Multivariate Calculus
&
Ordinary Differential Equations

Lecture 6



The story so far... 📖

- Definition of ODE and IVP.

- We learned how to solve $\ds \frac{dy}{dx}= f(x)\,$ or $\, \ds \frac{d^2y}{dt^2}= g(t).$

- Definition of Slope fields and Equilibrium solutions:

$\ds \frac{dy}{dx}= f(x,y)\;\;\; \text{or}\;\;\; \frac{dy}{dt}= f(t,y).$

- Euler's method. 🚀

- Identify and solve separable ODEs:

$\ds \frac{1}{g(y)}\frac{dy}{dx}= f(x)\;\;\; \text{or}\;\;\; \frac{1}{g(y)}\frac{dy}{dt}= f(t).$



1 Ordinary Differential Equations

1.5 Applications: Law of Cooling, Population Growth

Here we look at using some simple ODEs to model cooling and population dynamics. See Stewart Section 9.4



1 Ordinary Differential Equations

1.5.1 Newton's law of cooling

Newton's law of cooling states that the rate at which a "body" cools is proportional to the temperature difference between the body and its surrounding medium.



1 Ordinary Differential Equations

1.5.1 Newton's law of cooling

Newton's law of cooling states that the rate at which a "body" cools is proportional to the temperature difference between the body and its surrounding medium.

If $T$ is the temperature of the body and $T_m$ the temperature of the surrounding medium then, according to Newton,

$ \ds \dif{T}{t}=-k(T-T_m), \quad k>0. $



1 Ordinary Differential Equations

1.5.1 Newton's law of cooling

Newton's law of cooling states that the rate at which a "body" cools is proportional to the temperature difference between the body and its surrounding medium.

If $T$ is the temperature of the body and $T_m$ the temperature of the surrounding medium then, according to Newton,

$ \ds \dif{T}{t}=-k(T-T_m), \quad k>0. $

Here the constant $k$ is chosen such that if $T>T_m,$ $T'(t)$ is negative, describing cooling.



1 Ordinary Differential Equations

1.5.1 Newton's law of cooling



1 Ordinary Differential Equations

1.5.1 Newton's law of cooling



1 Ordinary Differential Equations

1.5.1 Newton's law of cooling

$ \ds \dif{T}{t}=-k(T-T_m), \quad k>0. $

Example: Solve this ODE with the intial condition $T(0)=T_0$.

Note that the ODE is separable.

The equilibrium solution is $T= T_m$.



1 Ordinary Differential Equations

1.5.1 Newton's law of cooling

$ \ds \dif{T}{t}=-k(T-T_m), \quad k>0. $

Example: Solve this ODE with the intial condition $T(0)=T_0$.

So we can write $\, \ds \frac{1}{T-T_m}\frac{dT}{dt}$ $= \ds - k,$ provided $T\neq T_m.$

$\Ra \,\ds \int \frac{1}{T-T_m}\frac{dT}{dt}dt$ $= \ds \int - k~dt$

$\Ra\, \ds \ln \abs{T-T_m} = - kt + C$



1 Ordinary Differential Equations

1.5.1 Newton's law of cooling

Example: Solve this ODE with the intial condition $T(0)=T_0$.

$\Ra\, \ds \ln \abs{T-T_m}= - kt + C$

$\Ra \,\ds\abs{T-T_m }= e^{-kt}e^{C}$ $\, \Ra\, T-T_m = A e^{-kt}.$

Then we have $\,T(t) = A e^{-kt} + T_m\,$ and

$\,T(0) = A + T_m$ $ =T_0$ $\, \Ra A =T_0-T_m.$

Hence $\, T(t) = \left(T_0-T_m\right) e^{-kt} + T_m.$



1 Ordinary Differential Equations

1.5.1 Newton's law of cooling

In real-life problems the cooling constant $k$ is usually not known and you will have to infer it from measurements.

Example: CSI Victoria Carl W.'s body was discovered in his cell at Barwon Prison, Victoria on the 19th of April 2010. The coroner's report stated that:

  • C.W.'s body discovered at 11am on 19 April 2010.
  • Body temperature at 11am: 34.8$^\circ$C.
  • Body temperature at 12.30pm: 34.1$^\circ$C.
  • Temperature in C.W.'s cell: 21.1$^\circ$C.

Use the above information to determine the time of Carl's death, assuming his body temperature at the time of death was 37$^\circ$C.



1 Ordinary Differential Equations

1.5.1 Newton's law of cooling

  • C.W.'s body discovered at 11am on 19 April 2010.
  • Body temperature at 11am: 34.8$^\circ$C.
  • Body temperature at 12.30pm: 34.1$^\circ$C.
  • Temperature in C.W.'s cell: 21.1$^\circ$C.

Use the above information to determine the time of Carl's death, assuming his body temperature at the time of death was 37$^\circ$C.

Newton's law of cooling: $\, T(t) = \left(T_0-T_m\right) e^{-kt} + T_m$

Let $t$ be time in hours after 11 am.

Then $T_0 = 34.8 ^{\circ}C$ is the temperature of the body when $t=0$. Also, $T_m = 21.1^{\circ}C.$



1 Ordinary Differential Equations

1.5.1 Newton's law of cooling

  • C.W.'s body discovered at 11am on 19 April 2010.
  • Body temperature at 12.30pm: 34.1$^\circ$C.

Use the above information to determine the time of Carl's death, assuming his body temperature at the time of death was 37$^\circ$C.

Thus we have $\, T(t) = \left(34.8-21.1\right) e^{-kt} + 21.1$

That is, $\, T(t) = 13.7 e^{-kt} + 21.1$

We need to find $k$ using the equation $T(1.5)= 34.1.$

So $\,T(1.5)$ $= 13.7 e^{-k(1.5)} + 21.1$ $= 34.1 $

$\Ra e^{-k(1.5)}= \ds \frac{13}{13.7} $ $\Ra k = \ds \frac{\ln\left(13/13.7\right)}{-1.5} $ $\approx 0.035 $



1 Ordinary Differential Equations

1.5.1 Newton's law of cooling

Use the above information to determine the time of Carl's death, assuming his body temperature at the time of death was 37$^\circ$C.

Thus we have $\, T(t) \approx 13.7 e^{-0.035t} + 21.1$

Assume that the time of death is $\, \tau,$ that is $\, T(\tau) = 37.$

Then $T(\tau) = 37$ $ \approx 13.7 e^{-0.035\tau} + 21.1$

$\Ra \tau \approx \dfrac{\ln \left(\dfrac{37-21.1}{13.7}\right)}{-0.035}$ $\approx -4.255$

$\;\;\;\quad \approx$ $4$ hrs, $15$ mins before 11 am.   ⏰💀 $\approx$ 6:45 am.



Credits