Applications of Differential Equations in Engineering and Mathematics

 
K S SCHOOL OF ENGINEERING AND MANAGEMENT
DEPARTMENT OF MATHEMATICS
 
TOPIC : Calculus And Linear Algebra
Module-4 : Applications of Differential
Equation
 
NAME : DIVYA .R.
DESIGNATION : Assistant Professor
SUBJECT CODE : 18MAT11
 
 
 
 
 
 
 
 
1.Newton’s Law of Cooling
2.LCR circuits
3.Problems on beams
4.Orthogonal Trajectories
Differential Equation
Exact Differential Equations:
The differential equation M(x , y) dx + N(x , y)dy=0 to be an
exact equation is
M/ 
y 
 =
N/ 
x
Further the solution of the exact equation is given by
M dx 
+ ò
 N 
(
 y 
)
dy
=
 c
 
 We  integrate M(x,y) w.r.t x keeping y fixed and N(y) indicate
the terms in N does not contain x
 
A
p
p
l
i
c
a
t
i
o
n
 
 
:
 
N
e
w
t
o
n
'
s
 
L
a
w
 
o
f
 
C
o
o
l
i
n
g
It is a model that describes, mathematically, the change in
temperature of an object in a given environment. The law states
that the rate of change (in time) of the temperature is proportional
to the difference between the temperature T of the object and the
temperature Te of the environment surrounding the object.
               d T / d t = - k (T - Te)
Let x = T – Te       so that dx / dt = dT / dt
Using the above change of variable, the above differential equation
becomes
d x / d t = - k x
The solution to the above differential equation is given by
x = A e 
- k t
substitute x by T - Te
T - Te = A e 
- k t
Assume that at t = 0 the temperature T = To
To - Te = A e 
0
which gives A = To - Te
The final expression for T(t) i given by
T(t) = Te + (To - Te)e 
- k t
This last expression shows how the temperature T of
the object changes with time.
Real life examples
A coffee  of 98
 degree centigrade will take
30 minutes to come down to 45  degree
centigrade in a room of temperature 30
degree centigrade
A dosa tava will take at teast 20 minutes
to come down from 240 degree
centigrade to 80 degree centigrade so
that we can put it in wash basin.
Orthogonal Trajectories(OT)
Trajectory: A curve which cuts every member of a
given family of curves according to some definite
law .
Orthogonal Trajectory: A curve which cuts every
member of a given family of curves at right angles .
 
Examples
 
Orthogonal trajecteries for  concentric circles is
family of stright lines passing through origin
Orthogonal trajecteries for cardioids
R=a(1+cos t) is r=a(1+sint)
 
W
o
r
k
i
n
g
 
r
u
l
e
 
t
o
 
f
i
n
d
 
t
h
e
 
e
q
u
a
t
i
o
n
 
o
f
 
O
T
 
i
n
C
a
r
t
e
s
i
a
n
 
f
o
r
m
Cartesian curves f(x,y,c)=0
1.Differentiate and eliminate the arbitrary constant c
and the resulting equation if f(x,y,dy/dx)=0
2.Replace dy/dx by – dx/dy
3. The diff equation of the form f(x,y,-dx/dy)=0
4.Solve the diff. equation to get the required O.T
 
Working rule to find the equation
of OT in Polar form
Polar curves f(r,
,c)=0
1.Diff and eliminate the arbitrary constant c and
the resulting equation if f(r,
,
dr/d
 
)=0
2.Replace dr/d
 
 
by – r2d
/dr
3. 
The diff equation of the form f(r,
,
r2d
/dr
)=0
4.Solve the diff. equation to get the required O.T
 
https://nptel.ac.in/courses/111/106/111106100
/
This NPTEL video link contains differential equations of first
order and higher order with applications.
 
THANK YOU
Slide Note
Embed
Share

Exploring the practical applications of differential equations, specifically focusing on Newton's Law of Cooling and Orthogonal Trajectories. The concept of exact differential equations and their solutions, along with real-life examples demonstrating temperature changes over time, are discussed. Understanding how differential equations are utilized in various scenarios within the realms of engineering and mathematics.

