09-10-2014 (Calculus Lecture 5th semester BBA)

09/10/14

Example Page No 685:

 

S

S

d = f(t) = 8t2 + 8t: Solution:

At f(0) initially at rest è f(0) = 8(0)² + 8(0) è                                                                                     f(0) = 0

At f(1) After 1 hour è f(1) = 8(1)² + 8(1) è                                                                                          f(1) = 16

ARC = ∆f(t)/∆t è (16-0)/(1-0) è 16/1 è                                                                    ARC = 16

At f(5) after 5 hour è f(5) = 8(5)² + 8(5) è                                                                                                 f(0) = 240

ARC = ∆f(t)/∆t è (240-0)/(5-0) è 240/5 è                                                                               ARC = 48

Diffcrence Quotient

 

 

 

 

∆y/∆x = {f( x + ∆x) – f(x)}/x +∆x –x è                    {f( x + ∆x) – f(x)}/∆x  Diffcrence Quotient               

 

Example 24:

Solution

1.       Y = x² è f(x) = {f(x + ∆x) – f(x)}/∆x  è f(x + ∆x) = (x + ∆x)² è so then (x + ∆x)²– x²/ ∆x è x² + (∆x)² + 2x ∆x – x²/ ∆x è ∆x (∆x+ 2x)/ ∆x è                                                                                                                                                                                                                           2x + ∆x

2.       Points (-2, 4) (3, 9) find the slop by using 2 point formula:

m = ∆y/∆x è = (9-4)/(3-(-2)) è = 5/5 = 1                                                                                                                               m = 1

3.       Find the slop: by using (1 & 2)’s points given we can say that x = -2 – (i) and  x + ∆x = 3 –(ii)

Putting x value from (i) into (ii) as : (-2) + ∆x = 3 ∆x = 3 + 2  so ∆x  = 5

by putting values of x and ∆x into 2x + ∆x 2(-2) + 5 = 1

Ans = 1