<< Chapter < Page Chapter >> Page >

Rate of change problems

Two concepts were discussed in this chapter: Average rate of change = f ( b ) - f ( a ) b - a and Instantaneous rate of change = lim h 0 f ( x + h ) - f ( x ) h . When we mention rate of change , the latter is implied. Instantaneous rate of change is the derivative . When Average rate of change is required, it will be specifically refer to as average rate of change.

Velocity is one of the most common forms of rate of change. Again, average velocity = average rate of change and instantaneous velocity = instantaneous rate of change = derivative . Velocity refers to the increase of distance(s) for a corresponding increade in time (t). The notation commonly used for this is: v ( t ) = d s d t = s ' ( t )

where s ' ( t ) is the position function. Acceleration is the change in velocity for a corersponding increase in time. Therefore, acceleration is the derivative of velocity a ( t ) = v ' ( t ) This implies that acceleration is the second derivative of the distance(s).

The height (in metres) of a golf ball that is hit into the air after t seconds, is given by h ( t ) = 20 t - 5 t 2 . Determine

  1. the average velocity of the ball during the first two seconds
  2. the velocity of the ball after 1,5 seconds
  3. when the velocity is zero
  4. the velocity at which the ball hits the ground
  5. the acceleration of the ball
  1. v a v e = h ( 2 ) - h ( 0 ) 2 - 0 = [ 20 ( 2 ) - 5 ( 2 ) 2 ] - [ 20 ( 0 ) - 5 ( 0 ) 2 ] 2 = 40 - 20 2 = 10 m s - 1
  2. v ( t ) = d h d t = 20 - 10 t

    Velocity after 1,5 seconds:

    v ( 1 , 5 ) = 20 - 10 ( 1 , 5 ) = 5 m s - 1
  3. v ( t ) = 0 20 - 10 t = 0 10 t = 20 t = 2

    Therefore the velocity is zero after 2 seconds

  4. The ball hits the ground when h ( t ) = 0

    20 t - 5 t 2 = 0 5 t ( 4 - t ) = 0 t = 0 o r t = 4

    The ball hits the ground after 4 seconds. The velocity after 4 seconds will be:

    v ( 4 ) = h ' ( 4 ) = 20 - 10 ( 4 ) = - 20 m s - 1

    The ball hits the gound at a speed of 20 m s - 1 . Notice that the sign of the velocity is negative which means that the ball is moving downward (the reverse of upward, which is when the velocity is positive).

  5. a = v ' ( t ) = - 10 m s - 1

    Just because gravity is constant does not mean we should think of acceleration as a constant. We should still consider it a function.

End of chapter exercises

  1. Determine f ' ( x ) from first principles if:
    1. f ( x ) = x 2 - 6 x
    2. f ( x ) = 2 x - x 2
  2. Given: f ( x ) = - x 2 + 3 x , find f ' ( x ) using first principles .
  3. Determine d x d y if:
    1. y = ( 2 x ) 2 - 1 3 x
    2. y = 2 x - 5 x
  4. Given: f ( x ) = x 3 - 3 x 2 + 4
    1. Calculate f ( - 1 ) , and hence solve the equation f ( x ) = 0
    2. Determine f ' ( x )
    3. Sketch the graph of f neatly and clearly, showing the co-ordinates of the turning points as well as the intercepts on both axes.
    4. Determine the co-ordinates of the points on the graph of f where the gradient is 9.
  5. Given: f ( x ) = 2 x 3 - 5 x 2 - 4 x + 3 . The x -intercepts of f are: (-1;0) ( 1 2 ;0) and (3;0).
    1. Determine the co-ordinates of the turning points of f .
    2. Draw a neat sketch graph of f . Clearly indicate the co-ordinates of the intercepts with the axes, as well as the co-ordinates of the turning points.
    3. For which values of k will the equation f ( x ) = k , have exactly two real roots?
    4. Determine the equation of the tangent to the graph of f ( x ) = 2 x 3 - 5 x 2 - 4 x + 3 at the point where x = 1 .
  6. Answer the following questions:
    1. Sketch the graph of f ( x ) = x 3 - 9 x 2 + 24 x - 20 , showing all intercepts with the axes and turning points.
    2. Find the equation of the tangent to f ( x ) at x = 4 .
  7. Calculate: lim x 1 1 - x 3 1 - x
  8. Given: f ( x ) = 2 x 2 - x
    1. Use the definition of the derivative to calculate f ' ( x ) .
    2. Hence, calculate the co-ordinates of the point at which the gradient of the tangent to the graph of f is 7.
  9. If x y - 5 = x 3 , determine d x d y
  10. Given: g ( x ) = ( x - 2 + x 2 ) 2 . Calculate g ' ( 2 ) .
  11. Given:        f ( x ) = 2 x - 3
    1. Find:        f - 1 ( x )
    2. Solve:        f - 1 ( x ) = 3 f ' ( x )
  12. Find f ' ( x ) for each of the following:
    1. f ( x ) = x 3 5 3 + 10
    2. f ( x ) = ( 2 x 2 - 5 ) ( 3 x + 2 ) x 2
  13. Determine the minimum value of the sum of a positive number and its reciprocal.
  14. If the displacement s (in metres) of a particle at time t (in seconds) is governed by the equation s = 1 2 t 3 - 2 t , find its acceleration after 2 seconds. (Acceleration is the rate of change of velocity, and velocity is the rate of change of displacement.)
  15. After doing some research, a transport company has determined that the rate at which petrol is consumed by one of its large carriers, travelling at an average speed of x km per hour, is given by: P ( x ) = 55 2 x + x 200 litres per kilometre
    1. Assume that the petrol costs R4,00 per litre and the driver earns R18,00 per hour (travelling time). Now deduce that the total cost, C , in Rands, for a 2 000 km trip is given by: C ( x ) = 256000 x + 40 x
    2. Hence determine the average speed to be maintained to effect a minimum cost for a 2 000 km trip.
  16. During an experiment the temperature T (in degrees Celsius), varies with time t (in hours), according to the formula: T ( t ) = 30 + 4 t - 1 2 t 2 t [ 1 ; 10 ]
    1. Determine an expression for the rate of change of temperature with time.
    2. During which time interval was the temperature dropping?
  17. The depth, d , of water in a kettle t minutes after it starts to boil, is given by d = 86 - 1 8 t - 1 4 t 3 , where d is measured in millimetres.
    1. How many millimetres of water are there in the kettle just before it starts to boil?
    2. As the water boils, the level in the kettle drops. Find the rate at which the water level is decreasing when t = 2 minutes.
    3. How many minutes after the kettle starts boiling will the water level be dropping at a rate of 12 1 8 mm/minute?

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Siyavula textbooks: grade 12 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11242/1.2
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Siyavula textbooks: grade 12 maths' conversation and receive update notifications?

Ask