Results 1 to 7 of 7

Math Help - rate of change and extrema

  1. #1
    Junior Member
    Joined
    Dec 2010
    Posts
    47

    Thumbs up rate of change and extrema

    Hi, I have a strangely worded problem that I'm having problems interpreting/solving

    (calculator active) The rate of change R, in km/hr, of the altitude of a hot air balloon is given by R(t) = t^3 - 4t^2 + 6 for time 0<(or equal to) t <(or equal to) 4, where t is measured in hours. Assume the balloon is initially at ground level.

    a. what is the max altitude of the balloon during the interval given above?

    Well, since the equation given above is the rate of change I plugged it into my calculator to find the critical points, I got t=1.572 and t=3.514. I then took the derivative of that equation and plugged in my critical points to determine the max. I got a max from t=1.572

    I know there is a max altitude at time = 1.572 but how do I figure out what the altitude is at this time? I know I have to come up with a position equation but I'm a little lost on that.

    All help is appreciated, thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,326
    Thanks
    1298

    Re: rate of change and extrema

    The rate of change of a function is its derivative. Saying that "The rate of change R, in km/hr, of the altitude of a hot air balloon is given by R(t)= t^3= 4t^2+ 6" means that, taking h(t) to be the altitude (in km) at time t (in hours), \frac{dh}{dt}= t^3- 4t^2+ 6. Take the "anti-derivative" to find h(t). Of course, that will involve a arbitrary "constant of integration". Use "Assume the balloon is initially at ground level" to find that constant.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2010
    Posts
    47

    Re: rate of change and extrema

    Quote Originally Posted by HallsofIvy View Post
    The rate of change of a function is its derivative. Saying that "The rate of change R, in km/hr, of the altitude of a hot air balloon is given by R(t)= t^3= 4t^2+ 6" means that, taking h(t) to be the altitude (in km) at time t (in hours), \frac{dh}{dt}= t^3- 4t^2+ 6. Take the "anti-derivative" to find h(t). Of course, that will involve a arbitrary "constant of integration". Use "Assume the balloon is initially at ground level" to find that constant.
    Oh ok, I know about anti differntiation, just didn't know what to use as a constant. Using your help, I came up with an altitude equation:

    h(t) = (1/4)t^4 - (4/3)t^3 + 6t + 0

    Now to find the max altitude I plugged in the critical point t = 1.572 which is a "max" into my altitude equation and I get:

    5.779 km, does this seem correct?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Dec 2010
    Posts
    47

    Re: rate of change and extrema

    also, there is another question for this equation:

    b. at what time is the altitude of the balloon increasing most rapidly?

    I used the derivative of the rate of change to find this, for wherever the value of the derivative of the rate of change is greatest should be where the altitude is increasing most rapidly? Am i correct on this?

    If so, then the answer should be t = 4
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,506
    Thanks
    765

    Re: rate of change and extrema

    Quote Originally Posted by TacticalPro View Post
    5.779 km, does this seem correct?
    Yes.

    Quote Originally Posted by TacticalPro View Post
    b. at what time is the altitude of the balloon increasing most rapidly?

    I used the derivative of the rate of change to find this, for wherever the value of the derivative of the rate of change is greatest should be where the altitude is increasing most rapidly?
    No, when the range of change (the velocity) is the greatest, the altitude is increasing most rapidly.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Dec 2010
    Posts
    47

    Re: rate of change and extrema

    Quote Originally Posted by emakarov View Post
    Yes.

    No, when the range of change (the velocity) is the greatest, the altitude is increasing most rapidly.
    Oh, well then wouldn't it still be t = 4? I plugged the "velocity" equation into my calc and at t=4 it is increasing the most
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4

    Re: rate of change and extrema

    Quote Originally Posted by TacticalPro View Post
    Oh, well then wouldn't it still be t = 4? I plugged the "velocity" equation into my calc and at t=4 it is increasing the most
    When the rate of change of altitude is at a maximum either dR/dt=0 or t is an end point of the interval of interest (in this case 0 or 4 hours). To answer this you need to find the roots of dR/dt=0 and evaluate R there and at t=0 and t=4, the maximum you seek is the largest of these values of R, and the time/s is the corresponding time/s.

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: April 12th 2011, 09:51 AM
  2. Rate change
    Posted in the Business Math Forum
    Replies: 2
    Last Post: July 4th 2010, 04:14 AM
  3. Related Rate and EXtrema
    Posted in the Calculus Forum
    Replies: 5
    Last Post: March 8th 2009, 12:12 PM
  4. Rate of change?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 29th 2008, 05:31 PM
  5. rate of change
    Posted in the Calculus Forum
    Replies: 9
    Last Post: October 23rd 2007, 03:32 AM

Search Tags


/mathhelpforum @mathhelpforum