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Math Help - solve with vectors

  1. #1
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    solve with vectors

    At 12 noon the position vectors r and the velocity vectors v of two ships A and B are

    rA = (2i+j)km vA = (3i+j)kmh^-1
    rB = (-i-4j)km vB = (11i+3j)kmh^-1

    (a) Show that at time t h after noon the position vector of B relative to A is given by:
    [(8t-3)i+(2t-5)j]km
    (b) Show that the distance d km between the vessels, at this time, is given by
    d^2=68[t^2 - t+1/2]
    (c) Hence show that the ships are nearest together at 12.30 p.m and find the distance between them at this time.
    Last edited by outlaw; April 22nd 2008 at 09:06 AM.
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  2. #2
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    Quote Originally Posted by outlaw View Post
    At 12 noon the position vectors r and the velocity vectors v of two ships A and B are

    rA = (2i+j)km vA = (3i+j)kmh^-1
    rB = (-i-4j)km vB = (11i+3j)kmh^-1

    (a) Show that at time t h after noon the position vector of B relative to A is given by:
    [(8t-3)i+(2t-5)j]km
    (b) Show that the distance d km between the vessels, at this time, is given by
    d^2=68[t^2 - t+1/2]
    (c) Hence show that the ships are nearest together at 12.30 p.m and find the distance between them at this time.
    (a) r_{BA}(t) = t.v_{BA} + r_{BA}(0)

    (b) d^2 = |r_{BA}(t)|^2

    (c) Differentiate d^2 and equate it to zero to find 't' and subsequently d. Or you can observe that  t^2 - t + \frac12 = (t - \frac12)^2 + \frac14. Clearly the minimum for this function is at t = \frac12, which is 30 minutes......
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  3. #3
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    Hello, outlaw!

    At 12 noon the position vectors \vec{r} and the velocity vectors \vec{v} of two ships A and B are:

    . . . \begin{array}{ccccccc}\vec{r}_A &= & 2{\bf i} + {\bf j}\text{ km} & & \vec{v}_A &=& 3i + j\text{ km/hr} \\<br />
\vec{r}_B &= &  \text{-}{\bf i} -4{\bf j} \text{ km} & & \vec{v}_B &=& 11{\bf i} + 3{\bf j}\text{ km/hr}<br />
\end{array}

    (a) Show that t hours after noon, the position vector of B relative to A is given by:

    . . . \overrightarrow{AB} \;=\;(8t-3){\bf i}+ (2t-5){\bf j}\;\text{km}
    In t hours, the position vectors are:

    . . \begin{array}{ccccc}\vec{r}_A &=& (2{\bf i} + {\bf j}) + (3{\bf i}+{\bf j})t &=& (2+3t){\bf i} + (1 + t){\bf j} \\<br />
\vec{r}_B &=& (\text{-}{\bf i}-4{\bf j}) + (11{\bf i} + 3{\bf j})t &=& (11t-1){\bf i} + (3t-4){\bf j} \end{array}

    Then: . \overrightarrow{AB} \;=\;\bigg[(11t - 1){\bf i} - (2+3t){\bf j}\bigg] + \bigg[(3t-4){\bf i}-(1+t){\bf j}\bigg]

    Therefore: . \boxed{\overrightarrow{AB} \;=\;(8t-3){\bf i} + (2t-5){\bf j}}



    (b) Show that the distance d between the vessels at time t
    . . . is given by: . d^2 \;=\;68\left(t^2 - t+\frac{1}{2}\right)
    d^2 \;=\;|\overrightarrow{AB}|^2\;=\;(8t-3)^2 + (2t-5)^2\;=\;68t^2-68t + 34

    Therefore: . \boxed{d^2\;=\;68\left(t^2 - t + \frac{1}{2}\right)}



    (c) Hence show that the ships are nearest together at 12:30 pm
    and find the distance between them at this time.
    We have . d^2 \;=\;68t^2 - 68t + 34

    Note that d is at a minimum when d^2 is a minimum.

    The graph for d^2 is an up-opening parabola.
    . . Its minimum occurs at its vertex.
    The vertex is at: . t \:=\:\frac{\text{-}b}{2a}

    We have: . a = 68,\;b = \text{-}68
    . . Hence, the vertex is at: . t \:=\:\frac{\text{-}(\text{-}68)}{2(68)} \:=\:\frac{1}{2}


    Since the distance is a minimum when t \:=\:\frac{1}{2}\text{ hour} \:=\:30\text{ minutes},

    . . the ships are at their nearest at \boxed{12:30\text{ pm}}


    When t = \frac{1}{2}\!:\;\;d^2 \;=\;68\left(\frac{1}{2}\right)^2 - 68\left(\frac{1}{2}\right) + 34 \;=\;17

    . . Therefore, their minimum distance is . d \;=\;\boxed{\sqrt{17}\text{ km}}

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