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Asadkarim7

  • 5 years ago

the tangent and normal to the curve y= sin x at the point P where x= π/3 cut the x-axis at A and B respectively (a). show that AB = (5√3)/4. (b). find the area of the triangle PAB.

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  1. amistre64
    • 5 years ago
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    the tangent to the curve is what..cos(pi/3) right?

  2. amistre64
    • 5 years ago
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    the normal is 90 from that....

  3. amistre64
    • 5 years ago
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    i dont quite understand the cuts the x axis at A and B

  4. anonymous
    • 5 years ago
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    maybe x and y axis at A and B?

  5. amistre64
    • 5 years ago
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  6. anonymous
    • 5 years ago
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    no

  7. anonymous
    • 5 years ago
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    they mean the tangent cuts at a, and the normal cuts at b

  8. Asadkarim7
    • 5 years ago
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    let me post the digaram

  9. anonymous
    • 5 years ago
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    its very simple find the eqns of the lines , find the intercepts of those lines , then use distance formula

  10. amistre64
    • 5 years ago
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    lol.... yeah, if you post it that would be more helpful :)

  11. amistre64
    • 5 years ago
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    simple yes, once you understand the question ;)

  12. anonymous
    • 5 years ago
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    just distance formula and area of a triangle = (1/2) bh , thats all there is , you dont even need a picture for this, should be able to draw one yourself, or visualise in your mind pretty easily

  13. Asadkarim7
    • 5 years ago
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  14. amistre64
    • 5 years ago
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    well the tangent is simply its derivative; cos(pi/3)

  15. amistre64
    • 5 years ago
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    the equation of the line then becomes: y-(sin(pi/3)) = cos(pi/3)(x-(pi/3))

  16. amistre64
    • 5 years ago
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    cos(pi/3) = 1/2 right?

  17. amistre64
    • 5 years ago
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    the normal will have a slope of -2 then

  18. amistre64
    • 5 years ago
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    y-(sin(pi/3)) = -2(x-(pi/3)) for the normal equation

  19. Asadkarim7
    • 5 years ago
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    Guys i still cant understand it , can sun1 explain it to me step by step ? plz

  20. amistre64
    • 5 years ago
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    they give you a point (pi/3, sin(pi/3)); or rather (pi/3, sqrt(3)/2) right?

  21. anonymous
    • 5 years ago
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    first take the derivative of sine, which is cosine. then plug in \[\frac{\pi}{3}\] into the derivative to get the slope. \[cos(\frac{\pi}{3})=\frac{1}{2}\] so that is the slope of the tangent line

  22. amistre64
    • 5 years ago
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    at this level you should have some redimentary skills at a graph id assume :)

  23. amistre64
    • 5 years ago
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    use the poit-slope form of a line to get the euation for the tangent and the norml

  24. anonymous
    • 5 years ago
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    amistre is in right direction,find the eqn of tangent and normal to the curve, find the x intercepts of both to find out the coordiantes of the two points, point P is given, it forms the triangle ABP

  25. amistre64
    • 5 years ago
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    someones been rearragnging my keyboard lol

  26. anonymous
    • 5 years ago
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    the equation of the line use the point slope formula. the point is \[(\frac{\pi}{3},\frac{\sqrt{3}}{2})\] and the slope is \[\frac{1}{2}\] so the equation is \[y-\frac{\sqrt{3}}{2}=\frac{1}{2}(x-\frac{\pi}{3})\]

  27. anonymous
    • 5 years ago
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    this crosses the x axis where \[y=0\] so set \[y=0\] and solve to get \[x=\frac{\pi}{3}-\sqrt{3}\] if i did the algebra right

  28. anonymous
    • 5 years ago
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    repeat the process for the line with slope -2 through \[(\frac{\pi}{3},\frac{\sqrt{3}}{2})\]

  29. amistre64
    • 5 years ago
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    x/2 - pi/6 = -sqrt.3/2 x = -sqrt3 + pi/3 its good ;)

  30. anonymous
    • 5 years ago
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    amistre did i screw this up? it looks good to me

  31. anonymous
    • 5 years ago
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    thnx

  32. amistre64
    • 5 years ago
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    the normal just uses -2 for th eslope then

  33. anonymous
    • 5 years ago
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    \[y-\frac{\sqrt{3}}{2}=-2(x-\frac{\pi}{3})\] is the normal line yes?

