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Calculus on vectors

  1. Oct 1, 2011 #1
    1. The problem statement, all variables and given/known data
    1)Find parametric equations of the line tangent to the graph of r(t) at the point where t=t0
    r(t)=ln(t) i + e^-t j +t^3 k; t0=2

    2) solve the vector initial-value problem for y(t) by integrating and using the initial conditions to find the constants of integration.
    y''(t) = 12T^2 i - 2t j, y(0)=2i-4j y'(0)=0


    3. The attempt at a solution
    I just wanted to make sure i am doing these right
    1) x=ln(2)+1 y=0 z=20
    2) (T^4 +2)i +(-(T^3/3)-4)j
     
  2. jcsd
  3. Oct 1, 2011 #2

    LCKurtz

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    The first answer describes a point, not the equation of a line.
    The second can be checked by plugging 0 into the answer and its derivatives and checking it.
     
  4. Oct 1, 2011 #3
    1) Do you know the equation for tangent here?
     
  5. Oct 1, 2011 #4
    1) yea its just the derivative,
    @Lckurtz do you mean this? x=ln(T) + 1 y=e^-T - e^-T z=T^3 +2*T^2

    2) not sure what you mean, if i plug in zero into the answer then i just get 2i -4j?
     
  6. Oct 1, 2011 #5
    No it's a line, you have a lot of nonlinear terms
     
  7. Oct 1, 2011 #6

    SammyS

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    LCKurtz is not online right now so I'll answer.

    No. As the flyingpig said, does NOT describe a line. However, it is parametric.



    Yes, you do get 2i -4j for y(0).

    What do you get for y'(0) >
     
  8. Oct 1, 2011 #7
    huh thats all i thought the question asked for, for the parametric equations? So what is the question looking for?

    you get 0 for y'.
     
  9. Oct 1, 2011 #8
    What you have don't even look like a line

    Look what you have

    That's y = 0!!!
     
  10. Oct 1, 2011 #9
    oh, i think i was simplying it too much, so do you mean this?
    r=(ln(T) +1) i +(e^-T-e^-T) j +(T^3 +3T^2)k
     
  11. Oct 1, 2011 #10
    No, don't even use r(t) again.

    T^3 + 3T^2 what is this? This is a curve, you know, that thing which isn't straight?
     
  12. Oct 1, 2011 #11

    SammyS

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    What does r'(t) give you when t = 2 ?
     
    Last edited: Oct 1, 2011
  13. Oct 1, 2011 #12
    oh i think i get what you guys are saying, i plugged into r(2) and solved for that and then r'(2) and got
    r(t) = <ln(2),e^-2,8> +T<.5,e^-2,12>, which is an equation of a line and is parametric.
     
  14. Oct 1, 2011 #13

    SammyS

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    Yes, that's more like it !
     
  15. Oct 2, 2011 #14
    cool finally understand what I was doing wrong, and as for number 2 you said to plug in 0 into the equation and its derivatives which i did, but im not sure how that tells me if i did it right
    2i -4j for y(0).
    0 for y'.
    and 0 for y''
    ?
     
  16. Oct 2, 2011 #15

    SammyS

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    One of the conditions to be satisfied for y is that y'(0) = 0. So from that point of view, you solution checks out.
     
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