Question

Cake inside an Oven Reread Examplein Section on the cooling of a cake that is taken out of an oven.

a. Devise a mathematical model for the temperature of a cake while it is inside the oven based on the following assumptions: At $t=0$the cake mixture is at the room temperature of 70 degree; the oven is not preheated, so at $t=0$, when the cake mixture is placed into the oven, the temperature inside

The oven is also 70degree; the temperature of the oven increases linearly until $t=4$minutes, when the desired temperature of 300 degree is attained; the oven temperature is a constant degree is 300 attained; the oven temperature is a constant degree for$t>4$ .

b. Use the Laplace transforms to solve the initial-valueProblem in part (a)

Short answer

a) The model for the temperature of the cake while it is inside the oven is

$\frac{\text{d}T}{\text{d}t}=k(T-70-57.5t+(70+57.5t\left)U\right(t-4)-300U(t-4\left)\right),T\left(0\right)=70$

b) The Laplace transform for the given initial value problem is

$$\begin{gathered} t=70-\frac{115}{2k}-\frac{115}{2}t+\frac{115}{2k}{e}^{kt}-230U(t-4) \\ +230e^{k(t-4)}U(t-4)-\frac{115}{2k}U(t-4)-\frac{115}{2}(t-4)U(t-4)+115 \end{gathered}$$

Step-by-step solution

Define Laplace transform.

When specific initial conditions are supplied, especially when the initial values are zero, the Laplace transform is a handy method of solving certain types of differential equations. The Laplace transform $\mathcal{L}$of a function is defined as

$\mathcal{L}\left\{f\right(t\left)\right\}=\backslash {\mathrm{smallint}}_0^{\mathrm{\mathbb{N}}}{e}^{-st}f\left(t\right)dt$

In words, we can describe this expression as the Laplace transform of equals function F of s.

$\mathcal{L}\left\{f\right(t\left)\right\}=F\left(s\right)$

(a) Find the piecewise function.

(a) We can model the temperature of a cake inside an oven with Newton's law of cooling

$\frac{\text{d}T}{\text{d}t}=k\left(T-T_m\right)\text{(1)}$

where $T_m$is the ambient temperature of the oven.

$$\begin{gathered} T_m=70+\frac{300-70}{4-0}t \\ =70+\frac{115}{2}t \end{gathered}$$

This is only for first four minutes. The ambient temperature for any time is,

$$\begin{gathered} T_m=\left\{70+\frac{115}{2}t,0⩽t<4\right\} \\ {T}_m=\left\{300,t⩾4\right\} \end{gathered}$$

A piecewise function is defined as,

$$\begin{gathered} f\left(t\right)=\left\{g\left(t\right),0⩽t<a\right\} \\ f\left(t\right)=\left\{h\left(t\right),t⩾a\right\} \end{gathered}$$

It can also be expressed as

$$\begin{gathered} f\left(t\right)=g\left(t\right)-g\left(t\right)U(t-a)+h\left(t\right)U(t-a) \\ {T}_m=70+\frac{115}{2}t-\left(70+\frac{110}{2}t\right)U(t-4)+300U(t-4) \end{gathered}$$

Hence the piecewise function is $T_m=70+\frac{115}{2}t-\left(70+\frac{110}{2}t\right)U(t-4)+300U(t-4)$.

Find the initial value problem.

Plugging in the ambient temperature into equation

$\frac{\text{d}T}{\text{d}t}=k\left(T-70-\frac{115}{2}t+\left(70+\frac{115}{2}t\right)U(t-4)-300U(t-4)\right)$

The initial value problem is then,

$\frac{\text{d}T}{\text{d}t}=k(T-70-57.5t+(70+57.5t\left)U\right(t-4)-300U(t-4\left)\right),T\left(0\right)=70$

Hence the initial value problem is

$\frac{\text{d}T}{\text{d}t}=k(T-70-57.5t+(70+57.5t\left)U\right(t-4)-300U(t-4\left)\right),T\left(0\right)=70$

(b) Find the Laplace transform.

Take Laplace transform of both sides.

$$\begin{gathered} \text{If}F\left(s\right)=\mathcal{L}\left\{f\right(t\left)\right\}\text{and}a>0,\text{then}\mathcal{L}\left\{f\right(t-a\left)U\right(t-a\left)\right\}=e^{-as}F\left(s\right) \\ \mathcal{L}\left\{t^n\right\}=\frac{n!}{s^{n+1}},\mathcal{L}\left\{1\right\}=\frac{1}{s} \\ \mathcal{L}\left\{T\right\}-70\mathcal{L}\left\{1\right\}+\frac{115}{2}\mathcal{L}\left\{t\right\}-\mathcal{L}\left\{\left(70+\frac{115}{2}t\right)U(t-4)\right\}+300\mathcal{L}\left\{U\right(t-4\left)\right\} \\ \frac{1}{k}\mathcal{L}\left\{\frac{\text{d}T}{\text{d}t}\right\}\stackrel{(*)}{=}\mathcal{L}\left\{T\right\}-70\frac{1}{s}+\frac{115}{2}\frac{1}{s^2}-70\frac{1}{s}{e}^{-4s}+\frac{115}{2}\frac{1}{s^2}{e}^{-4s}+300\frac{1}{s}{e}^{-4s} \end{gathered}$$

