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Mimi_x3 Group Title

Two straight roads intersect at right angles. Two men, A and B, are 100 km from the intersection, one on each road. They drive towards the intersection at 30km^-1 and 40 khmh^-1 respectively. Find the distance of each driver from the intersection as a function of t, the time in hours for which they are driving. Hence find their distance apart, d(t), at any time. For what value of t is the distance apart least?

  • 2 years ago
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  1. Ishaan94 Group Title
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    ah you posted it again

    • 2 years ago
  2. Mimi_x3 Group Title
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    Yup, i was staring it for hours i didn't even know how to do it -_-

    • 2 years ago
  3. Mimi_x3 Group Title
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    I tried working backwards but it didn't work. Or using the pythagorus theorem *sigh*

    • 2 years ago
  4. saifoo.khan Group Title
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    that looks soo scary.

    • 2 years ago
  5. Mimi_x3 Group Title
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    LOL, I know its killing me =/

    • 2 years ago
  6. Ishaan94 Group Title
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    |dw:1317388650593:dw|

    • 2 years ago
  7. Mimi_x3 Group Title
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    I know, I already have the diagram, but how can I apply the pythagorus theorem ? I tried it nd I got huge numbers

    • 2 years ago
  8. Ishaan94 Group Title
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    |dw:1317388745649:dw|

    • 2 years ago
  9. Mimi_x3 Group Title
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    Like this : 3999.23 / dt = 2000a + 2000b / dt ?

    • 2 years ago
  10. IanT Group Title
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    |dw:1317388825437:dw| this is the graph of time against distance. Very badly drawn.

    • 2 years ago
  11. Ishaan94 Group Title
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    \[H^2 = (100 - 30t)^2 +(100-40t)^2\]

    • 2 years ago
  12. Ishaan94 Group Title
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    \(t\) is in hours

    • 2 years ago
  13. Mimi_x3 Group Title
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    Don't you find H as well ? That is 141.42 ?

    • 2 years ago
  14. SimonS Group Title
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    Let A be the location of the driver driving at 40 km/hr and B be the location of the driver driving at 30 km/hr A and B are measured by distance from the intersection. Then the distance you want is \[d = \sqrt{A^2 + B^2}\] Now initially of course, i.e., when t = 0, A = B = 100 km and hence d= 100.sqrt(2) which is approximately 141 km. Now, what about for times t > 0. Well, A drives towards the intersection so if t is in hours: A(t) = 100 - 40t Making sense so far? If so. What is B(t) ?

    • 2 years ago
  15. IanT Group Title
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    from the graph, the slope is -0.333... The equation of the line is t = (-0.0333...)d + 3.333...

    • 2 years ago
  16. JamesJ Group Title
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    SimonS is trying to explain what Ishaan is saying and exactly right. So Mimi, what is B(t)?

    • 2 years ago
  17. JamesJ Group Title
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    and for the record what Ian is saying is unfortunately wrong.

    • 2 years ago
  18. Mimi_x3 Group Title
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    B(t) is the distance B from the intersection ?

    • 2 years ago
  19. SimonS Group Title
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    Yes

    • 2 years ago
  20. JamesJ Group Title
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    The distance from the intersection of the driver driving at 30 km/hr

    • 2 years ago
  21. IanT Group Title
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    For my sake, can you point out the error. Reading the other posts, I realize I'm not helping with the problem, but can you show me where I went wrong?

    • 2 years ago
  22. JamesJ Group Title
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    @mimi, do you see where the formula for A(t) came from?

    • 2 years ago
  23. Mimi_x3 Group Title
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    Yup, do i expand (100-30t)^2 ?

    • 2 years ago
  24. JamesJ Group Title
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    No.

    • 2 years ago
  25. JamesJ Group Title
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    Or not yet. So A(t) = 100 - 40t and hence B(t) = ... ?