  34. amistre64
    • 5 years ago
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    yes

  35. anonymous
    • 5 years ago
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    let \[y=0\] get \[x=\sqrt{3}+\frac{\pi}{3}\]

  36. anonymous
    • 5 years ago
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    at least that is what i got

  37. amistre64
    • 5 years ago
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    / -2 to begin with to ease the process

  38. amistre64
    • 5 years ago
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    sqrt.3/4 + pi/3 i think

  39. anonymous
    • 5 years ago
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    \[-\frac{\sqrt{3}}{2}=-2(x-\frac{\pi}{3})\] \[\sqrt{3}=x-\frac{\pi}{3}\] \[x=\sqrt{3}+\frac{\pi}{3}\]

  40. anonymous
    • 5 years ago
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    whoooooooooooooooooops

  41. anonymous
    • 5 years ago
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    lorda mercy

  42. anonymous
    • 5 years ago
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    \[\frac{\sqrt{3}}{4}+\frac{\pi}{3}\]

  43. anonymous
    • 5 years ago
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    another dumb guy award for me.

  44. anonymous
    • 5 years ago
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    thnx

  45. amistre64
    • 5 years ago
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    lol.... thats what all my medals are for ;)

  46. anonymous
    • 5 years ago
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    ok fine. now that we have the x-intercepts, after some elementary algebra mistakes on my part, what are we supposed to do with them?

  47. amistre64
    • 5 years ago
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    throw tham at moving trains :)

  48. anonymous
    • 5 years ago
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    u both deserve for it :)

  49. anonymous
    • 5 years ago
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    i forgot what we were supposed to do with these. multiply? add? find the length?

  50. anonymous
    • 5 years ago
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    i mean for medals ;)..not throwing at the moving train :P

  51. amistre64
    • 5 years ago
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    the distance between them is just large - small

  52. anonymous
    • 5 years ago
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    now the final task, area of the triangle

  53. amistre64
    • 5 years ago
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    since they are on the number line; distance between them is large - small

  54. amistre64
    • 5 years ago
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    distance between 6 and -3 is 6--3 = 9 ;)

  55. amistre64
    • 5 years ago
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    height = sqrt3/2 ; distance = l-s --------------------------- = area 2

  56. anonymous
    • 5 years ago
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    meant to find an area lol but you guys have done it all over the place and its highly likely that the OP has just giving up on trying to follow what you guys have wrote

  57. amistre64
    • 5 years ago
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    the journey taken is more important the the final destination....... grasshopper

  58. amistre64
    • 5 years ago
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    sats been consistently and systematically finding the points... its good

  59. anonymous
    • 5 years ago
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    ooh some arithmetic i can do! \[\frac{\sqrt{3}}{4}+\frac{\pi}{3}-(\frac{\pi}{3}-\sqrt{3})=\frac{\sqrt{3}}{4}+\sqrt{3}=\frac{5\sqrt{3}}{4}\]

  60. anonymous
    • 5 years ago
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    wow just what it said it was!

  61. anonymous
    • 5 years ago
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    very east to set out working out to make the solution of these questions look really easy instead of making it look so complicated with 1000s of posts etc.

  62. anonymous
    • 5 years ago
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    true but very hard to type all in in one box. specially if typesetting.

  63. amistre64
    • 5 years ago
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    depends on if you wanna read a novel or have a conversation type environment; there isnt only one way to do things :)

  64. anonymous
    • 5 years ago
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    hi yo man

  65. amistre64
    • 5 years ago
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    this type of environment is suited best for interaction between questioneer and answeer

  66. amistre64
    • 5 years ago
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    if they really want a static read thru answer; theres always the internet lol

  67. anonymous
    • 5 years ago
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    Asadkarim7 i will be happy to work all this out on one paper and send it to you if this is confusing. the idea is this: find the slope of the tangent line, then find its equation. find the slope of the normal line, and its equation. then in each equation set y = 0 and solve for x. that is the whole idea. once you have both x intercepts subtract to get the distance. the rest is mechanics, but i will be happy to write it out in all its gory details without my algebra errors if you like

  68. amistre64
    • 5 years ago
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    asad is intelligent :)

  69. anonymous
    • 5 years ago
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    i am sure he is but i will be happy to do it to practice my latex

  70. anonymous
    • 5 years ago
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    make a pdf

  71. amistre64
    • 5 years ago
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    :) latex.... i perfer ascii lol

