The standard form for such an equation is dy/dx + p(x)y = q(x). Here, dy/dx is dT/dt, p(x) is -k, y is T and q(x) is kM. We can use an integrating factor to solve this equation. The integrating factor is e^(鈭玴(x) dx), which in this case is e^(鈭�-k dt) = e^(-kt). Multiplying every term in the differential equation by e^(-kt) gives e^(-kt)dT/dt + e^(-kt)(-k)T = e^(-kt)kM. The left-hand side of this equation is the derivative of (e^(-kt)T) with respect to t. So, 鈭玠/dt[e^(-kt)T] dt = 鈭玨M e^(-kt) dt. Integration of both sides with respect to t yields e^(-kt)T = -M e^(-kt) - C. Multiplying all terms by -1 and rearranging gives T = M + Ce^(kt) where C is the integration constant.
02

Substituting the Given Values

From the given scenario, T(0) = 100. Substituting these values into T = M + Ce^(kt) gives 100 = 70 + C. So, C = 30. Next, it is given that T(6) = 80. Substituting these values into T = M + Ce^(kt) gives 80 = 70 + 30e^(6k). Solving for k gives k = ln[(80 - 70) / 30] / 6 鈮� -0.0465.
03

Finding the Temperature After 20 Min

Now that we have found the values of C and k, we can substitute these into the formula T = M + Ce^(kt) to find the temperature at t = 20. This gives T(20) = 70 + 30e^(-0.0465 脳 20) which is approximately 70.48 degrees Fahrenheit

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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 relate a function with its derivatives. They express how the rate of change in one variable is related to other variables.

For example, Newton's Law of Cooling provides a differential equation that describes how the temperature of an object changes over time. This equation shows that the rate of temperature change is proportional to the difference between the object's temperature and the surrounding medium's temperature.
  • This type of equation is essential in modeling scenarios where change rates are important.
  • They help predict how a system behaves under various conditions.
First Order Linear Differential Equation
A first order linear differential equation is one where the function and its first derivative appear linearly. The standard form is: \( \frac{dy}{dx} + p(x)y = q(x) \).

In Newton's Law of Cooling, the equation \( \frac{dT}{dt} = k(M-T) \) fits this structure by setting \( p(x) = -k \) and \( q(x) = kM \).
  • "First order" means it involves only the first derivative (\( \frac{dT}{dt} \)).
  • "Linear" indicates that both the function and its derivative are not raised to any power other than one.
Solving these equations often involves finding an integrating factor to simplify the integration process.
Integrating Factor
The integrating factor is a key technique for solving first order linear differential equations. It helps convert the non-exact differential equation into an exact one, allowing for easier integration.

To find the integrating factor, use \( e^{\int p(x) \, dx} \). In our problem, this becomes \( e^{-kt} \).
  • Multiply the entire differential equation by the integrating factor.
  • This converts the left side into the derivative of a product.
After multiplying, you can integrate both sides to find a general solution.

When solving Newton鈥檚 Law of Cooling, this method transforms the problem into a more straightforward integral, ultimately yielding a solution for the temperature function \( T(t) = M + Ce^{kt} \).
Temperature Change Rate
In the context of Newton's Law of Cooling, the temperature change rate describes how quickly or slowly an object's temperature approaches that of the surrounding environment.

This rate is governed by the proportionality constant \( k \) in the equation \( \frac{dT}{dt} = k(M-T) \).
  • If \( k \) is negative, it indicates the temperature is decreasing.
  • The greater the magnitude of \( k \), the faster the temperature moves toward equilibrium.
For advancements in practical understanding, the value of \( k \) can be determined experimentally by observing temperature changes over time.

This concept is crucial in many real-world applications where temperature regulation and control are vital.

