]> Exercise 4

Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)

Chapter 2: Integration; Section 1: Indefinite Integrals; page 1

Definitions and Basics, Exercise 4


According to Newton's mechanics for one dimension along the x axis, a point with mass m has an acceleration a: a= F m whereF=force a= dv dt where{ v= dx dt =velocity t=time

In this particular case the force is: F=Csin( ωt ) where C and ω are known constants.

  1. The following rule is necessary for the solution of the exercise: If f( x ) dx=I( x ) then f( bx )dx= I( bx ) b where b is a constant. Prove it!
  2. Obtain the velocity as a function of the time!
  3. Obtain the position x as a function of the time!
  4. In the expression of x as a function of time that you found, prove that all the terms have the correct physical dimensions!


{ f( x )= dI dx I( x )= fdx } (

The function I(x) from ( is called the indefinite integral of fdx , and its meaning is that its derivative is the function f(x).

x p dx= x p+1 p+1 +C (
{ cosxdx =sinx+C sinxdx =cosx+C ( 1+ tan 2 x ) dx= dx cos 2 x =tanx+C ( 1+ cot 2 x ) dx= dx sin 2 x =cotx+C } (

Parts 1-2

Solution of question 1

  1. According to ( one has to prove that If dI( x ) dx =f( x ) then dI( bx ) bdx =f( bx )
  2. From the first expression of part 1 it follows that
    dI( u ) du =f( u ) where u is any variable.
    But by choosing u=bx and taking in account that du=bdx we obtain dI( bx ) bdx =f( bx ) and the statement that If f( x ) dx=I( x ) then f( bx )dx= I( bx ) b is proven.

Parts 3-5

Solution of question 2

  1. from the data presented in this exercise we have:
    dv dt =a= F m = Csin( ωt ) m and the velocity v is v= C m sin( ωt ) dt
  2. From ( formally: sin( t ) dt=cost+C and therefore from part 2 we obtain: sin( ωt ) dt= cosωt ω +C
  3. Finally we obtain from parts 3 and 4 that the velocity is:
    v= Ccosωt mω + C 1 where C1 is a constant of the integration.

Parts 6-7

Solution of question 3

  1. Since dx dt =v the position x should be obtained from x= vdt= C mω cos( ωt ) dt+ C 1 dt
  2. By following the footsteps of parts 4 and 5 we obtain finally
    x= C m ω 2 sin( ωt )+ C 1 t+ C 2
    where C2 is a constant of integration.

Parts 8-10

Solution of question 4

  1. For the solution of question 4 we'll use the notation of
    [ z ]=physical dimensions of    z In this case [ x ]=Λ( length ) [ t ]=Θ( time ) From the data supplied to this exercise: [ a ]=[ F m ]=[ C m ]=Λ Θ 2
  2. From the argument of sine and cosine: [ ω ]= Θ 1 From this together with part 8 one obtains: [ x ]=[ C m ω 2 sin( ωt ) ]=Λ which takes care of the correct dimensions of the first right term in part 7.
  3. From part 5 [ C 1 ]=Λ Θ 1 and from part 7 [ C 2 ]=Λ which take care of the remaining terms.


By questions:

Questions 1,3 give credit of 2 points each.
Questions 2,4 give credit of 3 points each.

By parts:

All parts are worth 1 point each.