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Expert-verified Found in: Page 415 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # The temperature of a hot cup of coffee can be modelled by the DE${\mathbit{T}}{\mathbf{\text{'}}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{-}}{\mathbit{k}}{\mathbf{\left(}}{\mathbit{T}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{\right)}}{\mathbf{-}}{\mathbit{A}}{\mathbf{\right)}}$ (a) What is the significance of the constants K and A? (b) Solve the DE for T (t) in terms of K, A and the initial temperature ${{\mathbit{T}}}_{{\mathbf{0}}}$

(a) If the constant A is positive then the temperature of the hot cup of coffee modelled by the DE is positive and A is negative then the temperature of the hot cup of coffee modelled by the DE is negative.

(b)The solution is $A={T}_{0}-C{e}^{-kt}$

See the step by step solution

## Step 1 Explanation for Linear Differential operators and linear differential equations

A transformation T from ${C}^{\infty }\mathrm{to}{C}^{\infty }$to of the form role="math" localid="1660801324664" $T\left(f\right)={f}^{\left(n\right)}+{a}_{n-1}{f}^{\left(n-1\right)}+...+{a}_{1}f\text{'}+{a}_{0}f$is called an nth-order linear differential operator.Here ${f}^{\left(k\right)}$denote the k-th derivative of function f and the coefficients ${a}_{k}$are complex scalars.

If T is an nth-order linear differential operator and g is a smooth function, then the equation becomes

$T\left(f\right)=g$

(or)

${f}^{\left(n\right)}+{a}_{n-1}{f}^{\left(n-1\right)}+...+{a}_{1}f\text{'}+{a}_{0}f=g$

Is called an nth-order linear differential equation (DE).The DE is called homogeneous if g = 0 and inhomogeneous otherwise.

## (a) Step 2: Explanation of the significance for the constants K and A

Consider the temperature of a hot cup of coffee can be modelled by the DE

$\mathbit{T}\mathbf{\text{'}}\mathbf{\left(}\mathbit{t}\mathbf{\right)}\mathbf{=}\mathbit{k}\mathbf{\left(}\mathbit{T}\mathbf{\left(}\mathbit{t}\mathbf{\right)}\mathbf{-}\mathbit{A}\mathbf{\right)}$

Here k and A be the constants in the differential equation.

The constant is called the continuous growth rate if it is positive and the continuous decay rate if it is negative.

If the constant A is positive then the temperature of the hot cup of coffee modelled by the DE is positive and A is negative then the temperature of the hot cup of coffee modelled by the DE is negative.

Since both the constants k and A be the significant terms in the temperature of the hot cup of coffee modelled by the DE

## (b) Step 3 : Solution for the DE for T(t) in terms k, A and the initial temperature T0

Consider the temperature of a hot cup of coffee can be modelled by the DE

$\mathbit{T}\mathbf{\text{'}}\mathbf{\left(}\mathbit{t}\mathbf{\right)}\mathbf{=}\mathbit{k}\mathbf{\left(}\mathbit{T}\mathbf{\left(}\mathbit{t}\mathbf{\right)}\mathbf{-}\mathbit{A}\mathbf{\right)}$

Here k and A be the constants in the differential equation.

$T\text{'}\left(t\right)=k\left(T\left(t\right)-A\right)\phantom{\rule{0ex}{0ex}}T\text{'}\left(t\right)=-k\left(T\left(t\right)\right)+kA\phantom{\rule{0ex}{0ex}}T\text{'}\left(t\right)=\left(A-t\right)k\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\because T\text{'}\left(t\right)=\frac{dT}{dt}\phantom{\rule{0ex}{0ex}}\frac{dT}{dt}=\left(A-T\right)k\phantom{\rule{0ex}{0ex}}\frac{dT}{dt}=-k\left(T-A\right)$

Integrating on both sides we get,

$\frac{dT}{dt}=-k\left(T-A\right)\phantom{\rule{0ex}{0ex}}\frac{dT}{\left(T-A\right)}=-kdt\phantom{\rule{0ex}{0ex}}\int \frac{dT}{\left(T-A\right)}=\int -kdt\phantom{\rule{0ex}{0ex}}\mathrm{ln}\left|T-A\right|=-kt+C\phantom{\rule{0ex}{0ex}}T-A={e}^{-kt}{e}^{c}\phantom{\rule{0ex}{0ex}}T-A=C{e}^{-kt}\phantom{\rule{0ex}{0ex}}T\left(t\right)=A+C{e}^{-kt}\phantom{\rule{0ex}{0ex}}\because A=T-C{e}^{-kt}$

Here initial temperature becomes ${T}_{0}$and the solution is $A={T}_{0}-C{e}^{-kt}$. ### Want to see more solutions like these? 