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anonymous

  • 5 years ago

ok, last question in the assignment that i can't seem to get, [derivative = hard ;_;], once again the question is attached in next post.

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  1. anonymous
    • 5 years ago
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  2. anonymous
    • 5 years ago
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    alright, what i know so far: after 3 seconds, the x[t] = 42ft and the s[t]=58.7 while the y[t] is given as 41ft. so now im not exactly sure what the question wants cause wouldnt the rate of s[t] remain the same after 3 seconds? o:

  3. anonymous
    • 5 years ago
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    this is the same as the problem you posted earlier, Instead of the diagonal of the rectangle, you have the hypotenuse of the triangle in this problem

  4. anonymous
    • 5 years ago
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    ya, i know that, but the question is asking "how fast is the distance s[t] between bike and balloon increasing 3 seconds later".... so i mean, won't the s[t] have the same rate increasing after 3 secs? why would the rate change?

  5. anonymous
    • 5 years ago
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    alright, i see a mistake i made... the 41ft i used to fine the s[t] i didnt account that after 3 secs it would also increase like the x[t] is... so the y[t] after 3 secs would be 56ft, the x[t] 42ft and the s[t] = 70

  6. anonymous
    • 5 years ago
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    yes.

  7. anonymous
    • 5 years ago
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    alright, so now i calculated the s[t], x[t] and y[t] would be after 6 secs and i got, y[t] = 86ft, the x[t] = 126ft and s[t] = 153. So using these, i did: 153-70/6-3 = 27.5ft/sec and i believe this is the rate s[t] is increasing by.

  8. anonymous
    • 5 years ago
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    soooo, is this correct? o:

  9. anonymous
    • 5 years ago
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    how did you get x(t) = 126 ft for t =6sec?

  10. anonymous
    • 5 years ago
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    um, at 3 sec the x[t] is = 42ft so at 6 sec its 14ft/sec times 6s = 84 ft + 42 ft = 126 ft

  11. anonymous
    • 5 years ago
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    Hold on. Do you mean to say that If i am travelling at a constant rate of 1ft/sec, I will cover 3 feet in 3 seconds and 3+6 = 9 feet in 6 seconds?

  12. anonymous
    • 5 years ago
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    OHHHHHHHHHHHHHHHHHH, ok ok ok... my bad. i see now, k. so the x[t] after 6 seconds will be 84ft not 126ft... ok. so now, 120-70/6-3 = 16.7 ft/sec.

  13. anonymous
    • 5 years ago
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    yes

  14. anonymous
    • 5 years ago
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    no, thats wrong

  15. anonymous
    • 5 years ago
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    how did you get y(t) = 86 feet at t = 6 seconds?

  16. anonymous
    • 5 years ago
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    alright, looking at the y[t] im stuck.... now they got 41ft at the rate of 5 ft/s which means that was after 8.2 seconds... so after 6 seconds would be AFTER that 8.2 seconds....? so wouldn't i add them or wait, would it be 8.2+6 = 14.2 and then i use 14.2 x 5 =71 ft... so i'd use 71 ft then?

  17. anonymous
    • 5 years ago
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    define when you start your timer. Suppose you had a timer to time these events. Tell me when you would start your timer.

  18. anonymous
    • 5 years ago
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    x, y and s are functions of time, correct? So you have to start your time at some point. When do you start your time?

  19. anonymous
    • 5 years ago
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    do you have an answer? or do you want me to explain?

  20. anonymous
    • 5 years ago
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    at the 41ft, when the cycle and balloon are in a vertical line to each other... i'd start my timer then.

  21. anonymous
    • 5 years ago
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    okay great!

  22. anonymous
    • 5 years ago
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    lets define y(t) then. Can you give an expression for y(t)?

  23. anonymous
    • 5 years ago
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    y[t] = 5t

  24. anonymous
    • 5 years ago
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    so at t = 6 seconds, y(t) = 30 feet?

  25. anonymous
    • 5 years ago
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    are you sure your definition is correct?

  26. anonymous
    • 5 years ago
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    lets look at this another way. When you start your timer, how high is the balloon?

  27. anonymous
    • 5 years ago
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    41ft.

  28. anonymous
    • 5 years ago
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    so, at t = 0, y(t) = 41 ft

  29. anonymous
    • 5 years ago
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    how high is the balloon 1 second after you started your timer?

  30. anonymous
    • 5 years ago
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    46ft

  31. anonymous
    • 5 years ago
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    how did you get 46 ft?

  32. anonymous
    • 5 years ago
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    46 = 41ft+5ft/second*1second. correct?

  33. anonymous
    • 5 years ago
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    so what is the general expression for y(t)?

  34. anonymous
    • 5 years ago
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    y[t] = 5t+41

  35. anonymous
    • 5 years ago
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    okay great!

  36. anonymous
    • 5 years ago
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    so 6 seconds later, what is y(t)?

  37. anonymous
    • 5 years ago
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    71ft

  38. anonymous
    • 5 years ago
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    great!

  39. anonymous
    • 5 years ago
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    have you done derivatives?

  40. anonymous
    • 5 years ago
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    yes

  41. anonymous
    • 5 years ago
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    this question is part of chapter on applications of derivatives.

  42. anonymous
    • 5 years ago
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    okay.so you are asked to find ds(t)/dt at t = 3 seconds

  43. anonymous
    • 5 years ago
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    lets define s(t). What is s(t)?

  44. anonymous
    • 5 years ago
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    s[t] = square root of y[t]^2 plus x[t]^2

  45. anonymous
    • 5 years ago
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    okay, so \[ds(t)/dt = d(y(t) ^{2}+x(t)^{2})^{0.5}/dt\]

  46. anonymous
    • 5 years ago
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    right? and you know that dy/dt = 5 ft/sec and dx/dt = 14 ft/sec

  47. anonymous
    • 5 years ago
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    solve the above expression in terms of dy/dt and dx/dt. Substitute for dy/dt and dx/dt and find the value of ds/dt at t = 3.

  48. anonymous
    • 5 years ago
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    did you understand?

  49. anonymous
    • 5 years ago
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    alright, i think i get it, lemme have a shot at it and ill post my answer in a bit.

  50. anonymous
    • 5 years ago
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    note that dy/dt = d(41+5t)/dt = 5 and dx/dt = d(14t)/dt =14

  51. anonymous
    • 5 years ago
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    note also that x(t) = 14t

  52. anonymous
    • 5 years ago
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    and y(t) = 41+5t

  53. anonymous
    • 5 years ago
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    alright, finding the derivative of teh first equation is: 0.5(y[t]^2 + x[t]^2)^-0.5 times (2y[t]+2x[t]) ok, using this differentiation, and knowing: y[t]^2=3136 2y[t]=112 x[t]^2=1264 2x[t]=84 substituting it all in, i get the answer of 1.4 ft/sec.

  54. anonymous
    • 5 years ago
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    okay great, I dont know what the answer is, but your differentiation seems correct.

  55. anonymous
    • 5 years ago
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    alriiiight, thanks SO MUCH, you really were a GREAT help :] thanks ^^

  56. anonymous
    • 5 years ago
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    you are welcome. You should do the other problem the same way too. the earlier one with the rectangle i mean.

  57. anonymous
    • 5 years ago
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    good luck!

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