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

Math Help - Derivative as a rate of change

  1. #1
    Member VitaX's Avatar
    Joined
    Feb 2009
    From
    Ohio
    Posts
    185

    Derivative as a rate of change

    s = -t^3 +3t^2 -3t with the interval 0\leq t \leq 3

    a. Find the body's displacement and average velocity for the given time interval.

    V=-3t^2 +6t - 3
    V(0) = - 3
    V(3) = -32
    Is this all thats needed on this part or do I need to do V(3)-V(0) to find the average velocity?

    b. Find the body's speed and acceleration at the endpoints of the interval.

    s = -t^3 +3t^2 - 3t
    s(0)= 0
    s(3)=9 is positive because speed is always absolute value
    A = -6t + 6
    A(0) = 6
    A(3) = -12

    c. When, if ever, during the interval does the body change direction.

    Use the Velocity equation for this I believe and when I plug in values between the interval, the output (v) is never changed into a positive number so can I conclude it does not change direction?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    May 2009
    Posts
    211
    for a) , average velocity means distance/time so find [s(3)-s(0)]/(3-0). And displacement will just be distance, i.e. s(3) - s(0)

    for b) they ask you for absolute value of v(t) at the endpoints. and also a(t)

    Don't forget that: s'(t) = v(t) ; s''(t) = a(t) = v'(t).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member VitaX's Avatar
    Joined
    Feb 2009
    From
    Ohio
    Posts
    185
    Quote Originally Posted by Arturo_026 View Post
    for a) , average velocity means distance/time so find [s(3)-s(0)]/(3-0). And displacement will just be distance, i.e. s(3) - s(0)

    for b) they ask you for absolute value of v(t) at the endpoints. and also a(t)

    Don't forget that: s'(t) = v(t) ; s''(t) = a(t) = v'(t).
    For a) wouldn't it be v as the variable and not s? But for b) im cofused at why you say they ask for the absolute value of v(t) because didn't I solve that in a) and doesn't it ask to just find the speed and acceleration at the end points?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    May 2009
    Posts
    211
    It seems that you are confusing what the s, v, and t stand for. Here:
    s(t), the function you were given is the position function, it denots where the point is with respect to time t.
    v(t) is s'(t) and that's the velocity of the point at that instant t. And when they ask you for speed, it simply means the absolute value of v(t).
    and a(t) denotes acceleration.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member VitaX's Avatar
    Joined
    Feb 2009
    From
    Ohio
    Posts
    185
    Quote Originally Posted by Arturo_026 View Post
    It seems that you are confusing what the s, v, and t stand for. Here:
    s(t), the function you were given is the position function, it denots where the point is with respect to time t.
    v(t) is s'(t) and that's the velocity of the point at that instant t. And when they ask you for speed, it simply means the absolute value of v(t).
    and a(t) denotes acceleration.
    Could you show me your answers for a, b, and c if its not too much trouble so I can fully understand what you are saying
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    May 2009
    Posts
    211
    Quote Originally Posted by VitaX View Post
    For a) wouldn't it be v as the variable and not s?
    v is the velocity and s is the position, so to find the average velocity, as I said is distance/time , think of it as driving from point a to b, and they are 20 miles apart, then it takes you 2 hours to get there, therefore your average speed was 10 miles per hour.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member VitaX's Avatar
    Joined
    Feb 2009
    From
    Ohio
    Posts
    185
    Quote Originally Posted by Arturo_026 View Post
    for a) , average velocity means distance/time so find [s(3)-s(0)]/(3-0). And displacement will just be distance, i.e. s(3) - s(0)

    for b) they ask you for absolute value of v(t) at the endpoints. and also a(t)

