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Q 3.3-4E

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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 107
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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Short Answer

A red wine is brought up from the wine cellar, which is a cool 10°C, and left to breathe in a room of temperature 23°C. If it takes 10 min for the wine to reach 15°C, when will the temperature of the wine reach 18°C?

The temperature of the wine will reach 18°C after 19.7 minutes.

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Step by Step Solution

Step 1: Analyzing the given statement

The initial temperature of red wine is 10°C and left to breathe in a room of temperature 23°C. It takes 10 min for the wine to reach 15°C. By using Newton’s law of cooling, we have to determine the time after which the temperature of the wine will reach 18°C.

Newton’s Law of Cooling is,

Tt=M0+T0-M0e-kt······1

Here, we will take the values as,

Initial temperature,T0=10oC,

The temperature of the room, M0=23oC

Temperature after 10 min,T10=15oC

Step 2: To find the value of k in the formula of Newton’s Law of cooling

Using the given values in equation (1), to find the value of k,

T10=23+10-23e-10k 15=23+-13e-10k15-23=-13e-10k -8=-13e-10k e10k=138 10k=ln1.625 k=ln1.62510 k=0.0485

One will use this value of k in next step to find the time after which the temperature of the wine will reach 18°C.

Step 3: To determine the time after which the temperature of the wine will reach 18°C

Substituting Tt=18oC in equation (1),

18=23+10-23e-(0.0485)t 18-23=-13e-(0.0485)te(0.0485)t=1350.0485t=ln2.6 t=ln2.60.0485 t=19.7min

Hence, the temperature of the wine will reach 18°C after 19.7 minutes.

Most popular questions for Math Textbooks

Local versus Global Error. In deriving formula (4) for Euler’s method, a rectangle was used to approximate the area under a curve (see Figure 3.14). With

\({\bf{g(t) = f(t,f(t))}}\) , this approximation can be written as \(\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {{\bf{g(t)dt}} \approx {\bf{hg(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} \)where \({\bf{h = }}{{\bf{x}}_{{\bf{n + 1}}}}{\bf{ - }}{{\bf{x}}_{\bf{n}}}\) .

  1. Show that if g has a continuous derivative that is bounded in absolute value by B, then the rectangle approximation has error\(\left( {\bf{O}} \right){{\bf{h}}^{\bf{2}}}\); that is, for some constant M, \(\left| {\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {{\bf{g(t)dt - hg(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} } \right| \le {\bf{M}}{{\bf{h}}^{\bf{2}}}\).This is called the local truncation error of the scheme. [Hint: Write \(\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {{\bf{g(t)dt - hg(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} {\bf{ = }}\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {\left[ {{\bf{g(t)dt - g(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} \right]{\bf{dt}}} \). Next, using the mean value theorem, show that\(\left| {{\bf{g(t)dt - g(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} \right| \le {\bf{B}}\left| {{\bf{t - }}{{\bf{x}}_{\bf{n}}}} \right|\) . Then integrate to obtain the error bound\(\left( {\frac{{\bf{B}}}{{\bf{2}}}} \right){{\bf{h}}^{\bf{2}}}\).]
  2. In applying Euler’s method, local truncation errors occur in each step of the process and are propagated throughout the further computations. Show that the sum of the local truncation errors in part (a) that arise after n steps is (O)h. This is the global error, which is the same as the [ss1] [m2] convergence rate of Euler’s method.

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