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Homework Help: Related rates equation problem

  1. Nov 25, 2007 #1
    1. The problem statement, all variables and given/known data

    A boy is standing 50 ft. from the end of a swimming pool when he sees a girl 25 ft. along the end. He can swim 3 ft/s and run 5 ft/s. If he runs x feet, set up an equation for time consumed.

    2. Relevant equations

    There is a visual aid: A rectangle with length labeled 50 ft and a dot B designating where the boy is, and with a width labeled 25 ft and a dot G designating where the girl is.

    3. The attempt at a solution

    I know I have to set up a triangle, with my y-axis being the length (where the girl is, making my dy/dt= 3 ft/s) and the x-axis being the width (dx/dt=5 ft/s) Past this, I really don't know. I tried this:

    Since he is running x ft, the 50 ft must be losing x, and the 25 ft must be gaining x. So y/x=(25+x)/(50-x) However, I need t in there. If I integrate dy/dt, I will be left with a c that I'm not sure how to get rid of. So yeah, I'm really stumped on this one.

    I know it's kinda hard to visualize, so any help at all would be appreciated.
  2. jcsd
  3. Nov 25, 2007 #2


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    You haven't given enough information! For example, if he is standing on the right of the swimming pool and the girl is also to the right of the swimming pool, I can see no reason for him swimming at all!
  4. Nov 25, 2007 #3
    He is standing in the lower right corner, she in the upper left.
  5. Nov 26, 2007 #4


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    That completely contradicts what you originally said:
    What you meant, I think is that the swimming pool is 50 feet long and 25 feet wide, the boy and girl are at opposite corners. That's quite different from saying "A boy is standing 50 ft. from the end of a swimming pool"!

    He runs x feet along the length of the pool leaving 50-x to the end. That does NOT "add" anything to the 25 foot width. He swims along the hypotenuse of a right triangle having one leg 50-x and the other 25. The distance he swims is [itex]\sqrt{(50-x)^2+ 25^2}=\sqrt{3125- 100x+ x^2}[/itex]. The time it takes to go a given distance is that distance divided by the speed.
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