We know that,

$\mathcal{L}\left\{\frac{\text{d}T}{\text{d}t}\right\}=s\mathcal{L}\left\{T\right\}-T\left(0\right),T\left(0\right)=70$

Then,

$$\begin{gathered} \frac{1}{k}s\mathcal{L}\left\{T\right\}-70\frac{1}{k}=\mathcal{L}\left\{T\right\}-70\frac{1}{s}+\frac{115}{2}\frac{1}{s^2}-70\frac{1}{s}{e}^{-4s}+\frac{115}{2}\frac{1}{s^2}{e}^{-4s}+300\frac{1}{s}{e}^{-4s} \\ \frac{1}{k}s\mathcal{L}\left\{T\right\}-\mathcal{L}\left\{T\right\}=70\frac{1}{k}-70\frac{1}{s}+\frac{115}{2}\frac{1}{s^2}-70\frac{1}{s}{e}^{-4s}+\frac{115}{2}\frac{1}{s^2}{e}^{-4s}+300\frac{1}{s}{e}^{-4s} \\ 70\frac{1}{k}-70\frac{1}{s}+\frac{115}{2}\frac{1}{s^2}-70\frac{1}{s}{e}^{-4s}+\frac{115}{2}\frac{1}{s^2}{e}^{-4s}+300\frac{1}{s}{e}^{-4s} \\ \mathcal{L}\left\{T\right\}=\frac{k}{s-k}70\frac{1}{k}-70\frac{k}{s-k}\frac{1}{s}+\frac{115}{2}\frac{k}{s-k}\frac{1}{s^2}+230\frac{k}{s-k}\frac{1}{s}{e}^{-4s}+\frac{115}{2}\frac{k}{s-k}\frac{1}{s^2}{e}^{-4s} \end{gathered}$$

Hence the Laplace transform is

$\mathcal{L}\left\{T\right\}=\frac{k}{s-k}70\frac{1}{k}-70\frac{k}{s-k}\frac{1}{s}+\frac{115}{2}\frac{k}{s-k}\frac{1}{s^2}+230\frac{k}{s-k}\frac{1}{s}{e}^{-4s}+\frac{115}{2}\frac{k}{s-k}\frac{1}{s^2}{e}^{-4s}$

Decompose the fraction using the partial fraction technique.

Decompose the fraction $\frac{1}{s(s-k)}$using the partial fraction technique.

$$\begin{gathered} \frac{1}{s(s-k)}=\frac{A}{s}+\frac{B}{s-k} \\ =\frac{A(s-k)+Bs}{s(s-k)}\text{(2)} \end{gathered}$$

Since the fractions in above equation have the same denominators, it takes their numerators to be equal.

$1=A(s-k)+Bs$

Comparing the coefficients.

$$\begin{gathered} 1=-Ak \\ A=-\frac{1}{k} \\ 1=Bk \\ B=\frac{1}{k} \end{gathered}$$

Substitute A, B in equation (1)

$\frac{1}{s(s-k)}=-\frac{1}{ks}+\frac{1}{k(s-k)}$

Hence the decomposed value of $\frac{1}{s(s-k)}is\frac{1}{s(s-k)}=-\frac{1}{ks}+\frac{1}{k(s-k)}$

Decompose the fraction using the partial fraction technique.

Decompose the fraction $\frac{1}{s^2(s-k)}$using the partial fraction technique.

$$\begin{gathered} \frac{1}{s^2(s-k)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s-k} \\ =\frac{A(s-k)s+B(s-k)+C{s}^2}{s^2(s-k)}\text{(3)} \end{gathered}$$

Since the fractions in above equation have the same denominators, it takes their numerators to be equal.

$$\begin{gathered} 1=A(s-k)s+B(s-k)+C{s}^2 \\ 1=A{s}^2-Aks+Bs-kB+C{s}^2 \\ 1=(A+C)s^2+(-Ak+B)s-kB \end{gathered}$$

Comparing the coefficients.

localid="1663855691057" $$\begin{gathered} 1=-Bk \\ B=-\frac{1}{k} \\ 1=C{k}^2 \\ C=\frac{1}{k^2} \\ A+C=0 \\ A=-C \\ A=-\frac{1}{k^2} \end{gathered}$$

Substitute A,B,C in equation 3

$\frac{1}{s^2(s-k)}=-\frac{1}{k^{2}s}-\frac{1}{k{s}^2}+\frac{1}{k^2(s-k)}$

Hence, the decomposed value of

$\frac{1}{s^2(s-k)}is\frac{1}{s^2(s-k)}=-\frac{1}{k^{2}s}-\frac{1}{k{s}^2}+\frac{1}{k^2(s-k)}$

Find the inverse Laplace.