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Most popular questions from this chapter

Consider the initial value problem $$ \frac{d y}{d x}+\sqrt{1+\sin ^{2} x} y=x, \quad y(0)=2 $$. (a) Using definite integration, show that the integrating factor for the differential equation can be written as $$ \mu(x)=\exp \left(\int_{0}^{x} \sqrt{1+\sin ^{2} t} d t\right) $$ and that the solution to the initial value problem is $$ y(x)=\frac{1}{\mu(x)} \int_{0}^{x} \mu(s) s d s+\frac{2}{\mu(x)} $$. (b) Obtain an approximation to the solution at \( x=1 \) by using numerical integration (such as Simpson's rule, Appendix C) in a nested loop to estimate values of \( \mu(x) \) and, thereby, the value of $$ \int_{0}^{1} \mu(s) s d s $$. [Hint: First, use Simpson's rule to approximate \( \mu(x) \) at \( x=0.1,0.2, \dots, 1 \). Then use these values and apply Simpson's rule again to approximate \( \int_{0}^{1} \mu(s) s d s . ] \) (c) Use Euler's method (Section 1.4) to approximate the solution at \( x=1 \), with step sizes \( h=0.1 \) and 0.05. [A direct comparison of the merits of the two numerical schemes in parts (b) and (c) is very complicated, since it should take into account the number of functional evaluations in each algorithm as well as the inherent accuracies.]

Riccati Equation. An equation of the form $$\frac{d y}{d x}=P(x) y^{2}+Q(x) y+R(x)$$ is called a generalized Riccati equation. (a) If one solution_say,$$u(x)-\text { of }(18)$$ is known, show that the substitution $$y=u+1 / v \text { reduces }(18)$$ to a linear equation in v. (b) Given that $$u(x)=x$$ is a solution to $$\frac{d y}{d x}=x^{3}(y-x)^{2}+\frac{y}{x}$$ use the result of part (a) to find all the other solutions to this equation. (The particular solution $$u(x)=x$$ can be found by inspection or by using a Taylor series method; see Section 8.1.)

As stated in this section, the separation of equation (2) on page 42 requires division by \( p(y) \), and this may disguise the fact that the roots of the equation \( p(y)=0 \) are actually constant solutions to the differential equation. (a) To explore this further, separate the equation $$ \frac{d y}{d x}=(x-3)(y+1)^{2 / 3} $$ to derive the solution, $$ y=-1+\left(x^{2} / 6-x+C\right)^{3} $$ (b) Show that \( y \equiv-1 \) satisfies the original equation \( d y / d x=(x-3)(y+1)^{2 / 3} \) (c) Show that there is no choice of the constant \( C \) that will make the solution in part (a) yield the solution \( y \equiv-1 \) Thus, we lost the solution \( y=-1 \) when we divided by \((y+1)^{2 / 3}\).

Uniqueness Questions. In Chapter 1 we indicated that in applications most initial value problems will have a unique solution. In fact, the existence of unique solutions was so important that we stated an existence and uniqueness theorem, Theorem 1, page 11. The method for separable equations can give us a solution, but it may not give us all the solutions (also see Problem 30). To illustrate this, consider the equation \( d y / d x=y^{1 / 3} \). (a) Use the method of separation of variables to show that $$ y=\left(\frac{2 x}{3}+C\right)^{3 / 2} $$ is a solution. (b) Show that the initial value problem \( d y / d x \) = \( y^{1 / 3} \) with \( y(0)=0 \) is satisfied for \( C=0 \) by \( y=(2 x / 3)^{3 / 2} \) for \( x \geq 0 \). (c) Now show that the constant function \( y \equiv 0 \) also satisfies the initial value problem given in part (b). Hence, this initial value problem does not have a unique solution. (d) Finally, show that the conditions of Theorem 1 on page 11 are not satisfied. (The solution \( y \equiv 0 \) was lost because of the division by zero in the separation process.)

Compound Interest. If \(P(t)\) is the amount of dollars in a savings bank account that pays a yearly interest rate of \(r \%\) compounded continuously, then $$\frac{d P}{d t}=\frac{r}{100} P, \quad t$$ in years. Assume the interest is 5\(\%\) annually, \(P(0)=\$ 1000,\) and no monies are withdrawn. (a) How much will be in the account after 2 yr? (b) When will the account account every 12 months, (c) If \(\$ 1000\) is added to the account every 12 months, how much will be in the account after 3\(\frac{1}{2}\) yr?
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