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anonymous

  • one year ago

I can't think of how to label my illustration for this question... I need someone to direct me in the right way. A baseball diamond is a square with sides of length 90 ft. A batter hits the ball and runs toward first base with a speed of 26 ft/s. At what rate is his distance from second base changing when he is halfway to first base?

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  1. anonymous
    • one year ago
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    |dw:1434166696015:dw|

  2. anonymous
    • one year ago
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    I've misplaced #1 and 3 but it shouldn't make a difference

  3. anonymous
    • one year ago
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    @ganeshie8 thank god you're here! Can you help me on this?

  4. ganeshie8
    • one year ago
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    |dw:1434167956990:dw|

  5. anonymous
    • one year ago
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    Mhmm and how do I label the info given in the question

  6. ganeshie8
    • one year ago
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    |dw:1434168228668:dw|

  7. ganeshie8
    • one year ago
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    apply pythagorean theorem, differentiate and do the usual stuff

  8. wolf1728
    • one year ago
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    Just to make things easier, most people visualize a baseball diamond with home plate at the bottom, 1st base on the left, etc.

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  9. anonymous
    • one year ago
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    So is this what I have to do... \[90^2+(90-x)^2=y^2\] \[2(90-x)=2yy'\] \[y'=\frac{ 2(90-x) }{ 2y }\]

  10. Michele_Laino
    • one year ago
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    hint: using the drawing of @ganeshie8 we can write: \[y = \sqrt {{{90}^2} + {x^2}} \] furthermore, please keep in mind that: \[x = x\left( t \right) = vt = 26t\]

  11. anonymous
    • one year ago
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    Wouldn't it be \[y=\sqrt{90^2+(90-x)^2}\]

  12. Michele_Laino
    • one year ago
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    the right triangle is: |dw:1434170152802:dw|

  13. anonymous
    • one year ago
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    I'm a bit confused

  14. Michele_Laino
    • one year ago
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    when I write: \[y\left( t \right) = \sqrt {{{90}^2} + x{{\left( t \right)}^2}} \] am I right?

  15. Michele_Laino
    • one year ago
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    @Tracy96

  16. anonymous
    • one year ago
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    yes

  17. Michele_Laino
    • one year ago
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    |dw:1434170615708:dw|

  18. Michele_Laino
    • one year ago
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    ok, now try to differentiate y(t) with respect to the time t, what do you get?

  19. Michele_Laino
    • one year ago
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    please keep in mind that you have to use the chain rule, since: \[x = x\left( t \right) = vt = 26t\]

  20. anonymous
    • one year ago
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    Just a quick question, why is it x(t) on one of the legs?

  21. Michele_Laino
    • one year ago
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    it is the distance between the current position of the player, from the first base

  22. Michele_Laino
    • one year ago
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    |dw:1434170805543:dw|

  23. anonymous
    • one year ago
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    Ok got it

  24. Michele_Laino
    • one year ago
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    ok! Now what is: \[\frac{{dy}}{{dt}} = y' = ...\]?

  25. anonymous
    • one year ago
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    Is it \[y'=\frac{ 1 }{ 2 }\sqrt{(x(t))^2 + 90^2} \times2x(t)x'(t)\] ??

  26. Michele_Laino
    • one year ago
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    use this rule: \[y = \sqrt x , \Rightarrow y' = \frac{1}{{2\sqrt x }}\]

  27. anonymous
    • one year ago
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    \[\frac{ 2x(t)x'(t) }{ 2\sqrt{x(t)^2+90^2} }\] like that?

  28. Michele_Laino
    • one year ago
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    that's right!

  29. anonymous
    • one year ago
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    Relief! haha ok now what?

  30. Michele_Laino
    • one year ago
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    now, as I wrote before, we have: \[x = x\left( t \right) = vt = 26t\]

  31. Michele_Laino
    • one year ago
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    so, what is: \[x' = \frac{{dx}}{{dt}} = ...?\]

  32. anonymous
    • one year ago
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    26 ?

