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Math Help - Application of maxima

  1. #1
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    Application of maxima



    In the diagram above, Ox and Oy are the rectangular axes and the straight line PAQ passes through the fixed point A ( 4,1 ). The angle QPO= Theta and 0 < Theta < pi/2

    Let T = theta ( Sorry , I do not know how to insert the symbol, I tried &Theta but no luck )

    a) Show thatOP + OQ = 5 + 4 tan T + cot T and show that as T changes, the least value of OP + OQ is 9.

    b) Show that PQ = 4 sec T + cosec T and find, correct to two significant figures, the minimum value of Q when T changes.
    Last edited by ose90; September 22nd 2008 at 04:21 AM. Reason: Sorry!
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  2. #2
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    Is O not the origin (0,0)?
    Then angle QOP should be 90 degrees or pi/2.

    The posted diagram and the text of the question don't tally.
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  3. #3
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    Quote Originally Posted by ticbol View Post
    Is O not the origin (0,0)?
    Then angle QOP should be 90 degrees or pi/2.

    The posted diagram and the text of the question don't tally.
    Yes, Sir , It is the origin, sorry for typing wrongly.
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  4. #4
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    Quote Originally Posted by ose90 View Post


    In the diagram above, Ox and Oy are the rectangular axes and the straight line PAQ passes through the fixed point A ( 4,1 ). The angle QPO= Theta and 0 < Theta < pi/2

    Let T = theta ( Sorry , I do not know how to insert the symbol, I tried &Theta but no luck )

    a) Show thatOP + OQ = 5 + 4 tan T + cot T and show that as T changes, the least value of OP + OQ is 9.

    b) Show that PQ = 4 sec T + cosec T and find, correct to two significant figures, the minimum value of Q when T changes.
    Let OP = u, and OQ = v

    In the right triangle below the point (4,1):
    vertical leg = 1
    horizontal leg = (u -4)
    tanT = 1 / (u -4)
    u -4 = 1 / tanT = cotT
    u = cotT +4 -----------**

    In the right triangle above the (4,1):
    vertical leg = (v -1)
    horizontal leg = 4
    tanT = (v -1) / 4
    4tanT = v -1
    v = 4tanT +1

    So, (OP +OQ) = (u +v) = cotT +4 +4tanT +1
    (oP +OQ) = 5 +4tanT +cotT -----------------------shown.

    For the least value of (OP +OQ), get the derivative of (OP +OQ) with respect to T. Then equate that to zero.
    You should arrive to 4 sec^2(T) = csc^2(T).
    Which is the same as cot^2(T) = 4
    Or, cotT = 2, and tanT = 1/2
    Then minimum (OP +OQ) = 9

    ---------------------------------------
    b) Show that PQ = 4 sec T + cosec T and find, correct to two significant figures, the minimum value of Q when T changes

    In the said right triangle below (4,1):
    hypotense = say, r
    sinT = 1/r
    r = 1/sinT = cscT

    In the right triangle above the (4,1):
    hypotenuse = say, s
    cosT = 4/s
    s = 4/cosT = 4secT

    Thus, PQ = r +s = cscT +4secT
    PQ = 4secT +cscT ----------------------shown.

    For the minimum value of Q.....?
    Q? or PQ?
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  5. #5
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    Quote Originally Posted by ticbol View Post
    Let OP = u, and OQ = v

    In the right triangle below the point (4,1):
    vertical leg = 1
    horizontal leg = (u -4)
    tanT = 1 / (u -4)
    u -4 = 1 / tanT = cotT
    u = cotT +4 -----------**

    In the right triangle above the (4,1):
    vertical leg = (v -1)
    horizontal leg = 4
    tanT = (v -1) / 4
    4tanT = v -1
    v = 4tanT +1

    So, (OP +OQ) = (u +v) = cotT +4 +4tanT +1
    (oP +OQ) = 5 +4tanT +cotT -----------------------shown.

    For the least value of (OP +OQ), get the derivative of (OP +OQ) with respect to T. Then equate that to zero.
    You should arrive to 4 sec^2(T) = csc^2(T).
    Which is the same as cot^2(T) = 4
    Or, cotT = 2, and tanT = 1/2
    Then minimum (OP +OQ) = 9

    ---------------------------------------
    b) Show that PQ = 4 sec T + cosec T and find, correct to two significant figures, the minimum value of Q when T changes

    In the said right triangle below (4,1):
    hypotense = say, r
    sinT = 1/r
    r = 1/sinT = cscT

    In the right triangle above the (4,1):
    hypotenuse = say, s
    cosT = 4/s
    s = 4/cosT = 4secT

    Thus, PQ = r +s = cscT +4secT
    PQ = 4secT +cscT ----------------------shown.

    For the minimum value of Q.....?
    Q? or PQ?
    Hello sir, thanks for big help , I'm so grateful.
    Yup, the last part is PQ , I will solve it on my own, thanks again, your help is much appreciated.
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