    • 2 years ago
  26. Mimi_x3 Group Title
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    B(t) = 100-30t

    • 2 years ago
  27. JamesJ Group Title
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    Right. Good.

    • 2 years ago
  28. Mimi_x3 Group Title
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    Then the pythagoras theorem xD

    • 2 years ago
  29. JamesJ Group Title
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    so d(t)^2 = A(t)^2 + B(t)^2 = .... Now expand.

    • 2 years ago
  30. JamesJ Group Title
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    @Ian: you'll see that the equation we ultimately arrive at for d is not linear in t. Where you went wrong was writing down an equation without having really thought through the problem.

    • 2 years ago
  31. Mimi_x3 Group Title
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    H^2 = 2500t^2 - 14000t + 20000 What do i do next ? differentiate it ?

    • 2 years ago
  32. phi Group Title
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    you want H(t)= sqrt(stuff) then differentiate, set to zero and solve for t

    • 2 years ago
  33. phi Group Title
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    And, if you are comfortable with the chain rule, it wasn't necessary to expand all that stuff. But you already did that work...

    • 2 years ago
  34. JamesJ Group Title
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    Yes. So now you follow the usual procedure to find the minimum of a function. By the way, H is at a minimum exactly when H^2 is at a minimum. So you can take the square root, but it's going to make the differentiation a little more difficult for no more insight. If I was doing this problem I would analyze H^2(t), the equation you have written down.

    • 2 years ago
  35. IanT Group Title
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    My graph is actually for t and the distance of A from the intersection. Not the final function. Am I correct, unhelpful as it may be?

    • 2 years ago
  36. JamesJ Group Title
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    ah, in which case, the graph was right.

    • 2 years ago
  37. Mimi_x3 Group Title
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    Can I differentiate it keeping h^2 ?

    • 2 years ago
  38. phi Group Title
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    @JamesJ True, but the question asks for h(t)

    • 2 years ago
  39. JamesJ Group Title
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    Let f(t) = h^2(t). Differentiate this function.

    • 2 years ago
  40. Mimi_x3 Group Title
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    f(t) ?

    • 2 years ago
  41. Ishaan94 Group Title
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    won't make any difference differentiate with h^2 so it is easy

    • 2 years ago
  42. JamesJ Group Title
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    I'm just saying it's easier to differentiate the function f(t) = h^2(t). For the ultimate answer you need h(T) for whatever T gives the minimum. But it's easier to deal with the differentiation if you keep it in the h^2 form.

    • 2 years ago
  43. JamesJ Group Title
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    But if this causes anyone confusion, differentiate the function h(t)

    • 2 years ago
  44. Mimi_x3 Group Title
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    h^2 = 5000t-1400 h = squareroot ( 5000t-1400) Is that right >

    • 2 years ago
  45. JamesJ Group Title
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    No. As you write above: H^2 = 2500t^2 - 14000t + 20000

    • 2 years ago
  46. Mimi_x3 Group Title
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    Don't I differentiate it >

    • 2 years ago
  47. SimonS Group Title
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    Yes, but you made a few mistakes and your notation is wrong. (h^2) ' = 5000t - 14000

    • 2 years ago
  48. Mimi_x3 Group Title
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    Okay, the its h' = squareroot (5000t-1400)

    • 2 years ago
  49. Mimi_x3 Group Title
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    then*

    • 2 years ago
  50. JamesJ Group Title
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    No, no, definitely not. (h^2) ' does not equal (h')^2

    • 2 years ago
  51. phi Group Title
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    you are dropping a zero on 14000

    • 2 years ago
  52. JamesJ Group Title
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    Just set (h^2) ' = 0 and solve for t

    • 2 years ago
  53. JamesJ Group Title
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    i.e., (h^2) ' = 5000t - 14000 = 0 means t = .... Now for that t, h(t) = ....

    • 2 years ago
  54. Mimi_x3 Group Title
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    Oh I got it 2.8 thank you (:

    • 2 years ago
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