  • Applications
  • Differential Equations
  • Engineering
  • Mathematics
  • Newtons Law

Uploaded on Sep 17, 2024 | 0 Views


Download Presentation

Please find below an Image/Link to download the presentation.

The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author. Download presentation by click this link. If you encounter any issues during the download, it is possible that the publisher has removed the file from their server.

E N D

Presentation Transcript


  1. K S SCHOOL OF ENGINEERING AND MANAGEMENT DEPARTMENT OF MATHEMATICS TOPIC : Calculus And Linear Algebra Module-4 : Applications of Differential Equation NAME : DIVYA .R. DESIGNATION : Assistant Professor SUBJECT CODE : 18MAT11

  2. 1.Newtons Law of Cooling 2.LCR circuits 3.Problems on beams 4.Orthogonal Trajectories

  3. Differential Equation Exact Differential Equations: The differential equation M(x , y) dx + N(x , y)dy=0 to be an exact equation is M/ y = N/ x Further the solution of the exact equation is given by M dx + N ( y )dy= c We integrate M(x,y) w.r.t x keeping y fixed and N(y) indicate the terms in N does not contain x

  4. Application : Newton's Law of Cooling Application : Newton's Law of Cooling It is a model that describes, mathematically, the change in temperature of an object in a given environment. The law states that the rate of change (in time) of the temperature is proportional to the difference between the temperature T of the object and the temperature Te of the environment surrounding the object. d T / d t = - k (T - Te) Let x = T Te so that dx / dt = dT / dt Using the above change of variable, the above differential equation becomes d x / d t = - k x The solution to the above differential equation is given by x = A e- k t

  5. substitute x by T - Te T - Te = A e- k t Assume that at t = 0 the temperature T = To To - Te = A e0 which gives A = To - Te The final expression for T(t) i given by T(t) = Te + (To - Te)e- k t This last expression shows how the temperature T of the object changes with time.

  6. Real life examples A coffee of 98 degree centigrade will take 30 minutes to come down to 45 degree centigrade in a room of temperature 30 degree centigrade A dosa tava will take at teast 20 minutes to come down from 240 degree centigrade to 80 degree centigrade so that we can put it in wash basin.

  7. Orthogonal Trajectories(OT) Trajectory: A curve which cuts every member of a given family of curves according to some definite law . Orthogonal Trajectory: A curve which cuts every member of a given family of curves at right angles .

  8. Examples Orthogonal trajecteries for concentric circles is family of stright lines passing through origin Orthogonal trajecteries for cardioids R=a(1+cos t) is r=a(1+sint)

  9. Working rule to find the equation of OT in Working rule to find the equation of OT in Cartesian form Cartesian form Cartesian curves f(x,y,c)=0 1.Differentiate and eliminate the arbitrary constant c and the resulting equation if f(x,y,dy/dx)=0 2.Replace dy/dx by dx/dy 3. The diff equation of the form f(x,y,-dx/dy)=0 4.Solve the diff. equation to get the required O.T

  10. Working rule to find the equation of OT in Polar form Polar curves f(r, ,c)=0 1.Diff and eliminate the arbitrary constant c and the resulting equation if f(r, ,dr/d )=0 2.Replace dr/d by r2d /dr 3. The diff equation of the form f(r, , r2d /dr)=0 4.Solve the diff. equation to get the required O.T

  11. https://nptel.ac.in/courses/111/106/111106100/ This NPTEL video link contains differential equations of first order and higher order with applications. THANK YOU

More Related Content

giItT1WQy@!-/#giItT1WQy@!-/#giItT1WQy@!-/#giItT1WQy@!-/#giItT1WQy@!-/#giItT1WQy@!-/#giItT1WQy@!-/#giItT1WQy@!-/#giItT1WQy@!-/#