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Pytels Dynamics 12.25: Plane tracking

  1. Jun 18, 2017 #1
    1. The problem statement, all variables and given/known data
    the plane C is being tracked by radar stations A and B. At the instant shown,
    the triangle ABC lies in the vertical plane, and the radar readings are θA=30o,
    θB = 22o, θA(dot) = 0.026rad/s and θΒ(dot) = 0.032rad/s. Determine (a) the altitude y; (b) the speed v; and (c) the climb angle α of the plane at this instant

    2. Relevant equations


    3. The attempt at a solution
    I guess I can find (a) from trogonometry. Any hints for part (b)?
     

    Attached Files:

  2. jcsd
  3. Jun 18, 2017 #2
    I'd draw a good picture and think long and hard about related rates.
     
  4. Jun 18, 2017 #3

    haruspex

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    Solve part a) keeping everything symbolic - do not plug any of the given numbers in.
    That will give you an equation for y in terms of the two angles and the distance between the stations.
    Get another equation for the x coordinate.
    With those two equations, how might you get an equation for velocities?
     
  5. Jun 18, 2017 #4

    SammyS

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    Posting the picture, rather than the thumbnail makes the problem clearer and may encourage more help,

    12_25-pytel-jpg.jpg
     
  6. Jun 19, 2017 #5
    x is the distance from the first station to the second station plus the distance to the point that the vertical line meet the plane? (my english are not that good)
     
  7. Jun 19, 2017 #6

    jbriggs444

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    You can choose any coordinate system you like. Placing the origin at station A is a good choice.

    Yes, if the origin is at A then the x coordinate of the plane is the distance from A to B plus the distance from B to a vertical line from plane to ground.
     
  8. Jun 19, 2017 #7
    so I got
    ##y=\frac {1000} {tanθ_A - tanθ_B}##

    and

    ##x=1000\frac {tanθ_A} {tanθ_A-tanθ_B}##

    I guess I deffirentiate to get vx and vy
    but I don't know how to proceed since tan is confusing me
     
  9. Jun 19, 2017 #8

    haruspex

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    It's just a matter of applying the chain rule. What is the derivative wrt time of tan(θ)?
     
  10. Jun 20, 2017 #9
    since we apply the chain rule it should be:
    if ##y=tanθ## then ##\frac {dy} {dt} = \frac {dy} {dθ} \frac {dθ} {dt} = sec^2θ\frac {dθ} {dt} = sec^2θ \dot θ##
    Any hint how to proceed with the x and y derivatives?
     
  11. Jun 20, 2017 #10

    haruspex

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    In post #7 you have an equation for x as a function of the two angles. So differentiate it.
     
  12. Jun 20, 2017 #11
     

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  13. Jun 20, 2017 #12

    haruspex

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    Looks right.
    What about ##\dot y##?
     
  14. Jun 21, 2017 #13
    I will proceed finding ##\dot y## later but
    for now I want some clarification with the notation.
    In Mathematics for Engineers by Croft 2nd edition in pg. 709 the quotient rule is been described.
    According to this:
    if
    ##y(x) = \frac {u(x)} {v(x)}##
    then

    ##\frac {dy} {dx} = \frac { v \frac {du} {dx} - u \frac {dv} {dx}} {v^2}##

    an example is given

    ##y=\frac {sinx} {x} = \frac {u} {v}## so ##u=sinx , v=x##
    and so
    ##\frac {du} {dx} = cosx, \frac {dv} {dx} = 1##
    applying the quotient rule gives
    ##\frac {dy} {dx} = \frac { v \frac {du} {dx} - u \frac {dv} {dx}} {v^2}##
    ##=\frac {xcosx - sinx(1)} {x^2} ##
    ##=\frac {xcosx-sinx} {x^2}##

