1. Applications OF DIFFERENTIAL CALCULUS
BY ERUM RIAZ
Department of Mathematics
D.A Degree College For Women Ph-VII (Ext)

2. Applications of differential calculus:

3. Analysis :
By the end of the lesson students will be able to explain the applications of derivative.

4. Applications of differential calculus
There are two major applications
Rate of Change
Extreme Values

5. 1. Rate of Change :
The derivative of function is a mathematical tool that is used to study rates at which quantity changes.
Rate of change of position is velocity.
Rate of change of velocity is acceleration.
Rate of change of momentum is force .
We have many quantities involve rate of change in time
shook intensity of earthquakes , inflation of currency , water height of Sea , Stock rates , rate of increase of volume of balloon etc.

6. Example: 1 Gas is escaping from a spherical balloon at the rate of 2 ft3 / min .How fast is the surface area shrinking when the radius is 12 ft ?
Solution:
A sphere of radius r has volume v
V = 4/3 π r3
As we know surface area of sphere S
S = 4 π r 2
We have to find dS /dt when r = 12 ft here we can see S is independent of time but we can find it in this way
dS dS dr
= × (1)
dt dr dt
here S = 4 π r 2
dS/ dr = 8 π r
Now for dr /dt :
dV / dt = (given)
dV dV dr
= × (2)
dt dr dt
V = 4/3 π r3
dV/ dr = π r2

7. Substitute the values in (2)
= π r 2 × dr /dt
- 1 /2 π r2 = dr / dt
Substitute the values of dS/dr & dr/dt in (1)
dS / dt = π r × -1/ 2 π r 2
dS / dt = / r
when r = 12 ft
dS/dt = / 3 ft2/ min
The surface of Spherical balloon shrinking at the rate of /3 square feet per minute.

8. Example : 2 A man 6 ft tall walks away from a lamp post 16 ft high at the rate of 5 miles per hour. How fast does the end of his shadow move?
Solution :
By similar triangle
s - x s =
16 s – 16 x = 6 s
10 s = x
Differentiate with respect to t
ds dx =
dt dt ds
= ( 5 ) dt
ds/dt = miles/ hour
His shadow move at 8 miles/hour
16ft
6ft x
s - x s

9. Example :3 One end of a 13 foot – Ladder is on the floor , and the other end rests on a vertical wall. If the bottom end is drawn away from the wall at 3 ft/sec how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 5 ft from the wall?
Solution :
At any given instant , let y be the height of the top of the ladder above the floor and x be the distance b/w the base of the wall and bottom of the ladder
By hypothesis
dx/dt = 3
We must find the value of dy/dt at t where t is the time for which x = 5 ft
By pythgoras theorem
13 feet y
x y 2 =
x y = 169
Differentiating with respect to t
2x dx / dt y dy/ dt = 0
dy / dt = - x / y dx /dt (3) x
Now at time t for which base of the ladder is 5 ft from wall
Substitute x = 5 in (2)
y 2 = 169
y 2 =
y = ± 12

10. substitute values of x , y and dx /dt in (3)
dy / dt = / 12 (3)
dy / dt = /4 ft /sec
Thus when the ladder is 5 ft from the wall the top is sliding down at the rate of 5/4 ft /sec.
Example :4
Water is poured into a conical paper cup at the rate of 2/3 inches3 /sec . if the cup is 6 inches tall and top of the cup has radius of 2 inches , how fast does the water level rise when the water is 4 inches deep ?
Solution :
At any time t let h be the height of water , V be volume of water and r is the radius of top surface of water
As we know volume of cone is V = 1/3 π r2 h
Since water enters the cup at the rate of 2/3 inches3/ sec
dV/dt = 2/ (1)
2 inches
We have to find dh/dt at time t for which h = 4
Using similar triangle in figure
r / h = 2 /6
r / h = 1 / 3
r = h / 3 r 2 r
6 inches h 6
4 inches h

11. V = 1 / 3 π r2 h
V = 1 / 3 π ( h/3)2 h
V = 1/27 π h 3
dv/dh = 3 π h2 /27
dv/dh = π h2 / (2)
Now dv /dt = dv /dh × dh /dt
Substitute the values from ( 1 ) and (2)
2/3 = πh2 / 9 × dh/dt
dh /dt = / 3 π h2
dh /dt = 6 / π h 2
when the water is 4 inches deep
dh / dt = 6 / π (4)2
dh / dt = 3/8π inches / sec
Thus the water level is rising at the rate of 3/8π inches/sec when the water is 4 inches deep

12. 2 .Extreme Values Relative Maximum : Relative Minimum :
A function f is said to have an extreme value at a if it has either a relative maximum or a relative minimum at a in such a situation a is called an extreme point of f .
Relative Maximum :
Let f be a function , defined on some neighbourhood of a point a ,then f is said to have
relative maximum value at a if there exist δ – neighbourhood A of a such that
f ( a ) > f ( x ) for all a € A
and as such f ( a ) is called a relative maximum of f .
Relative Minimum :
f is said to have a relative minimum value at a if there exist δ – neighbourhood A of a such that
f ( a ) < f ( x ) for all a € A
and such f ( a ) is called relative minimum of f.

