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mathcalculus

  • one year ago

Help with related rate problem please! (DISTANCE)

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

  3. mathcalculus
    • one year ago
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    which formula do i use?

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

  5. mathcalculus
    • one year ago
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    right.

  6. mathcalculus
    • one year ago
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    got that.

  7. Coolsector
    • one year ago
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    dy/dt = (dy/dx) * (dx/dt)

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

  9. mathcalculus
    • one year ago
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    i know.

  10. Coolsector
    • one year ago
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    so?

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

  12. mathcalculus
    • one year ago
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    distance formula?

  13. Coolsector
    • one year ago
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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)

  14. mathcalculus
    • one year ago
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    what does s symbolize?

  15. mathcalculus
    • one year ago
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    right.

  16. Coolsector
    • one year ago
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    s - distance

  17. Coolsector
    • one year ago
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    so that is it. now you have it all

  18. mathcalculus
    • one year ago
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    thanks let me try

  19. Coolsector
    • one year ago
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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

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

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

  22. mathcalculus
    • one year ago
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    and the whole distance formula too.

  23. Coolsector
    • one year ago
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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

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

  25. Coolsector
    • one year ago
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    you just need to find ds/dx

  26. mathcalculus
    • one year ago
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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

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

  28. Coolsector
    • one year ago
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    you dont need to find s. you need to find ds/dx

  29. mathcalculus
    • one year ago
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    ds/dx=10

  30. mathcalculus
    • one year ago
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    what about dy?

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

  32. mathcalculus
    • one year ago
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    ok

  33. Coolsector
    • one year ago
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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

  34. mathcalculus
    • one year ago
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    yeah im not sure what's happening.

  35. Coolsector
    • one year ago
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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)

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

  37. Coolsector
    • one year ago
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    understand what i did ?

  38. mathcalculus
    • one year ago
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    sqrt(x^2+50x+50) ?

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

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

  41. mathcalculus
    • one year ago
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    x^2+50x+50)

  42. Coolsector
    • one year ago
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    plugged y=5sqrt(2x+2)

  43. mathcalculus
    • one year ago
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    10=10

  44. mathcalculus
    • one year ago
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    50x+50?

  45. mathcalculus
    • one year ago
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    i plugged in y and x

  46. Coolsector
    • one year ago
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    what ?

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

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

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

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

  51. mathcalculus
    • one year ago
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    okay thanks alot.

  52. mathcalculus
    • one year ago
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    i got it

  53. Coolsector
    • one year ago
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    are you sure ?

  54. Coolsector
    • one year ago
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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)

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