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Q 3.3-4E

Expert-verifiedFound in: Page 107

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**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.**

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,

$T\left(t\right)={M}_{0}+\left({T}_{0}-{M}_{0}\right){e}^{-kt}\xb7\xb7\xb7\xb7\xb7\xb7\left(1\right)$

Here, we will take the values as,

Initial temperature,${T}_{0}={10}^{o}C$,

The temperature of the room, ${M}_{0}={23}^{o}C$

Temperature after 10 min,$T\left(10\right)={15}^{o}C$

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

$T\left(10\right)=23+\left(10-23\right){e}^{-10k}\phantom{\rule{0ex}{0ex}}15=23+\left(-13\right){e}^{-10k}\phantom{\rule{0ex}{0ex}}15-23=\left(-13\right){e}^{-10k}\phantom{\rule{0ex}{0ex}}-8=\left(-13\right){e}^{-10k}\phantom{\rule{0ex}{0ex}}{e}^{10k}=\frac{13}{8}\phantom{\rule{0ex}{0ex}}10k=ln\left(1.625\right)\phantom{\rule{0ex}{0ex}}k=\frac{ln\left(1.625\right)}{10}\phantom{\rule{0ex}{0ex}}k=0.0485\phantom{\rule{0ex}{0ex}}$

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.

Substituting $T\left(t\right)={18}^{o}C$ in equation (1),

$18=23+\left(10-23\right){e}^{-(0.0485)t}\phantom{\rule{0ex}{0ex}}18-23=\left(-13\right){e}^{-(0.0485)t}\phantom{\rule{0ex}{0ex}}{e}^{(0.0485)t}=\frac{13}{5}\phantom{\rule{0ex}{0ex}}\left(0.0485\right)t=ln\left(2.6\right)\phantom{\rule{0ex}{0ex}}t=\frac{ln\left(2.6\right)}{0.0485}\phantom{\rule{0ex}{0ex}}t=19.7min\phantom{\rule{0ex}{0ex}}$

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

**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 \({\bf{h = }}{{\bf{x}}_{{\bf{n + 1}}}}{\bf{ - }}{{\bf{x}}_{\bf{n}}}\) .**

**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\(\left( {\frac{{\bf{B}}}{{\bf{2}}}} \right){{\bf{h}}^{\bf{2}}}\).]****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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