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marigirl

  • one year ago

I managed to get the equations for this question but I need a little advice when it comes to differentiating. Question: As part of a school ironman competition, students have to swim across a river which is 30m wide and then run along the side the river until they reach the finishing ling 200 m along the bank. Bob knows that he can run twice as he can swim. If he begins his swim at point X, at what point along the river Y should he get out of the water in order to finish his run at Z in the shortest possible time?

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  1. marigirl
    • one year ago
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    My Answer: I used the formula Velocity= Distance/Time and since I need the shortest possible time I rearranged the equation so T= D/V I obtained the equation below: \[t=\frac{ \sqrt{x ^{2+900}} }{ v } + \frac{ 200-x }{ 2v }\] MY QUESTION: Since we want the best point for Bob to swim up to (x)- we differentiate x with respect to t (time) ..... then what about v- as in i dont differentiate it because i am not after optimum velocity

  2. marigirl
    • one year ago
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    |dw:1440103563755:dw|

  3. marigirl
    • one year ago
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    I guess what I am trying to ask is, is it possible to differentiate one variable and not the other IN ONE EQUATION, and then find the maximum/minimum by making it equal to zero....

  4. IrishBoy123
    • one year ago
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    \(\frac{dT}{dx} = 0 \) is the measure of what you are trying to answer go ahead and solve

  5. marigirl
    • one year ago
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    so i dont differentiate the v ???

  6. IrishBoy123
    • one year ago
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    no! v is a constant. it could be 10, or 6 or anything

  7. marigirl
    • one year ago
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    \[t=\frac{ \sqrt{x^2+900} }{ v } + \frac{ 200-x }{ 2v }\] so should i do quotient rule? or differentiate the x expression and leave the v where it is?

  8. IrishBoy123
    • one year ago
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    put the v on the LHS!

  9. marigirl
    • one year ago
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    the answer reads: \[\frac{ dt }{ dx }=\frac{ x }{v \sqrt{x^2+900} }-\frac{ 1 }{ 2v}\]

  10. IrishBoy123
    • one year ago
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    which looks good

  11. marigirl
    • one year ago
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    oh ok, because it is a constant and in this question it is not the velocity we are after, it is the distance ?

  12. IrishBoy123
    • one year ago
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    no, you are after t, first and foremost.

  13. marigirl
    • one year ago
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    but we still need to regard the velocity to obtain the best possible distance.. so we dont eliminate it but sort of influence x?

  14. IrishBoy123
    • one year ago
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    you are minimising t. wrt x so you set dt/dx = 0

  15. IrishBoy123
    • one year ago
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    v is just a number, it could be 9 or 774 who cares what it is . i don't !! i'm caling it v.

  16. jim_thompson5910
    • one year ago
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    @marigirl if you simplify the expression you got for dt/dx, you should get \[\Large \frac{ dt }{ dx }=\frac{ 2x- \sqrt{x^2+900} }{v \sqrt{x^2+900} }\] set dt/dx equal to 0 and solve for x. You'll find that v plays no part in the solution to dt/dx = 0. So this means that the velocity v does not alter the point where it's best to get out of the water (for the shortest time possible)

  17. jim_thompson5910
    • one year ago
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    sorry I meant to say \[\Large \frac{ dt }{ dx }=\frac{ 2x- \sqrt{x^2+900} }{2v \sqrt{x^2+900} }\]

  18. IrishBoy123
    • one year ago
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    @marigirl you have done it correctly now just finish it off i hate being talked over so i am off. but good luck!

  19. marigirl
    • one year ago
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    @IrishBoy123 i might have a question for u . give me a min

  20. marigirl
    • one year ago
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    okay i think i know where im tripping up,, Differentiating (200-x)(2v)

  21. marigirl
    • one year ago
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    @IrishBoy123

  22. marigirl
    • one year ago
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    @jim_thompson5910 When u differentiated the equation- did u differentiate each part then add it both together? please tell me what you got when you differentiated the second term .. im just not getting it \[t=\frac{ \sqrt{x^2+900} }{ v }+\frac{ 200-x }{ 2v}\] \[t'=\frac{ x }{ v \sqrt{x^2+900} } + ???\]

  23. jim_thompson5910
    • one year ago
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    v is a constant, so 2v is also a constant along with 1/(2v) pull out the 1/(2v) to just focus on the 200-x derive 200 - x to get ???

  24. marigirl
    • one year ago
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    -1/2v but then i applied quotient rule and got the first term ... why is it not working for the second term?

  25. jim_thompson5910
    • one year ago
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    I'm not sure what you mean, but we have this \[\large \frac{ dt }{ dx }=\frac{ x }{v \sqrt{x^2+900} }-\frac{ 1 }{ 2v}\] simplify the RHS to get \[\large \frac{ dt }{ dx }=\frac{ x }{v \sqrt{x^2+900} }-\frac{ 1 }{ 2v}\] \[\large \frac{ dt }{ dx }=\frac{ {\color{red}{2}}*x }{{\color{red}{2}}v \sqrt{x^2+900} }-\frac{ 1 }{ 2v}\] \[\large \frac{ dt }{ dx }=\frac{ 2x }{2v \sqrt{x^2+900} }-\frac{ 1 }{ 2v}\] \[\large \frac{ dt }{ dx }=\frac{ 2x }{2v \sqrt{x^2+900} }-\frac{ 1*{\color{red}{\sqrt{x^2+900}}} }{ 2v*{\color{red}{\sqrt{x^2+900}}}}\] \[\large \frac{ dt }{ dx }=\frac{ 2x }{2v \sqrt{x^2+900} }-\frac{ \sqrt{x^2+900} }{ 2v\sqrt{x^2+900}}\] \[\large \frac{ dt }{ dx }=\frac{ 2x- \sqrt{x^2+900} }{ 2v\sqrt{x^2+900}}\]

  26. marigirl
    • one year ago
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    i guess what you did was treated the V variable as a constant - but didnt eliminate it ...

  27. jim_thompson5910
    • one year ago
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    you mean apply the quotient rule to (200-x)/(2v) ?

  28. marigirl
    • one year ago
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    yes... i applied quotient rule to the first term .... it is not working for the second for some reason

  29. jim_thompson5910
    • one year ago
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    let's make f(x) = 200-x g(x) = 2v derive both functions to get f ' (x) = -1 g ' (x) = 0 So using the quotient rule to derive f/g, we get [ f ' (x) * g(x) - g ' (x) * f(x) ]/[ g ^2 ] [ -1 * 2v - 0 * (200-x) ]/[ (2v)^2 ] [ -2v - 0 ]/[ (2v)^2 ] (-2v)/(4v^2) -1/(2v) which is the same if we just pull out the 1/(2v) and ignore it temporarily

  30. marigirl
    • one year ago
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    omggg i was doing silly things all this time!

  31. jim_thompson5910
    • one year ago
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    that's ok

  32. marigirl
    • one year ago
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    i KNEW v was a constant but i kept differentiating it incorrectly .. omg i wasted so much paper... thanks jim

  33. jim_thompson5910
    • one year ago
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    it's not a waste if you learned something though, so no worries

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