    According to the above example and in relation to this post can we say:
    ##x(θ) = \frac {u(θ)} {v(θ)}##
    but I guess this is not what we want. We want ##x(t) = \frac {u(t)} {v(t)}## so ##\frac {dy} {dt} = \frac { v \frac {du} {dt} - u \frac {dv} {dt}} {v^2}##
    We can say then that u is a function of θ and θ is a function of t or ##u = u(θ) = u(θ(t))## (chain rule)
    So the first line of my attached solution of post #11 should be ##x=\frac {u(t)} {v(t)}## and not ##x=\frac {u(θ)} {v(θ)}##

    Also I forgot to multiply the solution at post #11 with 1000

    Please verify/comment the above

    Alexandros
     
  15. Jun 21, 2017 #14

    haruspex

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    ##x=\frac{u(\theta)}{v(\theta)}## is fine. ##\theta## is allowed to be a function of t. ##\dot x=\frac{d}{dt}\frac{u(\theta)}{v(\theta)}= \dot\theta\frac {d}{d\theta}\frac{u(\theta)}{v(\theta)}##.
    True.
     
  16. Jun 22, 2017 #15
    Is the following correct?

    ##\frac {dx} {dt} = \frac {dx} {dθ} \frac {dθ} {dt} = \dot θ \frac {dx} {dθ}##

    and since ##x(θ) = \frac {u(θ)} {v(θ)}## where ##u(θ) = tanθ_A## and ##v(θ) = tanθ_A - tanθ_B##

    then ##\frac {dx} {dθ} =1000 \frac {v \frac {du} {dθ} - u \frac {dv} {dθ}} {v^2}##

    so ##\frac {du} {dθ} = sec^2 θ_A ## and ##\frac {dv} {dθ} = sec^2 θ_Α - sec^2 θ_B##

    so ##\frac {dx} {dθ} =1000 \frac {(tanθ_A - tanθ_B)(sec^2 θ_A) - (tanθ_A)(sec^2 θ_Α - sec^2 θ_B)} {(tanθ_A - tanθ_B)^2}##
    ##=1000\frac {tanθ_Asec^2 θ_A - tanθ_Bsec^2 θ_A - tanθ_Asec^2 θ_Α + tanθ_Asec^2 θ_B} {(tanθ_A - tanθ_B)^2}##
    ##=1000\frac {- tanθ_Bsec^2 θ_A + tanθ_Asec^2 θ_B} {(tanθ_A - tanθ_B)^2}##
    so
    ##\frac {dx} {dt} = \dot θ 1000\frac {- tanθ_Bsec^2 θ_A + tanθ_Asec^2 θ_B} {(tanθ_A - tanθ_B)^2}##
     
  17. Jun 22, 2017 #16

    haruspex

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    I do not understand the direction you have taken. What is this angle θ? Your attachment in post #11 was correct.
     
  18. Jun 22, 2017 #17
    ok.

    for vy look the attachement
     

    Attached Files:

  19. Jun 22, 2017 #18

    haruspex

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    You have not applied the chain rule correctly this time. The derivative of 1/v is not ##-(\dot v)^2##.
     
  20. Jun 23, 2017 #19
    ok I am totally confused. any hint?
    should that be ##\frac {dy} {dt} = 1000\frac {d} {dt} \left(\frac {1} {v}\right) = 1000(-1) (\frac {1} {v^2}) ##
     
    Last edited: Jun 23, 2017
  21. Jun 23, 2017 #20
    what about this:

    ##y(v) =1000\frac {1} {v}## and ##v(θ) = tanθ_A-tanθ_B##

    ##\frac {dy} {dt} = \frac {dy} {dv} \frac {dv} {dt}##

    where ##\frac {dy} {dv} = 1000(-1) \frac {1} {v^2}## and ##\frac {dv} {dt} = sec^2θ_A \dot θ_A - sec^2θ_B \dot θ_B##

    so

    ##\frac {dy} {dt}= -1000\frac{sec^2θ_A \dot θ_A - sec^2θ_B \dot θ_B} {(tanθ_A-tanθ_B)^2}##
     
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