  72. Asadkarim7
    • 5 years ago
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    thanks :P amistre64

  73. anonymous
    • 5 years ago
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    luddite

  74. anonymous
    • 5 years ago
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    ^ "its very simple find the eqns of the lines , find the intercepts of those lines , then use distance formula" esessentially what I wrote like a whole hour ago

  75. amistre64
    • 5 years ago
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    ohhh..... so everything after that is useless and should be erased becasue its just utter garbage? .... comes across a bit pompuos dontcha think ;)

  76. amistre64
    • 5 years ago
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    in fact the "i still dont understand can you explain it to me step by step" kinda leads me to believe otherwise ... lol

  77. anonymous
    • 5 years ago
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    also, geez, way to spam the question guys, nw my computer is going extremely slow when I view this particular question because you guys spammed it so much :|

  78. amistre64
    • 5 years ago
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    youre prolly using internet explorer; i had to down firefox to avoid that

  79. amistre64
    • 5 years ago
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    the equation editor stuff tends to put a strain on the processor; which is why i tend to avoid it

  80. amistre64
    • 5 years ago
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    so asad; how can we help you with this problem?

  81. anonymous
    • 5 years ago
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    here it is

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  82. anonymous
    • 5 years ago
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    let me know if you have any questions

  83. amistre64
    • 5 years ago
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    you forgot to put in the area lol

  84. anonymous
    • 5 years ago
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    oh heavens. fine.

  85. amistre64
    • 5 years ago
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    and here my fine useless garbage contribution ;)

    1 Attachment
  86. Asadkarim7
    • 5 years ago
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    sattelite73 u made the equation y = sqrt 3/ 2 it should have been y- sqrt 3/2

  87. anonymous
    • 5 years ago
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    \[\frac{1}{2}bh=\frac{1}{2}\times \frac{5\sqrt{3}}{4}\times \frac{\sqrt{3}}{2}=\frac{15}{16}\]

  88. amistre64
    • 5 years ago
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    its his latex.... makes him go a little insane at times and starts putting = signs for - signs :)

  89. anonymous
    • 5 years ago
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    everyone is a critic. ok that was a typo and look, it is gone!

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  90. amistre64
    • 5 years ago
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    that ones just blank.....

  91. Asadkarim7
    • 5 years ago
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    u r genius sattelite 73

  92. anonymous
    • 5 years ago
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    what lovely font! what nice equations!

  93. anonymous
    • 5 years ago
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    how didi u make pdf file

  94. anonymous
    • 5 years ago
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    eventually i will learn this latex ... eventually.

  95. amistre64
    • 5 years ago
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    he starts out with some fine kindling, then adds a bit of tinder to feed the flame..

  96. anonymous
    • 5 years ago
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    i am using texmaker as an editor. just converts for you if you ask, and then can ask for external viewer. really quite friendly. i tried to use this on micordemon windows whatever and it took me maybe a day and a half to download what i needed. miktex and all that garbage. never worked right. left the darkside for linux a few months ago, to maybe 2 minutes do download and works like a charm.

  97. anonymous
    • 5 years ago
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    it is true i start with the trees. used to roll my own when i smoked. also doing html by hand if i can (with editor of course)

  98. amistre64
    • 5 years ago
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    <html> <body> <input type=button value="Click me!!" onClick="qwe()"> </body> <script language=javascript> function qwe() {alert("Quit it!")} </script> </html>

  99. anonymous
    • 5 years ago
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    also spelled "point" and 'derivative' wrong. perils of not using wysiwyg

  100. anonymous
    • 5 years ago
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    armistre will this make my computer explode?

  101. amistre64
    • 5 years ago
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    1 Attachment
  102. amistre64
    • 5 years ago
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    hopefully lol

  103. anonymous
    • 5 years ago
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    this will keep me busy for at least an hour. click, click, ...

  104. amistre64
    • 5 years ago
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    lol :)

  105. Asadkarim7
    • 5 years ago
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    Thanks Everyone for your contributions , and giving your precious time to my question <3

  106. amistre64
    • 5 years ago
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    i made a 'flip book' javascript movie maker for my neices

  107. amistre64
    • 5 years ago
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    youre welcome :) and thank you elec for making it all possible lol

  108. anonymous
    • 5 years ago
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    welcome. hope it is clear enough even with all this garbage thrown in.

  109. Asadkarim7
    • 5 years ago
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    yeahh satellite its crystal clear . :)

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