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anonymous

  • 5 years ago

curvilinear motion...

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  1. anonymous
    • 5 years ago
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    parametric equations: y(t)=-4.9t^2 +4.864t+1.2192 and x(t)=5.08t... i found the equations for velocity individually etc etc. i got s(t)=-1.9291t+.9574 for the speed of the particle at any time, t. and the next question is asking to find dy/dx?

  2. anonymous
    • 5 years ago
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    You can find dy/dx using the chain rule. That is,\[\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}\]All you need to do is take the derivatives of your equations with respect to time and multiply them appropriately.

  3. anonymous
    • 5 years ago
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    that is what i did to get the speed function s(t)... if i found dy/dx as that, why is it asking for it again, or is it in reference to another derivative of something else?

  4. anonymous
    • 5 years ago
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    I'm thinking the speed is given by,\[s(t)=\sqrt{\left( \frac{dy}{dt} \right)^2+\left( \frac{dx}{dt} \right)^2}\]Is that how you found it?

  5. anonymous
    • 5 years ago
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    since you're looking for the magnitude of velocity at that point.

  6. anonymous
    • 5 years ago
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    nope but that does make sense, now that i reread the question, it just wants that formula, which then means that the next question wants me to take the derivative of that?

  7. anonymous
    • 5 years ago
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    Do what I wrote for the speed, and do what I wrote for dy/dx...you should get two different answers. Does that help?

  8. anonymous
    • 5 years ago
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    what does that s(t) formula tell me in terms of the problem and what does the dy/dx tell me?

  9. anonymous
    • 5 years ago
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    im sorry but im trying to understand the formulas

  10. anonymous
    • 5 years ago
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    s(t) is the speed - the magnitude of the velocity. dy/dx just gives you the rate of change of the y co-ordinate as x varies - so it's just looking at how vertical displacement varies with horizontal...nothing special.

  11. anonymous
    • 5 years ago
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    so finding dy/dx pretty much has nothing to do with the speed function and my teacher just put it in there randomly...?

  12. anonymous
    • 5 years ago
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    Sometimes it's useful to have motion problems in terms of displacements only, excluding time. It's easier for understanding the geometrical properties of the path taken.

  13. anonymous
    • 5 years ago
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    Well, speed is defined as the *time* rate of change of distance. dy/dx makes no mention of time. I'm not sure what your teacher intends.

  14. anonymous
    • 5 years ago
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    which one excludes time?

  15. anonymous
    • 5 years ago
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    ah, but if i divide dy/dt by dx/dt i still have dy/dx in terms of t

  16. anonymous
    • 5 years ago
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    Well, sometimes we solve one coordinate for time and sub. it into the other coordinate to combine the coordinates into a relationship between each other, that's all.

  17. anonymous
    • 5 years ago
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    Yes, you may have it in terms of t, and that's fine. That would just allow you to answer the question, "What is the path rate of change in y versus x at time t?"

  18. anonymous
    • 5 years ago
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    Or you could eliminate t using one of the equations and sub. it in and you'd have dy/dx in terms of x and y only...you can do anything! Okay, not *anything*.

  19. anonymous
    • 5 years ago
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    ok, how about the trajectory of the particle at time 0 and time 1.2? i dont think i even heard about what that means

  20. anonymous
    • 5 years ago
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    The trajectory is the path a moving object follows through space as a function of time. You have your displacement vector in terms of t, namely, \[(x(t),y(t))\]All you'd do is sub. in the times to get a displacement vector for the object at that time.

  21. anonymous
    • 5 years ago
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    The question is just asking, "Where is the object headed at time 0 and at time 1.2s?"

  22. anonymous
    • 5 years ago
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    its asking the angle that it left my hand (time 0) and the angle that it hit the ground (time 1.2)

  23. anonymous
    • 5 years ago
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    Actually, I should have said, "Where is the object?", not 'headed', since that would be velocity.

  24. anonymous
    • 5 years ago
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    Oh - so that's why you needed dy/dx in terms of t...remember what dy/dx means geometrically? It's the gradient of the function. dy/dx is telling you where the particle is headed.

  25. anonymous
    • 5 years ago
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    \[\frac{dy}{dx}(t)=\tan \theta\]

  26. anonymous
    • 5 years ago
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    gradient?

  27. anonymous
    • 5 years ago
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    I'm assuming from what you've said you're throwing something in this question. Draw a right-angled triangle where the hypotenuse is sloping up. Let theta be the angle...wait, I'll draw something.

  28. anonymous
    • 5 years ago
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    Look at this...