    Don't forget that: s'(t) = v(t) ; s''(t) = a(t) = v'(t).
    I understand what you are saying now about |v(t)| = speed i overlooked that part in the book and i was thinking that speed was just obtained by plugging in values into the original equation. But my final question is when you write to find the displacement and average velocity is that just plugging the endpoint values of the interval into the given equation or plugging them into v(t) (the derivative of the given equation)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,442
    Thanks
    1863
    The average velocity between t=0 and t= 3 is (s(3)- s(0))/(3- 0), the distanced moved between those two times, divided by the change in time.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member VitaX's Avatar
    Joined
    Feb 2009
    From
    Ohio
    Posts
    185
    Quote Originally Posted by HallsofIvy View Post
    The average velocity between t=0 and t= 3 is (s(3)- s(0))/(3- 0), the distanced moved between those two times, divided by the change in time.
    Ok then I guess I just use the given equation for part a. then
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Oct 2009
    Posts
    77
    If it helps make it semantically easier to understand, the derivative/rate of change of displacement is velocity, and the derivative/rate of change of velocity is acceleration. That makes sense if you think about it.

    (A) Average displacement uses the original equation/3.

    Average velocity uses its derivative, but you miscalculated V(3). It should be -12.

    (B) Looks correct

    (C) Your assumption is correct. The body will change direction if the sign of the velocity function changes within the interval.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member VitaX's Avatar
    Joined
    Feb 2009
    From
    Ohio
    Posts
    185
    Quote Originally Posted by Open that Hampster! View Post
    If it helps make it semantically easier to understand, the derivative/rate of change of displacement is velocity, and the derivative/rate of change of velocity is acceleration. That makes sense if you think about it.

    (A) Average displacement uses the original equation/3.

    Average velocity uses its derivative, but you miscalculated V(3). It should be -12.

    (B) Looks correct

    (C) Your assumption is correct. The body will change direction if the sign of the velocity function changes within the interval.
    Isn't displacment just s(3) - s(0) and then average velocity \frac{v(3)-v(0)}{3-0} If this is correct then all is cleared up
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    May 2009
    Posts
    211
    Quote Originally Posted by VitaX View Post
    Isn't displacment just s(3) - s(0) and then average velocity \frac{v(3)-v(0)}{3-0} If this is correct then all is cleared up
    I understand that you are having a hard time understanding it, but try reading the chapter in your book that talks about that.

    You are correct about displacement, but average velocity is displacement/time ; that is (s(0)-s(3))/(0-3) ; I understand if it doesn't make sense to you since it says average velocity and you have distance in your numerator, but trust me that's what it is.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Member VitaX's Avatar
    Joined
    Feb 2009
    From
    Ohio
    Posts
    185
    Quote Originally Posted by Arturo_026 View Post
    I understand that you are having a hard time understanding it, but try reading the chapter in your book that talks about that.

    You are correct about displacement, but average velocity is displacement/time ; that is (s(0)-s(3))/(0-3) ; I understand if it doesn't make sense to you since it says average velocity and you have distance in your numerator, but trust me that's what it is.
    Ok I trust you. So average velocity uses the given equation. but when it says find the velocity at the end points of the interval means find the derivative of the given equation and go from there. If thats correct than everything is cleared up. Just some terminology has me confused. mainly part a.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    May 2009
    Posts
    211
    Quote Originally Posted by VitaX View Post
    Ok I trust you. So average velocity uses the given equation. but when it says find the velocity at the end points of the interval means find the derivative of the given equation and go from there. If thats correct than everything is cleared up. Just some terminology has me confused. mainly part a.
    Yes.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Member
    Joined
    Oct 2009
    Posts
    77
    Apologies for my earlier post; it's been awhile since I've taken Physics.

    The unit for velocity is in \frac{m}{s}

    Which is distance over time. Average velocity is just the average of the distance over the average of time.

    Though in this example you're lucky, because V(3)-V(0) = S(3) - S(0)
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: April 12th 2011, 10:51 AM
  2. Derivative to find rate of change/population.
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 18th 2010, 09:49 AM
  3. Derivative & Average Rate of Change
    Posted in the Calculus Forum
    Replies: 6
    Last Post: December 29th 2009, 09:26 AM
  4. Replies: 3
    Last Post: October 18th 2009, 07:09 PM
  5. The Derivative as a Rate of Change
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 10th 2009, 10:13 PM

Search Tags


/mathhelpforum @mathhelpforum