Page 1 of 2 12 LastLast
Results 1 to 15 of 16

Math Help - Application of calculus

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    65

    Application of calculus

    Could someone please tell me how to calculate the following...?

    The angular displacement Θ
    radians of a flywheel varies with time "t"seconds and follows the equation Θ  = 9t^2 - 2t^2

    Determine:

    a)The angular velocity and acceleration of the flywheel when time "t" = 1 Second

    b) The time when the angular acceleration is zero.

    Many thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,832
    Thanks
    1602
    You have the displacement. Surely you know that the velocity is the time derivative of displacement, and that acceleration is the time derivative of velocity...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,454
    Thanks
    1868
    The angular velocity is \frac{d\theta}{dt}= \frac{d(9t^2- 2t)}{dt} and the angular acceleration is the second derivative.

    Note: I have assumed that you meant \theta(t)= 9t^2- 2t rather than \theta(t)= 9t^2- 2t^2 as you wrote since that would be written as 7t^2.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2010
    Posts
    65
    It should actually read...

    The angular displacement Θ radians of a flywheel varies with time "t"seconds and follows the equation Θ  = 9t^2 - 2t^3

    Sorry for the Typo
    Last edited by HNCMATHS; January 29th 2011 at 10:19 AM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,969
    Thanks
    1788
    Awards
    1
    The velocity is \Theta '(t) and the acceleration is \Theta ''(t).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jan 2010
    Posts
    65
    Sorry I dont follow.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,454
    Thanks
    1868
    Do you know how to differentiate t^n with respect to t?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    The rate of change of the angular displacement is the angular velocity.

    The rate of change of the angular velocity is the angular acceleration
    (rate of change of velocity).

    Differentiate your displacement expression to get the velocity expression.
    Whatever new expression you get for velocity, differentiate that to get the acceleration expression.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jan 2010
    Posts
    65
    Right so I have made a start... So far I have worked out that the velocity = 18t -6t2

    Now I have to differentiate this
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Yes, then you will have 2 expressions for velocity and acceleration.

    You will then be able to set t=1 in both of these to answer part (a).

    Then onto part (b).
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Jan 2010
    Posts
    65
    Quote Originally Posted by Archie Meade View Post
    Yes, then you will have 2 expressions for velocity and acceleration.

    You will then be able to set t=1 in both of these to answer part (a).

    Then onto part (b).
    \frac{d}{dt} = 18t = 18 (Does the t stay or go?)


    \frac{d}{dt} = -6t^2 = -12t

    Thus the Velocity = 18 - 12t

    Is this ok so far?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by HNCMATHS View Post
    \frac{d}{dt}(18t )= 18 (Does the t stay or go?)


    \frac{d}{dt}\left(-2t^3\right) = -6t^2 = -12t

    Thus the Velocity = 18 - 12t

    Is this ok so far?
    You have a typo.
    Your second derivative is the acceleration.

    \frac{d}{dt}(18t)=18\frac{d}{dt}t^1=(18)(1)t^0

    t^0=1

    so you get

    acceleration=18-12t

    Now write your 2 expressions for velocity and acceleration and try parts (a) and (b) of the question.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Junior Member
    Joined
    Jan 2010
    Posts
    65
    Quote Originally Posted by Archie Meade View Post
    You have a typo.
    Your second derivative is the acceleration.

    \frac{d}{dt}(18t)=18\frac{d}{dt}t^1=(18)(1)t^0

    t^0=1

    so you get

    acceleration=18-12t

    Now write your 2 expressions for velocity and acceleration and try parts (a) and (b) of the question.
    Velocity when t = 1 second

    (18)(1)-((6)(1))^2 Thus Velocity = -18

    Acceleration when t = 1 second

    (18)-((12)(1)) = 6
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Try the velocity again.

    You have squared all of (6t)

    instead of just the "t" part.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Junior Member
    Joined
    Jan 2010
    Posts
    65
    Quote Originally Posted by Archie Meade View Post
    Try the velocity again.

    You have squared all of (6t)

    instead of just the "t" part.
    (18)(1)-((6)(1^2)) = 12
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Application of calculus
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 23rd 2011, 02:11 AM
  2. Application of integral calculus-3
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 18th 2009, 09:13 AM
  3. Application of calculus
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 12th 2009, 04:56 AM
  4. Application of calculus?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 29th 2008, 08:10 AM
  5. calculus application
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 30th 2006, 01:06 PM

Search Tags


/mathhelpforum @mathhelpforum