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Math Help - Need Calc 1 Help.

  1. #1
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    Need Calc 1 Help.

    1. A plane flying with a constant speed of 14 km/min passes over a ground radar station at an altitude of 8 km and climbs at an angle of 50 degrees. At what rate, in km/min is the distance from the plane to the radar station increasing 1 minutes later?


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    2. A particle is moving along the curve y=2*sqrt(5x+1). As the particle passes through the point (38), its x-coordinate increases at a rate of 4 units per second. Find the rate of change of the distance from the particle to the origin at this instant.
    Last edited by lmao; April 15th 2007 at 05:43 PM.
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  2. #2
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    This is a good set of problems.
    What have you done on them?
    If you try them then you are bound to learn a great deal.
    But if we do them for you, you learn little.
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  3. #3
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    I have done about 16 of these and these are the ones that have me stumped...I can get started and get an answer but its always incorrect.
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  4. #4
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    Hello, lmao!

    I'll walk through the first one . . .
    And I'm changing the units to miles and miles per hour, okay?


    1. At noon, ship A is 50 miles due west of ship B.
    Ship A is sailing west at 25 mph and ship B is sailing north at 18 mph.
    How fast is the distance between the ships changing at 7 PM?
    Code:
                                  * B
                              *   |
                     x    *       |
                      *           | 18t
                  *               |
              *                   |
          * - - - - * - - - - - - *
          A   25t   P     50      Q

    At noon, ship A starts at point P, 50 miles west of point Q.
    . . At 25 mph, in the next t hours, it moves 25t miles to point A.

    At noon, ship B start at point Q, sailing north at 18 mph.
    . . In the next t hours, it moves 18t miles to point B.

    Let x = the distance AB.

    Pythagorus says: .x .= .(25t + 50) + (18t)

    So we have: .x .= .949t + 2500t + 2500

    Differentiate with respect to time: .2x(dx/dt) .= .1898t + 2500

    . . and we have: .dx/dt .= .(1898t + 2500)/2x .[1]


    At 7 PM (t = 7): .x .= .9497 + 25007 + 2500 .= .66,501
    . . Hence: .x .= .√(66501)


    Substitute into [1]: .dx/dt .= .(18987 + 2500)/(2√66501) . .30.6 mph

    Now you can change it back to "knots" . . .

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  5. #5
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    Excellent explanation. Thank you so much!
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  6. #6
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    Re:

    I know I am not an Admin but I have seen this exact post twice already, whats up with the double posting?
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  7. #7
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    Yeah my bad, I'll delete the other one if it lets me. I posted there then realized there was a urgent help forum. Sorry again!
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  8. #8
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    Re:

    RE:

    I can tell this is WebWorks problem, thus I am going to help you with 1 of them.

    #4.
    Attached Thumbnails Attached Thumbnails Need Calc 1 Help.-related-rates-1.gif  
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  9. #9
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    I solved #2 so I'll take that one out. Thanks for understanding that webwork is a pain. So #1,2,4 are now done. Two more! Thanks guys!!
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lmao View Post
    1. A plane flying with a constant speed of 14 km/min passes over a ground radar station at an altitude of 8 km and climbs at an angle of 50 degrees. At what rate, in km/min is the distance from the plane to the radar station increasing 1 minutes later?
    Here. As I always say, ALWAYS DRAW A DIAGRAM WHEN DOING A RELATED RATES PROBLEM.

    related rates 1 is the solution
    related rates 1b is the diagram i used to think about a solution
    Attached Thumbnails Attached Thumbnails Need Calc 1 Help.-related-rates1.gif   Need Calc 1 Help.-related-rates-1b.gif  
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  11. #11
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    Re:

    RE: Sorry I made an error, its getting late...UPDATE
    Last edited by qbkr21; April 15th 2007 at 08:15 PM.
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  12. #12
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    All of them are complete! Thanks everyone for helping.
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