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anonymous

  • 5 years ago

in calculus, what does the d/dx and d/dt mean? what's the difference? i am starting integration and am confused as to what the problem is asking for

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  1. amistre64
    • 5 years ago
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    they voth mean basically the same theing, it is a notation that reads: the derivative with respect to "variable".

  2. nowhereman
    • 5 years ago
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    You could also say derivation in direction of some dimension.

  3. amistre64
    • 5 years ago
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    dy/dt tends to mean: the derivative of y with respect to t (for time)

  4. anonymous
    • 5 years ago
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    so what if it's dx/dt = 20/sqrt(t)? then it asks to find the antiderivative given x(1) = 40

  5. amistre64
    • 5 years ago
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    all it means is that the amount of change in the top part depends on the amount of change that occurs to the variable inthe bottom part

  6. amistre64
    • 5 years ago
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    initial conditions helpt to anchor the family of curves that is produced to a single curve.

  7. anonymous
    • 5 years ago
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    dy/d(something) just means the rate of change of 'y'(here) with respect to 'something'

  8. amistre64
    • 5 years ago
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    x' = 20/sqrt(t) means that there is a function X(t) that this came from.....and they want you to know where it came from...

  9. nowhereman
    • 5 years ago
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    Yes but as integration is the inverse operation of differentiation he is right, that in order to find x(t) he has to integrate dx/dt which in that example is 20/sqrt(t)

  10. amistre64
    • 5 years ago
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    take out the constant, and change the radical to an appropriate exponent

  11. anonymous
    • 5 years ago
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    so is dx/dt the derivative of a function called x? which in this case would be 20/sqrt(t)?

  12. nowhereman
    • 5 years ago
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    You should also know, that 1/sqrt(t) is the the derivative of 2\sqrt(t)

  13. amistre64
    • 5 years ago
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    20 times the integral of 1/t^(1/2)

  14. amistre64
    • 5 years ago
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    20 times the integral of t^(-1/2)

  15. amistre64
    • 5 years ago
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    20 (1/2)(sqrt(t) = 10sqrt(t)

  16. anonymous
    • 5 years ago
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    my book says the answer is 40*sqrt(t)

  17. amistre64
    • 5 years ago
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    poke: yes, dx/dt is the derivative of a funtion X(t) find X(t) and input your initial condition that x(1) = 40

  18. amistre64
    • 5 years ago
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    10sqrt(1) + C = 40

  19. amistre64
    • 5 years ago
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    if we derive 10sqrt(t) we get -> 10/2 sqrt(t)..right? so yeah, my mistake :)

  20. amistre64
    • 5 years ago
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    ahhh.... i see my mistake :) i did it in me head for starters. the reciprocal of 1/2 = 2..... 20(2) sqrt(t) = 40sqrt(t)

  21. anonymous
    • 5 years ago
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    correct

  22. amistre64
    • 5 years ago
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    now this X(t) function is floating around....it needs an anchor...it needs a +C to hold it in place.

  23. amistre64
    • 5 years ago
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    40sqrt(t) + C is what you want.... plug in your initial condition that X(1) = 40 and solve for "C" then you have the right answer

  24. anonymous
    • 5 years ago
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    so when you pull the 20 out originally you have 20*(1/sqrt(t))... then the antideriv of that is 20*(2/sqrt(t))?

  25. amistre64
    • 5 years ago
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    poke: yes..

  26. anonymous
    • 5 years ago
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    so why isnt the final answer 40/sqrt(t)??

  27. amistre64
    • 5 years ago
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    you add one to the exponent, then put that number as the denominator and also use that number as the new exponent. -1/2 + 2/2 = 1/2

  28. amistre64
    • 5 years ago
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    40t^(1/2) = 40sqrt(t)

  29. amistre64
    • 5 years ago
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    t^(-1/2 + 2/2) 20t^(1/2) 20 * ---------------- = --------- = 40t^(1/2) -1/2 + 2/2 1/2

  30. amistre64
    • 5 years ago
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    which equals 40sqrt(t)

  31. anonymous
    • 5 years ago
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    ok... you changed the 1/sqrt(t) to 2/sqrt(t) and i got lost at that step. it makes sense now.

  32. amistre64
    • 5 years ago
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    its just reversing the steps of taking the derivative

  33. anonymous
    • 5 years ago
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    so basically when it says like dy/dt that means there is a function called y and the variable used will be t? such as dy/dt = 5t-4?

  34. amistre64
    • 5 years ago
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    thats correct

  35. amistre64
    • 5 years ago
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    but we have to finish our problem here..... tell me, do these curves have the same derivative: y= 2x+4 and y= 2x-17

  36. anonymous
    • 5 years ago
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    no

  37. amistre64
    • 5 years ago
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    take the derivative of both of them...what do you get?

  38. anonymous
    • 5 years ago
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    one is x^2 +4x and the other is x^2-17x

  39. amistre64
    • 5 years ago
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    not the integral....the derivative...

  40. anonymous
    • 5 years ago
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    o... both would be 2

  41. amistre64
    • 5 years ago
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    thats right.... then how do we know which curve is our answer when we suit them up with the integrals?

  42. amistre64
    • 5 years ago
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    dy/dx = 2 y = 2x + C right?

  43. amistre64
    • 5 years ago
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    where C can be any constant, any number

  44. anonymous
    • 5 years ago
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    right

  45. amistre64
    • 5 years ago
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    our integral is floating around..it needs to be anchored to a specific point in order to help us out

  46. anonymous
    • 5 years ago
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    in the example i originally gave the C would be 0. right?

  47. amistre64
    • 5 years ago
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    we have your problem.. X = 40sqrt(t) + C ....... how do we find the right equation of the curve? when we are given the X(1) = 40?

  48. anonymous
    • 5 years ago
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    u would do 40sqrt(1) + C = 40

  49. amistre64
    • 5 years ago
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    40 = 40sqrt(1) + C 40 = 40 + C 0 = C .... so yes, in this probelm C would be zero

  50. amistre64
    • 5 years ago
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    what would our answer be if we wanted X(1) = 20?

  51. anonymous
    • 5 years ago
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    1/2?

  52. amistre64
    • 5 years ago
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    20 = 40sqrt(1) + C 20 = 40 + C 20 -40 = C -20 = C X = 40sqrt(t) - 20 is the curve wed want right? you see how I did that?

  53. anonymous
    • 5 years ago
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    i mean -20

  54. anonymous
    • 5 years ago
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    ya. i divided it instead

  55. amistre64
    • 5 years ago
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    :) happens to me all the time lol

  56. amistre64
    • 5 years ago
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    does this make better sense to you now?

  57. anonymous
    • 5 years ago
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    ya. thanks for your help. i have to run off to class and take a quiz over this stuff. i should be fine now. thanks again!

  58. amistre64
    • 5 years ago
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    good luck, Ciao! :)

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