13. Example :
The domain of f(x) = x2 is all real numbers and the range is all nonnegative real numbers. The graph in the figure below suggests that the function has no absolute maximum value and has an absolute minimum of 0, which occurs at x = 0.
The Absolute Extreme Values on a Restricted Domain
If the domain of f(x) = x2 is restricted to [-2, 3], the corresponding range is [0, 9]. As shown below, the graph on the interval [-2, 3] suggests that f has an absolute maximum of 9 at x = 3 and an absolute minimum of 0 at x = 0.
The two examples above show that the existence of absolute maxima and minima depends on the domain of the function.
[-5, 5, 1] x [-2, 10, 1]

14. Example : 6
Find absolute maximum or minimum of the function f ( x ) = (2 x ) / x in [ 1 , 5 ]
Solution :
To find absolute maximum and minimum , First we must find critical numbers by taking derivative of the function
x d/ dx (2x2 + 4 ) (2x2 +4 ) d/dx (x)
f ‘ (x) = x2
x (4x) (2x2 + 4) (1)
f ‘ (x) = 2x
f ‘ (x) =
Now for critical numbers f ‘ ( x ) = 0 2x
= 0 x2
2x = 0
2x2 = 4
x = ± √ 2
Since - √ 2 does not lie in [1 , 5] , we can not include as a critical number , thus the only critical number is √2.

15. For determining function has maximum or minimum value at √ 2 , we have to find the 2nd derivative
f ‘ (x) = x2
x2 d/dx (2x2 -4 ) (2x ) d/dx ( x2 )
f ‘’ (x) = x4
x2 (4x) (2x ) (2x) =
- 8
f ‘’ (x) = < 0 x3
Since f ‘’ (x) < 0 ,function has maximum value at √ 2
2 ( √ 2 )
f ( √ 2 ) =
( √ 2 )
= √ 2
4 √2 is the absolute maximum of the given function in [1 , 5 ].

16. Example : 7
What should be the shape of a rectangular field of a given area, if it is to be enclosed by the least amount of fencing?
Solution :
Let x be the length , y be width and A be area of rectangle
A = x . y
Differentiate with respect to x
0 = x . dy/dx + y
dy / dx = - y / x
Now let P be the perimeter of rectangle
P = 2x y
0 = dy/dx
- 2 = 2 dy/dx
- 2 = 2 ( - x / y )
1 = x / y
x = y
Rectangular field should be square for least amount of fencing. y x

17. Example : 8 Solution : x 9 9 -2x x x x 9 – 2x x 9 -2x 9 -2x 9
Example : 8
A box is to be made of a piece of cardboard 9 inches square by cutting equal squares out of the corners and turning up the sides. Find the volume of the largest box that can be made in this x 9
9 -2x x x x
9 – 2x x
9 -2x
9 -2x
Solution : 9
Let V be the volume of the box
V = (9 -2x )2 .x
V = 81x – 36x2 + 4x3
Differentiate with respect to x
dV / dx = x x 2
For finding maximum volume dV/dx = – 72x + 12 x 2 = 0
4x 2 – 24 x = 0

18. By solving above equation
x = or x = 1.5
use x = 1.5 ( for maximum volume )
V= ( 9 – 2 ( 1.5 ) )2 (1.5)
V = 54 cubic inches

19. Example : 9
A rectangular field of given area is to be fenced off along the bank of a river. If no fence is needed along the river, what is the shape of the rectangle requiring the least amount of fencing?
Solution :
Let x be the length and y be the width of the field
A be the area
A = x.y
Differentiate with respect to x
dA/dx = x. dy/dx + y
0 = x . dy/dx + y
dy / dx = - y / x (1)
River
Field
Let P be the perimeter of the field excluding the
River side
P = x y
Differentiate with respect to x
dP/dx = dy/dx
For least amount of fencing dP/dx =0
0 = dy/dx
0 = ( - y / x ) from (1)
y = 1 / 2 x
Width = ½ length for least amount of fencing

20. Example : 10 Solution : c y x
If the hypotenuse of the right triangle is given, show that the area is maximum when the triangle is isosceles.
Solution :
Let x , y be the base and perpendicular and c be the hypotenuse
of the triangle , by pythagoras theorem
x y 2 = c 2
Differentiate with respect to x
2x y .dy /dx = 0
dy/dx = - x /y (1) c y x
let A be the area of triangle A = ½ x . y
Differentiate with respect to x
dA / dx = ½ [ x . dy/dx + y ]
= ½ [ x . dy/dx + y ]
x . dy/dx + y = 0
x . ( - x / y ) + y = 0
y = x 2 / y
x 2 = x 2
y = x
This shows that area is maximum when the triangle is an isosceles.

21. Conclusion :
Above Examples shows that differential calculus is a mathematical tool that solve a lot of problems of daily life , Engineering and other sciences.

22. References : (1 ) Calculus by Robert Ellis & Denny Gulick
Applied Calculus
by Tan
Calculus
by Frank Ayres , JR. Elliott Mendelson
Mathematics XII
Sindh text book board Jamshoro
(5) www. Mathalino.com