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  29. anonymous
    • 5 years ago
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    so basically its the tennisball project in two directions. up and out. i threw the ball, from 1.2192m and it took 1.2s to land 6.096m away. i got all the equations above then it asked for the formula for speed of the particle at any time. and right after that said "step 6: find dy/dx."

  30. anonymous
    • 5 years ago
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    tan (theta) = (dy/dt)/(dx/dt) = dy/dx

  31. anonymous
    • 5 years ago
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    that makes sense... what do it do with that?

  32. anonymous
    • 5 years ago
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    Well, from that setup (picture), dy/dt is always vertical and dx/dt is always horizontal. Their sizes will change depending on time, and therefore, so will dy/dx. All that picture is doing is to show the relationship between the angle with the horizontal and the velocity vector. You're looking for the angle at a certain time. You have dy/dx as a function of t, and that picture shows you how dy/dx relates to that angle (i.e. tan(theta) = dy/dx). So, find dy/dx at your times and take the inverse tangent to find the angles. Check to see if the answers make sense.

  33. anonymous
    • 5 years ago
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    that makes complete sense, lemmy try that

  34. anonymous
    • 5 years ago
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    and so just making sure...: dy/dx (t) gives me both dy vertical height and dx horizontal distance as a function of time?

  35. anonymous
    • 5 years ago
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    at the same time?

  36. anonymous
    • 5 years ago
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    Nearly. What it's doing is giving you the *rate of change* of y with respect to x (as a function of time).

  37. anonymous
    • 5 years ago
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    So you're getting the rate of change in y wrt x given the time...you don't have to know anything about x (or y) explicitly (like you do when you're asked more elementary questions like, y=x^2 -> dy/dx = 2x. To fing dy/dx here, we need to know x).

  38. anonymous
    • 5 years ago
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    right, being the derivative, but yeah it does both in relevance to time... jeez im clearly too tired if im being this stupid

  39. anonymous
    • 5 years ago
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    so i got that it left at approx 137 degrees and landed at 83 degrees which looks right

  40. anonymous
    • 5 years ago
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    yes, you're right (not about being stupid!).

  41. anonymous
    • 5 years ago
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    nah man im definitely being stupid right now... idk where you are but here on the east coast its 4am and i got school first thing tomorrow

  42. anonymous
    • 5 years ago
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    alright so to get the greatest speed ill use the dy/dx formula?

  43. anonymous
    • 5 years ago
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    83 degrees...might be (180-83) degrees if you think about the fact the coordinate system and how the ball lands. I'll draw something.

  44. anonymous
    • 5 years ago
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    Well, to get the greatest speed, I would use s(t) and take the derivative wrt to t to find the time at which the speed was greatest. Then you can use that time to find position.

  45. anonymous
    • 5 years ago
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    yeah it would be about -263 or +97 degrees technically but same thing in this perspective... and i can understand why dy/dx is a speed formula but whats the whole root of dy/dt sqrd + dx/dt sqrd) ?

  46. anonymous
    • 5 years ago
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    what is wrt?

  47. anonymous
    • 5 years ago
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    wrt = with respect to

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  48. anonymous
    • 5 years ago
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    whats the difference between using s(t) and dy/dx, would they get me the same speed?... would dy/dx get me "speed" ?

  49. anonymous
    • 5 years ago
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    The whole 'root' thing is because of the following. The velocity vector is\[v=(x'(t),y'(t))\]and since the speed is the magnitude of the velocity vector, you have\[v^2=v.v=(x'(t),y'(t)).(x'(t),y'(t))=x'(t)^2+y'(t)^2\]

  50. anonymous
    • 5 years ago
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    But this is the square of speed, so we have to take the square root of both sides.

  51. anonymous
    • 5 years ago
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    \[v=\sqrt{x'(t)^2+y'(t)^2}\]

  52. anonymous
    • 5 years ago
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    Re. using s(t) and dy/dx...speed is the time rate of change, which is what s(t) is. dy/dx is *not* a time rate of change, it's a displacement rate of change; it measures the change in the y-coordinate as a function of the change in x-coordinate. It's different. It gives you information about the geometry of what's going on - as in the case where you have to find angles for where things fall.

  53. anonymous
    • 5 years ago
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    4am...are you getting up to study, or have you not gone to bed?

  54. anonymous
    • 5 years ago
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    not gone to bed

  55. anonymous
    • 5 years ago
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    Have you got a test or something?

  56. anonymous
    • 5 years ago
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    technically i dont actually take calculus

  57. anonymous
    • 5 years ago
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    Is this just for physics or something?

  58. anonymous
    • 5 years ago
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    cant take physics either, yet... im a freshman in hs

  59. anonymous
    • 5 years ago
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    So what's this for?

