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Math Help - Word Problem. Help please!

  1. #1
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    Word Problem. Help please!

    A steel pipe is being carried down a hallway 9 ft. wide. At the end of the hall there is a right-angled turn into a narrower hallway 6 ft. wide.

    a.) Show that the length of the pipe in the figure is modeled by the function.

    L(theta) = 9 csc (theta) + 6 sec (theta)

    b.) Graph the function L for 0< (theta)< pi/2.

    c.) Find the minimum value of the function L.

    d.) Explain why the value of L you found in part (c.) is the length of the longest pipe that can be carried around the corner.
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  2. #2
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    1) We can't see the figure.

    2) We are not going to do your homework for you. What have YOU done on this?
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  3. #3
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    ok... I am sorry, I don't understand how to go about figuring out the problem. I would like someone to help me so that I can understand this because I have a test next week. I am in college, not high school. Homework doesn't count anymore.
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  4. #4
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    http://math.emich.edu/~enadler/cours...5_sampleQs.pdf


    There is a website, if you scroll down to problem 20., thats what the problem in my book looks like.
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  5. #5
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    Hello, aligator0207!

    Here's part (a).


    A steel pipe is being carried down a hallway 9 ft. wide.
    At the end of the hall there is a right-angled turn into a narrower hallway 6 ft. wide.

    a.) Show that the length of the pipe in the figure is modeled by the function:

    . . . . . L(\theta) \:=\: 9\csc\theta + 6\sec\theta
    Code:
                                          C
          *-------------------------------o------
          |                           * θ :
          |                       *       :6
          |              B    *           :
          |               o---------------o------
          |           * θ |               E
          | θ     *       |
          |   *           |
        A o - - - - - - - o D
          |       9       |
          |               |

    The pipe is: . L \:=\:AC \:=\:AB+BC

    AD = 9,\;CE = 6

    Note that: . \angle ABD \,=\,\angle BCE \,=\,\theta


    In right triangle BDA\!:\;\;\csc\theta \:=\:\frac{AB}{9} \quad\Rightarrow\quad AB \:=\:9\csc\theta

    In right triangle CEB\!:\;\;\sec\theta \:=\:\frac{BC}{6} \quad\Rightarrow\quad BC \:=\:6\sec\theta


    Therefore: . L \;=\;9\csc\theta + 6\sec\theta

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  6. #6
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    Thank you, I understand that more clearly now . Now I think that I can get the next 3 problems. I'll let you know what I get and hopefully I will get them correct!
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