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anonymous

  • 4 years ago

y=10cosx at the point (π/3,5), what is the derivative

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  1. dumbcow
    • 4 years ago
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    derivative of cos is -sin --> dy/dx = -10sin x

  2. anonymous
    • 4 years ago
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    how will i find these two also The slope of the tangent line is m1= . The equation of the tangent line is y= .

  3. dumbcow
    • 4 years ago
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    slope = dy/dx evaluated at some x

  4. anonymous
    • 4 years ago
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    im not familiar with that method. would you mind demonstrating it please?

  5. dumbcow
    • 4 years ago
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    what method? its just the rule, derivative of sin is cos derivative of cos is -sin

  6. anonymous
    • 4 years ago
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    cos/-sin is the solution?

  7. dumbcow
    • 4 years ago
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    ?? your answer will have 3 parts a) derivative, dy/dx b) slope of tangent line, take dy/dx plug in x=pi/3 c) equation of tangent line, use slope and find y_intercept

  8. anonymous
    • 4 years ago
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    i dont understand how to do any of those. im working ahead on my homework b/c im going to busy with other things. so my teacher hasnt cover those yet and i am clueless. can you help me out please?

  9. dumbcow
    • 4 years ago
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    i already gave you the derivative, now just plug in pi/3 for x to get slope

  10. anonymous
    • 4 years ago
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    -10sin(pi/3) ok

  11. dumbcow
    • 4 years ago
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    yep...now use unit circle or calculator to find out what sin(pi/3) is

  12. anonymous
    • 4 years ago
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    60degrees (1/2,sqrt3/2

  13. anonymous
    • 4 years ago
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    what do i do with those numbers

  14. dumbcow
    • 4 years ago
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    which one refers to sin

  15. anonymous
    • 4 years ago
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    sqrt3/2

  16. dumbcow
    • 4 years ago
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    cos() --> similar to x coord sin() --> similar to y coord good so slope is -5sqrt(3)

  17. anonymous
    • 4 years ago
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    ok

  18. anonymous
    • 4 years ago
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    the line tangent?

  19. dumbcow
    • 4 years ago
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    y = mx+b m = -5sqrt3 (x,y) = (pi/3,5) find b

  20. anonymous
    • 4 years ago
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    wait, i didnt mention that my homework software was satisfied with -10sinsqrt(3) as the slope

  21. dumbcow
    • 4 years ago
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    you mean...-10sin(pi/3)

  22. anonymous
    • 4 years ago
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    haha yeah

  23. anonymous
    • 4 years ago
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    my homework software accepted that

  24. dumbcow
    • 4 years ago
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    ok, same thing...you can interchange sqrt3/2 and sin(pi/3) because they are equal y = mx+b b = y - mx b = 5 -(-10sin(pi/3)*pi/3)

  25. anonymous
    • 4 years ago
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    ok thanks now i have to do slope and tangent line for m2

  26. anonymous
    • 4 years ago
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    wait

  27. dumbcow
    • 4 years ago
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    m2?? haha, same function different point

  28. anonymous
    • 4 years ago
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    my software didnt accept the previous answer

  29. dumbcow
    • 4 years ago
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    what did you put in? did you enter it as y=mx+b form

  30. anonymous
    • 4 years ago
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    i entered it in this form 5 -(-10sin(pi/3)*pi/3)

  31. dumbcow
    • 4 years ago
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    well what does that number represent ?

  32. dumbcow
    • 4 years ago
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    i was trying to help you, i didn't give you the final answer

  33. anonymous
    • 4 years ago
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    hahaha

  34. anonymous
    • 4 years ago
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    5 -(-10sin(pi/3)*pi/3) i got 5+10sin(pi/3)pi/3

  35. anonymous
    • 4 years ago
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    when i simplify

  36. dumbcow
    • 4 years ago
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    good

  37. dumbcow
    • 4 years ago
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    not done yet...what did you just solve for? what answer are they looking for..."equation of tangent line"

  38. anonymous
    • 4 years ago
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    the line that runs diagonally and only touches a parabola on its sides

  39. dumbcow
    • 4 years ago
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    yes thats definition of tangent line. equation of tangent line is : y = mx+b so answer must be in that form: slope*x + y_intercept

  40. anonymous
    • 4 years ago
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    our slope is 10sin(pi/3) and y-int is 5 right?

  41. dumbcow
    • 4 years ago
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    no

  42. dumbcow
    • 4 years ago
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    y = mx+b b = y - mx m = -10sin(pi/3) (x,y) = (pi/3,5) b = 5 -(-10sin(pi/3)*pi/3)

  43. anonymous
    • 4 years ago
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    i dont see how it isnt. you have m=-10sin(pi/3) and y=5

  44. dumbcow
    • 4 years ago
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    the y_coord of the point is 5. don't confuse that with y_intercept, where line crosses y-axis we represent y_intercept as b

  45. anonymous
    • 4 years ago
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    ohhhhh ok

  46. anonymous
    • 4 years ago
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    im not sure how you solve for b here

  47. dumbcow
    • 4 years ago
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    b = y - mx b = 5 -(-10sin(pi/3)*pi/3)

  48. anonymous
    • 4 years ago
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    omg

  49. anonymous
    • 4 years ago
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    i didnt see the y-mx

  50. anonymous
    • 4 years ago
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    i got 45

  51. dumbcow
    • 4 years ago
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    really? after multiplying by pi....or did you round it to 45

  52. dumbcow
    • 4 years ago
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    i get 14.069

  53. anonymous
    • 4 years ago
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    my calc was in radians mode

  54. dumbcow
    • 4 years ago
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    yeah thats what you want you could probably leave it as: b= 5+10(pi/3)*sin(pi/3),

  55. anonymous
    • 4 years ago
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    ahh i didnt distribute the negative outside the parantheisis

  56. anonymous
    • 4 years ago
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    whats next

  57. dumbcow
    • 4 years ago
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    enter the equation of tangent line y = ?

  58. dumbcow
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=y+%3D+10cos%28x%29+find+tangent+line+at+x%3Dpi%2F3 here is a nice reference to check your answers

  59. anonymous
    • 4 years ago
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    ok thanks m2, do find the derivative of the derivative of m1

  60. dumbcow
    • 4 years ago
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    i dunno what they mean by m2

  61. dumbcow
    • 4 years ago
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    is their another function, or another point ?? does it look like dy^2/dx

  62. anonymous
    • 4 years ago
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    The slope of the normal line is m2 The equation of the normal line is y

  63. dumbcow
    • 4 years ago
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    oh ok, so m1 refers to tangent line m2 refers to normal line which is perpendicular to tangent line at point (pi/3, 5)

  64. dumbcow
    • 4 years ago
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    this will be easier for perpendicular lines: m1*m2 = -1 m2 = -1/m1

  65. dumbcow
    • 4 years ago
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    Then find equation of normal line same way solve for y_intercept b = y - m2*x

  66. dumbcow
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=y+%3D+10cos%28x%29+find+normal+line+at+x%3Dpi%2F3

  67. anonymous
    • 4 years ago
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    what is that link for?

  68. anonymous
    • 4 years ago
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    thanks for the help

  69. dumbcow
    • 4 years ago
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    your welcome link is to check your answer

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