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1018

  • one year ago

please help. numbers 24 & 25 (picture)

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  1. 1018
    • one year ago
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  2. madhu.mukherjee.946
    • one year ago
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    @ganeshie8

  3. 1018
    • one year ago
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    oh and i tried to answer #23 but my answer didnt show up among the choices. if anyone can answer that would be helpful too. thanks!

  4. welshfella
    • one year ago
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    |dw:1441468653415:dw|

  5. welshfella
    • one year ago
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    |dw:1441468744582:dw|

  6. welshfella
    • one year ago
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    - that's Q 23

  7. beginnersmind
    • one year ago
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    Shouldn't the integrand be \(2\pi\ x^2 dx\) ?

  8. welshfella
    • one year ago
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    no the area of a 'slice' is pi r^2

  9. welshfella
    • one year ago
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    and r = x^2

  10. beginnersmind
    • one year ago
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    Sorry meant \(2\pi\ (x^2)^2 dx \)

  11. beginnersmind
    • one year ago
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    Oh, right

  12. beginnersmind
    • one year ago
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    Area of a circle, lol

  13. welshfella
    • one year ago
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    - i dont think so - why the 2 ?

  14. welshfella
    • one year ago
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    circle yes

  15. welshfella
    • one year ago
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    no that is not the curve the curve is y - x^2

  16. sohailiftikhar
    • one year ago
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    as given y=x^2

  17. welshfella
    • one year ago
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    |dw:1441469338036:dw|

  18. welshfella
    • one year ago
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    we only dealing with x = 0 to x = 1

  19. welshfella
    • one year ago
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    |dw:1441469432075:dw|

  20. welshfella
    • one year ago
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    it is the sum of all the discs with height dx and radius x^2

  21. beginnersmind
    • one year ago
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    Anyway, for 24 and 25 use the definition of the total differential: for z = f(x,y) \[dz = \frac{\partial f }{\partial x} dx + \frac{\partial f }{\partial y} dy\]

  22. welshfella
    • one year ago
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    the cross section is a circle

  23. sohailiftikhar
    • one year ago
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    yes

  24. welshfella
    • one year ago
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    yes

  25. welshfella
    • one year ago
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    partial differentials

  26. 1018
    • one year ago
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    hey guys! sorry i lost my connection

  27. 1018
    • one year ago
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    what are your answers?

  28. beginnersmind
    • one year ago
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    @welshfella got the answer for 23 in his first two posts. You can ignore the discussion under that. For 24 and 25 use \[dz = \frac{\partial f }{\partial x} dx + \frac{\partial f }{\partial y} dy\]

  29. 1018
    • one year ago
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    oh yes! squared. i forgot x^2 still needs ^2. thanks @welshfella thanks @beginnersmind

  30. 1018
    • one year ago
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    ok ill try and see if i can get the answer

  31. 1018
    • one year ago
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    @beginnersmind hey can you show me your work, i think im close to it but i dont fully understand. thanks

  32. 1018
    • one year ago
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    i mean when im deriving already. i think im missing something

  33. beginnersmind
    • one year ago
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    You mean show why my formula is true? Or apply it?

  34. 1018
    • one year ago
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    apply

  35. beginnersmind
    • one year ago
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    ok, for 24 |dw:1441470707423:dw|

  36. beginnersmind
    • one year ago
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    Remember to take the partial derivative with respect to x you just treat y as a constant.

  37. 1018
    • one year ago
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    oh i should still write the dx and dy in my answer?

  38. 1018
    • one year ago
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    it's still part of it?

  39. beginnersmind
    • one year ago
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    Sure, they are asking for dz. The answer will be in a form: (something) * dx + (something else) * dy

  40. 1018
    • one year ago
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    ok thanks! Ill try the other one thanks again!

  41. beginnersmind
    • one year ago
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    No problem. Let me know if you want me to check your solution.

  42. 1018
    • one year ago
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    oh no wait can you help me out on 25, hahaha. that is a hard one

  43. beginnersmind
    • one year ago
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    Oh, 25 is different because x and y are a function of t.

  44. beginnersmind
    • one year ago
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    Ok, it's actually simpler than 24

  45. beginnersmind
    • one year ago
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    They are only asking for dx/dt. You can ignore everything except the definition of x(t).

  46. beginnersmind
    • one year ago
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    It's weird though, I'm getting \( \pi/2\), for 25. Paging @welshfella

  47. beginnersmind
    • one year ago
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    -pi/2 that is. But it's not one of the options.

  48. 1018
    • one year ago
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    im still solving haha

  49. 1018
    • one year ago
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    do i just ignore the e? i will just derive the exponent then sub to the orig value?

  50. beginnersmind
    • one year ago
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    IDK, I checked my work and I'm still getting -pi/2 Here it is in full if someone wants to go over it. x(t) = t*cost dx/dt = cost - t*sint evaluate at t=pi/2 dx/dt = 0 - (pi/2)*1 = -pi/2.

  51. beginnersmind
    • one year ago
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    @ganeshie8 can you check please?

  52. 1018
    • one year ago
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    that's my sol'n for x. haha at least i'm getting somewhere

  53. beginnersmind
    • one year ago
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    The only thing I can think of is that they actually meant to ask for dz/dt and dx/dt is a typo.

  54. beginnersmind
    • one year ago
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    I tried calculating dz/dt for t = pi/2 and got -[(pi)^3]/8. I think I'll just have to give up here. Something looks fishy, especially giving z(x(t),y(t)) but then only asking for dx/dt.

  55. 1018
    • one year ago
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    maybe there really is an error, hey thanks anyway. big help!

  56. beginnersmind
    • one year ago
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    No problem. Just for completeness's sake this is the formula for dz/dt|dw:1441473257074:dw|

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