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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 9 - Differential EquationsProblem 9.4.38a
Chapter 9, Problem 9.4.38a

Cooling time Suppose an object with an initial temperature of T₀ > 0 is put in surroundings with an ambient temperature of A, where A < T₀/2. Let t₁/₂ be the time required for the object to cool to T₀/2.


a. Show that t₁/₂ = −1/k ln((T₀ − 2A)/(2(T₀ − A))).

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1
Start with Newton's Law of Cooling, which states that the temperature \( T(t) \) of the object at time \( t \) satisfies the differential equation: \[\frac{dT}{dt} = -k (T - A)\] where \( k > 0 \) is a constant and \( A \) is the ambient temperature.
Solve the differential equation by separating variables or recognizing it as a first-order linear ODE. The general solution is: \[T(t) = A + (T_0 - A) e^{-k t}\] where \( T_0 \) is the initial temperature at \( t = 0 \).
Define \( t_{1/2} \) as the time when the temperature reaches half of the initial temperature, i.e., when \( T(t_{1/2}) = \frac{T_0}{2} \). Substitute this into the solution: \[\frac{T_0}{2} = A + (T_0 - A) e^{-k t_{1/2}}\]
Isolate the exponential term: \[\frac{T_0}{2} - A = (T_0 - A) e^{-k t_{1/2}}\] Then divide both sides by \( T_0 - A \): \[\frac{\frac{T_0}{2} - A}{T_0 - A} = e^{-k t_{1/2}}\]
Take the natural logarithm of both sides to solve for \( t_{1/2} \): \[-k t_{1/2} = \ln \left( \frac{\frac{T_0}{2} - A}{T_0 - A} \right)\] Rearranging gives: \[t_{1/2} = -\frac{1}{k} \ln \left( \frac{T_0 - 2A}{2 (T_0 - A)} \right)\] which is the desired expression.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Newton's Law of Cooling

Newton's Law of Cooling states that the rate of change of an object's temperature is proportional to the difference between its temperature and the ambient temperature. Mathematically, this is expressed as dT/dt = -k(T - A), where k > 0 is a constant. This law models how objects cool or warm over time toward the surrounding temperature.
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Newton's Law of Cooling

Solving First-Order Linear Differential Equations

The temperature function T(t) satisfies a first-order linear differential equation. Solving it involves separating variables or using an integrating factor to find T as a function of time. This solution typically involves an exponential decay term reflecting how temperature approaches ambient temperature over time.
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Logarithmic Manipulation in Solving for Time

To find the cooling time t₁/₂ when the temperature reaches T₀/2, one must solve the exponential equation for t. This requires isolating t by taking the natural logarithm of both sides, using properties of logarithms to simplify the expression, and expressing t₁/₂ in terms of known quantities like T₀, A, and k.
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Related Practice
Textbook Question

38–43. Equilibrium solutions A differential equation of the form y′(t)=f(y) is said to be autonomous (the function f depends only on y). The constant function y=y0 is an equilibrium solution of the equation provided f(y0)=0 (because then y'(t)=0 and the solution remains constant for all t). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations.

a. Find the equilibrium solutions. 


y′(t) = y(y - 3)

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Textbook Question

A second-order equation Consider the differential equation y''(t) - k²y(t) = 0 where k > 0 is a real number.


a. Verify by substitution that when k = 1, a solution of the equation is y(t) = C₁eᵗ + C₂e⁻ᵗ. You may assume this function is the general solution.

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Textbook Question

42–43. Implicit solutions for separable equations For the following separable equations, carry out the indicated analysis.

a. Find the general solution of the equation.


y'(t) = t²/(y² + 1); y(−1) = 1, y(0) = 0, y(−1) = −1


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Textbook Question

{Use of Tech} Endowment model An endowment is an investment account in which the balance ideally remains constant and withdrawals are made on the interest earned by the account. Such an account may be modeled by the initial value problem B′(t)=rB−m, for t≥0, with B(0)=B0. The constant r>0 reflects the annual interest rate, m>0 is the annual rate of withdrawal, B0 is the initial balance in the account, and t is measured in years.


a. Solve the initial value problem with r=0.05, m=\(1000/year, and B0=\)15,000 Does the balance in the account increase or decrease?

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Textbook Question

52-56. In this section, several models are presented and the solution of the associated differential equation is given. Later in the chapter, we present methods for solving these differential equations.


where P(t) is the population, for t ≥ 0, and r > 0 and K > 0 are given constants.


a. Verify by substitution that the general solution of the equation is P(t) = K/(1 + Ce⁻ʳᵗ), where C is an arbitrary constant.

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Textbook Question

23–26. Stirred tank reactions For each of the following stirred tank reactions, carry out the following analysis.

a. Write an initial value problem for the mass of the substance.


A 500-L tank is initially filled with pure water. A copper sulfate solution with a concentration of 20 g/L flows into the tank at a rate of 4 L/min. The thoroughly mixed solution is drained from the tank at a rate of 4 L/min.

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