1. Application of calculus in
everyday life.
Newton’s Law of Cooling.

2. What is the differential equation?
 A differential equation is an equation involving derivatives of an unknown
function and possibly the function itself as well as the independent
variable.
 Differential equations have many forms and its order is determined based on the
highest order of a derivative in it.
 First order differential equations are such equations that have the unknown
derivative is the first derivative and its own function.
 They are divided into separable and 1st order DFE linear.

3. First DFE
1st order DFE linear1st order DFE linear
𝑑𝑦
𝑑𝑥
= 𝐹(𝑥, 𝑦)
𝒅𝒚
𝒅𝒙
= 𝒇 𝒙 ∗ 𝒈 𝒚
So F(x, y) is simply f(x)*g(y)

4. How can we find the solution of the 1st ODE?
A first order linear differential equation is an equation of the form
( ) ( )
dy
P x y Q x
dx
 
( ) 0
dy
P x y
dx
 
Which can be solved by
separating the variables.
( )
dy
P x dx
y
   
ln ( )y P x dx c   
( )P x dx c
y e
  
( )P x dx c
y e e
 
( )P x dx
y Ce
 
( )P x dxd
ye
dx
 
( ) ( )
( )
P x dx P x dxdy
e yP x e
dx
 
( )
( )
P x dxdy
P x y e
dx
    
 

5. ( ) ( )
dy
P x y Q x
dx
  If we multiply both sides by
( )P x dx
e
( ) ( )
( )
P x dx P x dxd
ye Q x e
dx
 
Now integrate both sides.
( ) ( )
( )
P x dx P x dx
ye Q x e dx  
Returning to equation 1,

6. The change in temperature
• An object’s temperature over time will approach the
temperature of its surroundings (the medium).
• The greater the difference between the object’s temperature
and the medium’s temperature, the greater the rate of change
of the object’s temperature.
• This change is a form of exponential decay.
T0
Tm

7. Newton’s Law of Cooling
 It is a direct application for differential equations.
 Formulated by Sir Isaac Newton.
 Has many applications in our everyday life.
 Sir Isaac Newton found this equation behaves like what is called in Math
(differential equations) so his used some techniques to find its general solution.

8. Derivation of Newton’s law of Cooling
 Newton’s observations:
He observed that observed that the temperature of the body is proportional to the
difference between its own temperature and the temperature of the objects in
contact with it .
 Formulating:
First order separable DE
 Applying calculus:
𝑑𝑇
𝑑𝑡
= −𝑘(𝑇 − 𝑇𝑒)
Where k is the positive proportionality constant

9. Derivation of Newton’s law of Cooling (continued)
 By separation of variables we get
𝑑𝑇
(𝑇−𝑇𝑒)
= −𝑘 𝑑𝑡
 By integrating both sides we get
ln 𝑇 − 𝑇𝑒 + 𝐶 = −𝑘𝑡
 At time (t=0) the temperature is T0
−ln 𝑇0 − 𝑇𝑒 = 𝐶
 By substituting C with −ln 𝑇0 − 𝑇𝑒 we get
ln
(𝑇 − 𝑇𝑒)
(𝑇0 − 𝑇𝑒)
= −𝑘𝑡
𝑇 = 𝑇𝑒 + (𝑇0 − 𝑇𝑒)𝑒−𝑘𝑡

10. Applications on Newton’s Law of Cooling:
Investigations.
• It can be used to
determine the
time of death.
Computer
manufacturing.
• Processors.
• Cooling systems.
solar water
heater.
calculating the
surface area of
an object.

11. Expressing the
applications of
Newton’s law of cooling
through mathematical
problems
Investigations in a crime scene
Processor manufacturing

12. The police came to a house at 10:23 am were a murder had
taken place. The detective measured the temperature of the
victim’s body and found that it was 26.7℃. Then he used a
thermostat to measure the temperature of the room that
was found to be 20℃ through the last three days. After an
hour he measured the temperature of the body again and
found that the temperature was 25.8℃. Assuming that the
body temperature was normal (37℃), what is the time of
death?

13. Solution
T (t) = Te + (T0 − Te ) e – kt
Let the time at which the death took place be x hours before the arrival of the
police men.
Substitute by the given values
T ( x ) = 26.7 = 20 + (37 − 20) e-kx
T ( x+1) = 25.8 = 20 + (37 − 20) e - k ( x + 1)
Solve the 2 equations simultaneously
0.394= e-kx
0.341= e - k ( x + 1)
By taking the logarithmic function
ln (0.394)= -kx …(1)
ln (0.341)= -k(x+1) …(2)

14. Solution (continued)
By dividing (1) by (2)
ln(0.394)
ln 0.341
=
−𝑘𝑥
−𝑘 𝑥+1
0.8657 =
𝑥
𝑥+1
Thus x≃7 hours
Therefore the murder took place 7 hours before the arrival of the detective
which is at 3:23 pm

15. A global company such as Intel is willing to produce a new cooling system for their processors
that can cool the processors from a temperature of 50℃ to 27℃ in just half an hour when
the temperature outside is 20℃ but they don’t know what kind of materials they should use
or what the surface area and the geometry of the shape are. So what should they do ?
Simply they have to use the general formula of Newton’s law of cooling
T (t) = Te + (T0 − Te ) e – k
And by substituting the numbers they get
27 = 20 + (50 − 20) e-0.5k
Solving for k we get k =2.9
so they need a material with k=2.9 (k is a constant that is related to the heat capacity ,
thermodynamics of the material and also the shape and the geometry of the material)