Substitute the decomposed values in

$\mathcal{L}\left\{T\right\}=\frac{k}{s-k}70\frac{1}{k}-70\frac{k}{s-k}\frac{1}{s}+\frac{115}{2}\frac{k}{s-k}\frac{1}{s^2}+230\frac{k}{s-k}\frac{1}{s}{e}^{-4s}+\frac{115}{2}\frac{k}{s-k}\frac{1}{s^2}{e}^{-4s}$

The equation becomes,

$$\begin{gathered} \mathcal{L}\left\{T\right\}=\frac{1}{s-k}70-70k\left[-\frac{1}{ks}+\frac{1}{k(s-k)}\right]+\frac{115}{2}k\left[-\frac{1}{k^{2}s}-\frac{1}{k{s}^2}+\frac{1}{k^2(s-k)}\right] \\ +230k\left[-\frac{1}{ks}+\frac{1}{k(s-k)}\right]e^{-4s}+\frac{115}{2}k\left[-\frac{1}{k^{2}s}-\frac{1}{k{s}^2}+\frac{1}{k^2(s-k)}\right]e^{-4s} \end{gathered}$$

Take inverse Laplace transform on both sides.

$$\begin{gathered} {\mathcal{L}}^{-1}\left\{T\right\}=70{\mathcal{L}}^{-1}\left\{\frac{1}{s-k}\right\}-70k\left[-{\mathcal{L}}^{-1}\left\{\frac{1}{ks}\right\}+{\mathcal{L}}^{-1}\left\{\frac{1}{k(s-k)}\right\}\right] \\ +\frac{115}{2}k\left[-{\mathcal{L}}^{-1}\left\{\frac{1}{k^{2}s}\right\}-{\mathcal{L}}^{-1}\left\{\frac{1}{k{s}^2}\right\}+{\mathcal{L}}^{-1}\left\{\frac{1}{k^2(s-k)}\right\}\right] \end{gathered}$$

$$\begin{gathered} +230k\left[-{\mathcal{L}}^{-1}\left\{\frac{1}{ks}\right\}+{\mathcal{L}}^{-1}\left\{\frac{1}{k(s-k)}\right\}\right]e^{-4s} \\ +\frac{115}{2}k\left[-{\mathcal{L}}^{-1}\left\{\frac{1}{k^{2}s}\right\}-{\mathcal{L}}^{-1}\left\{\frac{1}{k{s}^2}\right\}+{\mathcal{L}}^{-1}\left\{\frac{1}{k^2(s-k)}\right\}\right]e^{-4s} \end{gathered}$$

$$\begin{gathered} =70-\frac{115}{2k}-\frac{115}{2}t+\frac{115}{2k}{e}^{kt}-230U(t-4) \\ +230e^{k(t-4)}U(t-4)-\frac{115}{2k}U(t-4)-\frac{115}{2}(t-4)U(t-4)+\frac{115}{2k}{e}^{k(t-4)}U(t-4) \end{gathered}$$

$$\begin{gathered} \text{If}{\mathcal{L}}^{-1}\left\{F\right(s\left)\right\}=f\left(t\right)\text{and}a>0,\text{then}f(t-a)U(t-a)={\mathcal{L}}^{-1}\left\{e^{-as}F\left(s\right)\right\} \\ {t}^n={\mathcal{L}}^{-1}\left\{\frac{n!}{s^{n+1}}\right\},1={\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\} \\ U(t-a)={\mathcal{L}}^{-1}\left\{\frac{e^{-as}}{s}\right\} \end{gathered}$$

So,

$$\begin{gathered} t=70-\frac{115}{2k}-\frac{115}{2}t+\frac{115}{2k}{e}^{kt}-230U(t-4) \\ +230e^{k(t-4)}U(t-4)-\frac{115}{2k}U(t-4)-\frac{115}{2}(t-4)U(t-4)+\frac{115}{2k}{e}^{k(t-4)}U(t-4) \end{gathered}$$

Hence the Laplace transform for the given initial value problem is

$$\begin{gathered} t=70-\frac{115}{2k}-\frac{115}{2}t+\frac{115}{2k}{e}^{kt}-230U(t-4) \\ +230e^{k(t-4)}U(t-4)-\frac{115}{2k}U(t-4)-\frac{115}{2}(t-4)U(t-4)+\frac{115}{2k}{e}^{k(t-4)}U(t-4) \end{gathered}$$

$$\begin{gathered} t=70-\frac{115}{2k}-\frac{115}{2}t+\frac{115}{2k}{e}^{kt}-230U(t-4) \\ +230e^{k(t-4)}U(t-4)-\frac{115}{2k}U(t-4)-\frac{115}{2}(t-4)U(t-4)+\frac{115}{2k}{e}^{k(t-4)}U(t-4) \end{gathered}$$