  60. anonymous
    • 5 years ago
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    Torture?

  61. anonymous
    • 5 years ago
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    torture to who?

  62. anonymous
    • 5 years ago
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    You.

  63. anonymous
    • 5 years ago
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    I wasn't being rude :)

  64. anonymous
    • 5 years ago
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    haha no, this is my fun

  65. anonymous
    • 5 years ago
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    i got half an hour and about 7 speed questions that i know are wrong now

  66. anonymous
    • 5 years ago
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    so i really have no option but taking the root of that big equation? :(

  67. anonymous
    • 5 years ago
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    If you want the speed, no.

  68. anonymous
    • 5 years ago
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    first thing i want is speed at time 0

  69. anonymous
    • 5 years ago
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    dx/dt will be 0 and dy/dt will be 4.864... so i sqare that then root it? which is still 4.864...

  70. anonymous
    • 5 years ago
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    I would just take the derivative of x(t) and y(t) and sub. the times in to find numerical values, square each, add, take square root of sum (if your aim is computation only).

  71. anonymous
    • 5 years ago
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    what would be another aim? than computation?

  72. anonymous
    • 5 years ago
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    Well, if you're asked to find the actual formula...if you're not asked to, and only asked to find values (and you're in an exam), it would take up unnecessary time.

  73. anonymous
    • 5 years ago
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    \[y(t)=-4.9t^2+4.864t+1.2192 \rightarrow y(0)=1.2192\]\[x(t)=5.08t \rightarrow x(0)=0\]So,\[s(0)=\sqrt{1.2192^2+0^2}=1.2192\]

  74. anonymous
    • 5 years ago
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    dy/dt or y(t) ??

  75. anonymous
    • 5 years ago
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    oh

  76. anonymous
    • 5 years ago
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    yeah - I'm tired too

  77. anonymous
    • 5 years ago
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    lol you on the eat cost too?

  78. anonymous
    • 5 years ago
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    No, other side of the world. Sydney. Off sick.

  79. anonymous
    • 5 years ago
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    are you a physicist or something or just enjoy math?

  80. anonymous
    • 5 years ago
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    I'm trained in pure maths and physics. I'm a UK citizen, worked with people in physics from Oxford, Cambridge, Yale.

  81. anonymous
    • 5 years ago
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    wait a sec... it asked for the speed as the ball approached the max height.. if the max height is when y' is 0 then ill end up with dy/dt is 0 and dx/dt is 5.08?

  82. anonymous
    • 5 years ago
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    that is freaking awesome.

  83. anonymous
    • 5 years ago
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    :)

  84. anonymous
    • 5 years ago
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    that makes me wonder, how many teens have you worked with that your first answer is "torture" to math :P

  85. anonymous
    • 5 years ago
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    and i need help with this one: on what intervals is the speed increasing and decreasing? (i dont want to use an analysis -.- )

  86. anonymous
    • 5 years ago
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    lol, I've worked with people of all ages and abilities. All I care about is the attitude.

  87. anonymous
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    You don't want to use an analysis? What do you mean?

  88. anonymous
    • 5 years ago
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    logic says that its decreasing till the max height and increasing therein after

  89. anonymous
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    i dont want to use any derivative analysiss

  90. anonymous
    • 5 years ago
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    oh

  91. anonymous
    • 5 years ago
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    that makes no sense, my calculations say that it had a speed of approx, 3 when i let go... 8 when it hit the ground, and 5.08 at the max height

  92. anonymous
    • 5 years ago
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    One sec...

  93. anonymous
    • 5 years ago
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    oh i see my mistake... omg this is pathetically horrible ><

  94. anonymous
    • 5 years ago
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    i got 7 this time :DDD

  95. anonymous
    • 5 years ago
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    the speed should be fastest when it hits the ground, right?

  96. anonymous
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    Well, you could find the maximum height by understanding that the vertical component of velocity must be zero there...so \[\frac{dy}{dt}=0\]should give you the time, which you can use to find the height, y. I know you said no calculus, but I figured you already had y'(t).

  97. anonymous
    • 5 years ago
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    Sorry, I think I've answered a question you didn't ask.

  98. anonymous
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    Yes, the speed should be fastest when it hits the ground. You get that from energy considerations.

  99. anonymous
    • 5 years ago
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    yeah i already got everything else

  100. anonymous
    • 5 years ago
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    niceee finally i can glue this crap together lol

  101. anonymous
    • 5 years ago
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    Good! I can't believe you're up at ~4am and doing this...lol.

  102. anonymous
    • 5 years ago
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    Do you want to study math/science after school?

