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mathcalculus

Help with related rate problem please! (DISTANCE)

  • 6 months ago
  • 6 months ago

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  1. mathcalculus
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    • 6 months ago
  2. Coolsector
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    hint -> you want dy/dt so: dy/dt = (dy/dx) * (dx/dt)

    • 6 months ago
  3. mathcalculus
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    which formula do i use?

    • 6 months ago
  4. Coolsector
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    you have dx/dt at this point and you have the point itself. that should help you

    • 6 months ago
  5. mathcalculus
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    right.

    • 6 months ago
  6. mathcalculus
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    got that.

    • 6 months ago
  7. Coolsector
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    dy/dt = (dy/dx) * (dx/dt)

    • 6 months ago
  8. Coolsector
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    dy/dx means derivative of y with respect to x dx/dt is given at this point

    • 6 months ago
  9. mathcalculus
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    i know.

    • 6 months ago
  10. Coolsector
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    so?

    • 6 months ago
  11. mathcalculus
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    but what formula should i use, for example there are problems for volume, circle, triangle, etc.

    • 6 months ago
  12. mathcalculus
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    distance formula?

    • 6 months ago
  13. Coolsector
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    ok so distance from the origin is s = sqrt(x^2 + y^2) so you want ds/dt = (ds/dy) * (dy/dt) + (ds/dx) * (dx/dt)

    • 6 months ago
  14. mathcalculus
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    what does s symbolize?

    • 6 months ago
  15. mathcalculus
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    right.

    • 6 months ago
  16. Coolsector
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    s - distance

    • 6 months ago
  17. Coolsector
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    so that is it. now you have it all

    • 6 months ago
  18. mathcalculus
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    thanks let me try

    • 6 months ago
  19. Coolsector
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    you can, as well, to express y in terms of x in the distance formula and then find ds/dt = (ds/dx) * (dx/dt) it will be much more simple

    • 6 months ago
  20. mathcalculus
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    right the whole 5*sqrt(2x+2) is throwing me off though,

    • 6 months ago
  21. mathcalculus
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    im not sure if i should find the derivative of that function with respect to time.

    • 6 months ago
  22. mathcalculus
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    and the whole distance formula too.

    • 6 months ago
  23. Coolsector
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    ok so if we do it the way i said at the end : "you can, as well, to express y in terms of x in the distance formula and then find ds/dt = (ds/dx) * (dx/dt) it will be much more simple" then you dont have to do much

    • 6 months ago
  24. Coolsector
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    cause s=sqrt(x^2 + y^2) = sqrt(x^2+50x+50) now ds/dt = (ds/dx) * (dx/dt)

    • 6 months ago
  25. Coolsector
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    you just need to find ds/dx

    • 6 months ago
  26. mathcalculus
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    i did this: i found s=10.0499 then i found the derivative which i got 5/sqrt(2x+2) *dx/dt

    • 6 months ago
  27. mathcalculus
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    i got ds/dt as 10 but i know it's wrong.

    • 6 months ago
  28. Coolsector
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    you dont need to find s. you need to find ds/dx

    • 6 months ago
  29. mathcalculus
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    ds/dx=10

    • 6 months ago
  30. mathcalculus
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    what about dy?

    • 6 months ago
  31. Coolsector
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    since we express y in terms of x we dont need to worry about it anymore.

    • 6 months ago
  32. mathcalculus
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    ok

    • 6 months ago
  33. Coolsector
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    ds/dx = (x + 25)/sqrt(x^2+50x+50) at the point x=1 it is ds/dx = 26 / sqrt(101) so ds/dt = (26 / sqrt(101))* 4 i might done some mistake though

    • 6 months ago
  34. mathcalculus
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    yeah im not sure what's happening.

    • 6 months ago
  35. Coolsector
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    why? s = sqrt(x^2+y^2) = sqrt(x^2+50x+50) so we want ds/dt ds/dt = ds/dx * dx/dt ds/dx = (x + 25)/sqrt(x^2+50x+50) ds/dx at this point = 26/sqrt(101) so ds/dt at this point = 26 * 4 /sqrt(101)

    • 6 months ago
  36. Coolsector
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    plugged into the derivative x=1 and dx/dt = 4

    • 6 months ago
  37. Coolsector
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    understand what i did ?

    • 6 months ago
  38. mathcalculus
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    sqrt(x^2+50x+50) ?

    • 6 months ago
  39. Coolsector
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    this is the distance.. s=sqrt(x^2 + y^2) = sqrt(x^2 + 50x + 50)

    • 6 months ago
  40. mathcalculus
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    i know, how'd you come up with this sqrt(x^2+50x+50)

    • 6 months ago
  41. mathcalculus
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    x^2+50x+50)

    • 6 months ago
  42. Coolsector
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    plugged y=5sqrt(2x+2)

    • 6 months ago
  43. mathcalculus
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    10=10

    • 6 months ago
  44. mathcalculus
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    50x+50?

    • 6 months ago
  45. mathcalculus
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    i plugged in y and x

    • 6 months ago
  46. Coolsector
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    what ?

    • 6 months ago
  47. Coolsector
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    y=5sqrt(2x+2) y^2 = 50x+50

    • 6 months ago
  48. mathcalculus
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    plugged y=5sqrt(2x+2). i plugged in y and x. 10=10

    • 6 months ago
  49. Coolsector
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    so now s = sqrt(x^2 + y^2) but since y^2 = 50x+50 s = sqrt(x^2+50x+50)

    • 6 months ago
  50. Coolsector
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    not the value of y at the point. plug y as a function of x

    • 6 months ago
  51. mathcalculus
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    okay thanks alot.

    • 6 months ago
  52. mathcalculus
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    i got it

    • 6 months ago
  53. Coolsector
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    are you sure ?

    • 6 months ago
  54. Coolsector
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    s = sqrt(x^2+y^2) = sqrt(x^2+50x+50) so we want ds/dt ds/dt = ds/dx * dx/dt when we calculate ds/dx we calculate is using s as a function of x. not plugging numerical values yet. ds/dx = (x + 25)/sqrt(x^2+50x+50) now plug x=1 ds/dx at this point = 26/sqrt(101) so ds/dt at this point = 26 * 4 /sqrt(101)

    • 6 months ago
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