Wednesday 17 Nov:

 

Target: Understanding

Der's & Integration

 

L# 22: 1- 5, 7- 9, 12, 14 - 20   Due Today Wed 17 Nov

(Calculus: Derivative & Integration )

L# 23: 1-4, 8, 13, 15, 16, 18, 20   Due Fri 19 Nov

(Center of Mass)

 

Review Monday 22 Nov

Test  L# 19 - 23  Tues 23 Nov

 

 

 

 

 

 

 

 

 

 

Simple Machine Project:

 

Requirements:

1. Individual

 

2. Build and its due Wed 24 Nov

(must be made and not just a store example)

 

2. Any size & any of the 6 types you want.

(as long as you can carry it by your self through the doorway)

 

3. One page paper explaining your simple machine and its use in History

 

 

 

 

 

 

 

. Bridge Contest:

  • Due by Friday Dec 17

 

 

 

 

Tower Contest:

  • Due by Friday Jan 7th

 

 

 

L#22

 

 

Review Derivative:

The old exponent is brought down as a coefficient and the new exponent is the old one reduced by 1. Note: constants are = 0

 

 

 

 

 

 

The x position in meters of a particle (along the x axis) is given by the equation (function of time):

Find the 1st (velocity)

& 2nd Derivative (acceleration) :

 

 

Ex: 5t4  + 4t2  + 7      where t = 3

 

 

 

 

 

 

 

 

 

1st Derivative-->

    X(t) =  5t4  + 4t2  + 7 

          dx/dt  =   4(5t3)  + 2(4t1)  + 0

     dx/dt   =     20t3  + 8t

 

               =  20(27) + 8(3)

               =    540   +   24 = 564 m/s

 

 

 

2nd Derivative (Acceleration) of

 

 

 

1st Derivative  -->  20t3  + 8t  

 

 

 

2nd: V(t)       =    20t3   +   8t1  

       dV/dt  =  3(20t2)  +  1(8t0)

 

      dV/dt  =   60(32)  + 8 = 548 m/s2

 

 

 

 

 

 

Review Integral:  (t = 3)

Increase the exponent by one and divide by new exponent. Note: constants are t0

The velocity of a particle as a function of time is:

                  V(t) =  2t3  - 3t2    -2t   + 40

 

Find the distance traveled in 4 seconds:

 

 

  4
   = (2t3  - 3t2    -2t   + 40)dt
  0         
      =   [(2t4) / 4] – [(3t3)/ 3] - [2t2 /2] + 40t     
      =   [2(44)/ 4] – [3(43)/ 3] - [2(42)/2] + 40(4)

               =   [128] –  [64]   -  [16]   +   160  

               =   208m

   

 

 

 

 

The velocity of a particle as a function of time is:

                 V(t) =  3t2    -2t   + 40)

 

Find the distance traveled

     in 2 to 4 seconds:

 

 4
  = (3t2    -2t   + 40)dt
  2        

 

 

=  {[(3t3)/ 3] - [2t2 /2] + 40t}  -  {[(3t3)/ 3] - [2t2 /2] + 40t}

 

=   {(43)  -  (42) + 40(4)} - {(23)  -  (22)  +  40(2)}

 

 

      =   {64 - 16 + 160} -  {8 - 4 + 80}

 

      =  {208}  -  {84} =  124m