Results 1 to 4 of 4

Math Help - Calc 3 question: initial value problem

  1. #1
    Newbie
    Joined
    Oct 2006
    Posts
    15

    Calc 3 question: initial value problem

    Hello,

    I'm not sure how to get started on this:

    Solve initial value problem for r as a function of t

    Differential equation: d^2r/dt^2 = -(i + j + k)
    Initial conditions: r(0) = 10i +10j +10k and (dr/dt) evaluated at t=0 =0

    I'd appreciate all the help I can get; truthfully, I'm pretty lost in this course and regretting taking it at all.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by riverjib View Post
    Hello,

    I'm not sure how to get started on this:

    Solve initial value problem for r as a function of t

    Differential equation: d^2r/dt^2 = -(i + j + k)
    Initial conditions: r(0) = 10i +10j +10k and (dr/dt) evaluated at t=0 =0

    I'd appreciate all the help I can get; truthfully, I'm pretty lost in this course and regretting taking it at all.
    what are the initial conditions for r'(0) ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    778
    Hello, riverjib!

    Solve initial value problem for r as a function of t.

    Differential equation: . \frac{d^2r}{dt^2} \:=\:-(i + j + k)
    Initial conditions: . r(0) \:=\:10i +10j +10k,\quad  \frac{dr}{dt}(0) \:=\:0

    We have: . \frac{d^2r}{dt^2} \;=\;-(i+j+k)

    Integrate: . \frac{dr}{dt} \;=\;-(i+j+k)t + C_1

    We are told that, when t = 0,\:\frac{dr}{dt} = 0

    . . So we have: . -(i+j+k)0 +C_1 \:=\:0\quad\Rightarrow\quad C_1 = 0

    Hence: . \frac{dr}{dt} \:=\:-(i+j+k)t


    Integrate: . r(t) \;=\;-\frac{1}{2}(i+j+k)t^2 + C_2

    We are told that, when t = 0,\:r = 10i + 10j +10k

    . . So we have: . -\frac{1}{2}(i+j+k)0^2 + C_2 \:=\:10i + 10j + 10k\quad\Rightarrow\quad C_2\:=\:10i + 10j + 10k


    Therefore: . \boxed{r(t) \;=\;-\frac{1}{2}(i+j+k)t^2 + 10(i+j+k)}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Soroban View Post
    Hello, riverjib!


    We have: . \frac{d^2r}{dt^2} \;=\;-(i+j+k)

    Integrate: . \frac{dr}{dt} \;=\;-(i+j+k)t + C_1

    We are told that, when t = 0,\:\frac{dr}{dt} = 0

    . . So we have: . -(i+j+k)0 +C_1 \:=\:0\quad\Rightarrow\quad C_1 = 0

    Hence: . \frac{dr}{dt} \:=\:-(i+j+k)t


    Integrate: . r(t) \;=\;-\frac{1}{2}(i+j+k)t^2 + C_2

    We are told that, when t = 0,\:r = 10i + 10j +10k

    . . So we have: . -\frac{1}{2}(i+j+k)0^2 + C_2 \:=\:10i + 10j + 10k\quad\Rightarrow\quad C_2\:=\:10i + 10j + 10k


    Therefore: . \boxed{r(t) \;=\;-\frac{1}{2}(i+j+k)t^2 + 10(i+j+k)}

    oh! i didn't notice the part where it said dr/dt = 0
    Last edited by Jhevon; December 8th 2007 at 05:09 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. vector calc 1 problem solving question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 30th 2009, 12:23 AM
  2. Initial Value Problem General Question
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: January 14th 2009, 09:05 AM
  3. Replies: 2
    Last Post: October 12th 2008, 05:33 AM
  4. Replies: 1
    Last Post: September 30th 2008, 03:10 PM
  5. Problem getting a Trig question check in My Calc
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: December 3rd 2006, 06:58 PM

Search Tags


/mathhelpforum @mathhelpforum