  103. anonymous
    • 5 years ago
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    ?

  104. anonymous
    • 5 years ago
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    like after i graduate, keep studying?

  105. anonymous
    • 5 years ago
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    Yeah

  106. anonymous
    • 5 years ago
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    i mean, i intend to get a Ph.D in both mathematics and probably EE... ill be doing this till the day i die

  107. anonymous
    • 5 years ago
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    XD

  108. anonymous
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    good

  109. anonymous
    • 5 years ago
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    ( when you can answer) why did you get into math?

  110. anonymous
    • 5 years ago
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    School...I accelerated my studies. After school, I had offers for medicine and law, but stuck to mathematics and physics...to my father's grief. I've worked in and biophysics with a chemist from Ox. and nuclear physicist (turned biophysicist) from Yale. Now, I'm taking a break.

  111. anonymous
    • 5 years ago
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    a break?

  112. anonymous
    • 5 years ago
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    Sabbatical.

  113. anonymous
    • 5 years ago
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    ?

  114. anonymous
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    Just a formal break from university stuff.

  115. anonymous
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    are you a professor or a researcher?

  116. anonymous
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    I was a researcher. I'm just teaching for a while without the research.

  117. anonymous
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    i can tell that you come on here often haha

  118. anonymous
    • 5 years ago
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    Lol, I know...it's a nice distraction from other things I should be doing.

  119. anonymous
    • 5 years ago
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    i could probably use a little more help on a related rates problem DX

  120. anonymous
    • 5 years ago
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    i'll see what i can do

  121. anonymous
    • 5 years ago
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    whether baloon is launched on a windless day, goes straight up... rises at a rate of 1000ft/minutes, observer stands 10ft away, his eyes are 6 ft off the ground at what rate is the angle of elevation changing when the baloon is 10000 ft high

  122. anonymous
    • 5 years ago
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    i can do related rates, i just cant figure out how to set this one up ...

  123. anonymous
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    well, you won't have, by definition, an angle of elevation until the balloon hits 6 ft. So I would consider finding the vertical velocity and other 'initial' conditions at the y=6ft mark and work from there...

  124. anonymous
    • 5 years ago
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    ...whaa??

  125. anonymous
    • 5 years ago
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    one sec.

  126. anonymous
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    no i get what youre saying... nvrmnd that "whaa?" but we are taking the calculations mostly at 10000ft where that 6ft will have been very much covered

  127. anonymous
    • 5 years ago
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    and yeah i tried that, but i still cant set up the equation right

  128. anonymous
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    with the +6 or -6 or even without it, its the trig that i cant get down ><

  129. anonymous
    • 5 years ago
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    Okay, I'm going to draw something.

  130. anonymous
    • 5 years ago
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    actually, given that it is 6 o clock now.. its probably best if you dont... i need to get my 15 minutes of sleep XDDD

  131. anonymous
    • 5 years ago
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    oh, i'll continue and you can go...

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  132. anonymous
    • 5 years ago
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    ill prolly be back tonight, i always liked helping people with their math

  133. anonymous
    • 5 years ago
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    iight im out- thank you for everything btw

  134. anonymous
    • 5 years ago
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    no probs.

  135. anonymous
    • 5 years ago
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    Had a break...so, you need to find \[\frac{d \theta }{dt}\] as some function of height. From the pic. I sent, you should see that\[\tan \theta = \frac{h}{10} \rightarrow \frac{d}{dt}\tan \theta = \frac{d}{dt}\frac{h}{10} \rightarrow \sec^2 \theta \frac{d \theta}{dt}=\frac{1}{10}\frac{dh}{dt}\]That is,\[\frac{d \theta }{dt}=\frac{dh/dt}{10\sec^2 \theta}=\frac{dh/dt}{10(1+\tan^2 \theta)}=\frac{dh/dt}{10(1+h^2/100)}\]\[\frac{d \theta}{dt}=\frac{10(dh/dt)}{100+ h^2}\]You're given\[\frac{dh}{dt}=1000ft/\min\]and your height at this point is (10,000 - 6) ft since our calculations for angle of elevation begin at 6ft. Then\[\frac{d \theta }{dt }|_{h=9994}=\frac{10 \times 1000}{100+(10000-6)^2}\approx 1 \times 10^{-4} radians/\min\]

  136. anonymous
    • 5 years ago
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    From your question, I assume any acceleration on the balloon in the vertical is ignored.

  137. anonymous
    • 5 years ago
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    thats precisely what i got! and yes, there is no acceleration in this case

  138. anonymous
    • 5 years ago
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    [thumbs up]

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