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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 9, Problema 9.4.30

27–30. Newton’s Law of Cooling Solve the differential equation for Newton’s Law of Cooling to find the temperature function in the following cases. Then answer any additional questions.


A pot of boiling soup (100°C) is put in a cellar with a temperature of 10°C. After 30 minutes, the soup has cooled to 80°C. When will the temperature of the soup reach 30°C 

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Identify the variables and constants in Newton's Law of Cooling. Let \(T(t)\) be the temperature of the soup at time \(t\) (in minutes), \(T_a\) be the ambient temperature (10°C), and \(T_0\) be the initial temperature of the soup (100°C).
Write the differential equation representing Newton's Law of Cooling: \(\frac{dT}{dt} = -k (T - T_a)\), where \(k\) is a positive constant representing the cooling rate.
Solve the differential equation by separating variables or using an integrating factor. The general solution has the form: \(T(t) = T_a + (T_0 - T_a) e^{-k t}\).
Use the given information that after 30 minutes, the temperature is 80°C to find the constant \(k\). Substitute \(t=30\), \(T(30) = 80\), \(T_a = 10\), and \(T_0 = 100\) into the solution and solve for \(k\).
Once \(k\) is found, set \(T(t) = 30\) and solve for \(t\) to find when the soup reaches 30°C.

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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, it is expressed as a first-order differential equation, which models how the temperature approaches the surrounding temperature over time.
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Newton's Law of Cooling

Solving First-Order Differential Equations

To find the temperature function, you solve the first-order linear differential equation derived from Newton’s Law of Cooling. This involves separating variables or using an integrating factor to obtain a general solution that describes temperature as a function of time.
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06:06
Solving Separable Differential Equations

Applying Initial Conditions and Solving for Time

After finding the general temperature function, initial conditions like the starting temperature and temperature at a specific time are used to determine constants. Then, to find when the temperature reaches a certain value, you solve the equation for time by substituting the target temperature.
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Percorso guidato
05:03
Initial Value Problems
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25–28. Two steps of Euler’s method For the following initial value problems, compute the first two approximations u1 and u2 given by Euler’s method using the given time step.


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12–16. Sketching direction fields Use the window [-2, 2] x [-2, 2] to sketch a direction field for the following equations. Then sketch the solution curve that corresponds to the given initial condition. A detailed direction field is not needed.

y(x) = sin y, y(−2) = 1/2

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5–16. Solving separable equations Find the general solution of the following equations. Express the solution explicitly as a function of the independent variable.

u'(x) = e²ˣ⁻ᵘ

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33–42. Solving initial value problems Solve the following initial value problems.

y'(x) = 4 sec² 2x, y(0) = 8

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39–42. Special equations A special class of first-order linear equations have the form a(t)y'(t)+a'(t)y(t)=f(t), where a and f are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form

a(t)y'(t) + a'(t)y(t) = d/dt (a(t)y(t)) = f(t). 

Therefore, the equation can be solved by integrating both sides with respect to t. Use this idea to solve the following initial value problems. 


t³y′(t) + 3t²y = (1 + t)/t, y(1) = 6

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