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

The temperature T (in units of 100 F) of a university classroom on a cold winter day varies with time t (in hours) as $\frac{dT}{dt} = \{1 - T, if heating units is On - T, if heating units is OFF .$
$T = 0$ Suppose at 9:00 a.m., the heating unit is ON from 9-10 a.m., OFF from 10-11 a.m., ON again from 11 a.m.–noon, and so on for the rest of the day. How warm will the classroom be at noon? At 5:00 p.m.?

The temperature is 71.8-degree Fahrenheit at 12 noon and 26.9-degree Fahrenheit at 5pm and so on.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Find the solution when heat is on

Here heat is on at time 9 a.m, 11a.m and 1 p.m. $T_{0}$ is the temperature at room.

The differential equation when heat is on.

$\frac{dT}{dt} + T = 1$

The integrating factor is $e^{t}$ .then

$T . e^{t} = \int e^{t} dt T = 1 + {ce}^{- t}$

Apply the initial conditions then

$T = 1 + (T_{o} - 1) e^{- (t - t_{o})}$

Step 2: Determine the solution when heat is off

The differential equation when heat is off at time 10 a.m 12 noon and so on.

$\frac{dT}{dt} + T = 0$

The integrating factor is $e^{t}$ . And apply the initial conditions then

$T = T_{o} e^{- (t - t_{0})}$

Step 3: Find temperature

Let now $t = 0$ corresponding to 9 a.m. The temperature is 0 degree. And the heat is turned off.

Then $T = 1 - e^{- t}$ .

Now at 10 a.m , $t = 1$ then $T = (1 - e^{- t}) e^{- (t - 1)}$

Now at 11 a.m, $t = 2$ then $T = 1 + ((1 - e^{- 1}) e^{- 1} - 1) e^{- (t - 2)}$

Proceeding like this for heat is on then $T (t_{1}) = \sum_{n = 1}^{t_{1}} {(- 1)}^{n + 1} e^{- n}$ .

And proceeding for heat Is off then $T (t_{2}) = \sum_{n = 1}^{t_{2}} {(- 1)}^{n} e^{- n}$

At noon $t = 3$ then $T (3) = \sum_{n = 1}^{3} {(- 1)}^{n} e^{- n} = 1 - e^{- 1} + e^{- 2} - e^{- 3} = 0.718$

Thus, the temperature at this time is 71.8-degree Fahrenheit.

Similarly at 5pm then $T (8) = 26 . 9$

Thus, the temperature at this time is 26.9-degree Fahrenheit.

Therefore the temperature is 71.8-degree Fahrenheit at 12 noon and 26.9-degree Fahrenheit at 5pm and so on.

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Step by Step Solution

Step 1: Find the solution when heat is on

Step 2: Determine the solution when heat is off

Step 3: Find temperature

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Find the solution when heat is on

Step 2: Determine the solution when heat is off

Step 3: Find temperature

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