  33. Michele_Laino
    • one year ago
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    perfect!

  34. Michele_Laino
    • one year ago
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    next, we have to substitute the expression of x and its derivative, into the above firtst derivative of y(t)

  35. anonymous
    • one year ago
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    Wait so why is x(t) = vt ? Is that coming from ---> distance = velocity * time

  36. Michele_Laino
    • one year ago
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    yes! I think that in your exercise, it is supposed that the motion of the player is an uniform motion

  37. anonymous
    • one year ago
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    Ohh gotcha! So we first find y' and then x' and then substitute x' into the equation we got from y'... am i following?

  38. Michele_Laino
    • one year ago
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    yes!

  39. anonymous
    • one year ago
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    Alrighty! So then \[y'=\frac{ 52x(t) }{ 2\sqrt{x(t)^2 +90^2} }\] right?

  40. Michele_Laino
    • one year ago
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    ok! Now simplify: 52/2=26 and substitute x= vt

  41. anonymous
    • one year ago
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    Ohh so \[y'=\frac{ 26(26t) }{ \sqrt{(26t)^2+90^2} }\]

  42. Michele_Laino
    • one year ago
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    that's right!

  43. anonymous
    • one year ago
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    Great!

  44. Michele_Laino
    • one year ago
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    next we have to compute the time at which the player is at midpoint between the first base and the home plate

  45. Michele_Laino
    • one year ago
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    in other words we have to solve this equation: \[45 = 26t\] what is t?

  46. anonymous
    • one year ago
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    t = 1.73

  47. Michele_Laino
    • one year ago
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    ok!

  48. anonymous
    • one year ago
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    or should i leave it as a fraction 45/26 ?

  49. Michele_Laino
    • one year ago
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    finally, replace t with 1.73 into your last formula for y' (t), and you will get your answer

  50. Michele_Laino
    • one year ago
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    better is the fraction form

  51. anonymous
    • one year ago
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    Right cuz then I can cancel things out and at the end i get 25/90

  52. Michele_Laino
    • one year ago
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    please wait I'm checking your computation

  53. anonymous
    • one year ago
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    Oh wait i made a mistake

  54. anonymous
    • one year ago
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    \[\frac{ 25(45) }{ \sqrt{45^2 +90^2} }\]

  55. Michele_Laino
    • one year ago
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    ok!

  56. anonymous
    • one year ago
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    Is that correct?

  57. Michele_Laino
    • one year ago
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    yes!

  58. anonymous
    • one year ago
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    Ahhh awesome!! I really bugged you didn't I?

  59. Michele_Laino
    • one year ago
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    It is all right! :)

  60. anonymous
    • one year ago
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    Thanks so much!! I have one more question about this problem xD

  61. Michele_Laino
    • one year ago
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    ok! I can help you!

  62. anonymous
    • one year ago
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    Oh my goodness I love you so much! :D It's also asking to find at what rate the player's distance from third base is changing at the same moment... would I have to start all over?

  63. Michele_Laino
    • one year ago
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    yes! I think so!

  64. anonymous
    • one year ago
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    Oh no! :S

  65. Michele_Laino
    • one year ago
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    it is necessary, since we have to rewrite the function y(t), after that the procedure is the same as before

  66. anonymous
    • one year ago
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    |dw:1434173455402:dw| is this drawing right?

  67. Michele_Laino
    • one year ago
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    if we pick as origin of our x-axis the first base, then we have: |dw:1434173776450:dw| so: \[y\left( t \right) = \sqrt {{{90}^2} + {{\left\{ {90 - x\left( t \right)} \right\}}^2}} \]

  68. anonymous
    • one year ago
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    Ohhh so now I get what's happening!

  69. anonymous
    • one year ago
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    That's gonna be the only difference correct?

  70. Michele_Laino
    • one year ago
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    correct!

  71. anonymous
    • one year ago
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    Otherwise I just do the same thing as before

  72. Michele_Laino
    • one year ago
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    yes! the procedure is the same as before, only the function y(t) is different

  73. anonymous
    • one year ago
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    I honestly don't know how to thank you!! You're a lifesaver! Many thanks!

  74. Michele_Laino
    • one year ago
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    :)

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