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inkyvoyd

  • 3 years ago

differentiation of vector valued functions

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  1. inkyvoyd
    • 3 years ago
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    Find r'(1) if r(t)=<cos^-1 (1-2t),6,5^(t^2-1)>

  2. inkyvoyd
    • 3 years ago
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    r(t) is a vector valued function, by the way.

  3. inkyvoyd
    • 3 years ago
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    I have differentiated individual components, getting r'(t)=<2/(sqrt(1-(1-2t)^2),0,5*ln(5)*2t*5^(t^2)>

  4. eseidl
    • 3 years ago
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    Take the derivative of each component separately. The input t=1 into the results.

  5. inkyvoyd
    • 3 years ago
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    the only problem is that r'(1) is undefined for 2/(sqrt(1-(1-2t)^2), @eseidl

  6. inkyvoyd
    • 3 years ago
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    I mean resubstituting in results in a 0 in the denominator...

  7. eseidl
    • 3 years ago
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    yeah good call. then that's your answer. r'(1) is undefined.

  8. inkyvoyd
    • 3 years ago
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    so it must be defined on for all components?

  9. eseidl
    • 3 years ago
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    yeah

  10. eseidl
    • 3 years ago
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    but don't you get: \[\frac{1}{2t^2+2t}\]for the x component of r'(t)?

  11. eseidl
    • 3 years ago
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    thus r'(1)=<1/4, y'(1),z'(1)>

  12. inkyvoyd
    • 3 years ago
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    @eseidl , I thought d/dx arccos(u)=1/sqrt(1-u^2)*du/dx?

  13. eseidl
    • 3 years ago
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    oh, I mean\[\frac{1}{2t-2t^2}\]missed a neg sign

  14. eseidl
    • 3 years ago
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    @inkyvoyd yes that's what I used

  15. eseidl
    • 3 years ago
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    got to go to work...good luck :)

  16. inkyvoyd
    • 3 years ago
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    alright >.< thanks for helping @amistre64 , helps pwease? :)

  17. sirm3d
    • 3 years ago
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    \[\large \frac {d}{dx} \cos^{-1}(1-2t)=-\frac{ 1 }{ \sqrt{1-(1-2t)^2} }(-2)=\frac{ 2 }{ \sqrt{4t-4t^2} }\]

  18. sirm3d
    • 3 years ago
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    clearly, when t = 1, the denominator of the first component is zero

  19. inkyvoyd
    • 3 years ago
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    That's my question - what should I do with that 0 in the denominator, @sirm3d ?

  20. sirm3d
    • 3 years ago
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    the vector r'(1) does not exist.

  21. inkyvoyd
    • 3 years ago
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    Alright, thanks!

  22. sirm3d
    • 3 years ago
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    welcome.

  23. amistre64
    • 3 years ago
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    what does an undefined derivative tell us?

  24. inkyvoyd
    • 3 years ago
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    the x component increases such that it is infinite?

  25. amistre64
    • 3 years ago
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    the slope of the line in the x direction is vertical

  26. inkyvoyd
    • 3 years ago
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    so does the vector exist?

  27. amistre64
    • 3 years ago
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    |dw:1354543586939:dw| i dont see why not, but id have to do some review to make sure

  28. inkyvoyd
    • 3 years ago
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    but I mean in a cartesian plane when you have dy/dx has 0 in denom, it means that the line is vertical because x does not change when y changes - here t does not change when x changes...

  29. amistre64
    • 3 years ago
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    x does not change with respect to t that is ....

  30. inkyvoyd
    • 3 years ago
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    I thought it was dx/dt though?

  31. amistre64
    • 3 years ago
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    dx/dt means the rate of change of x with respect to t

  32. inkyvoyd
    • 3 years ago
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    that's what we are calculating right? <dx/dt,dy/dt,dz/dt>

  33. amistre64
    • 3 years ago
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    les try this out as an example y = x^(1/3) ; is continuous y' = 1/3x^(-2/3) ; what is the derivative at x=0, what does it tell us about the slope at x=0?

  34. amistre64
    • 3 years ago
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    and, yes that is what we are trying to determine can we write y=x^(1/3) in vector form ?

  35. inkyvoyd
    • 3 years ago
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    ?

  36. inkyvoyd
    • 3 years ago
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    ugh I have to go, bell rang for next class - I'll try to read over this around 6 hours later, hopefully you (or someone else) will be there. Thanks!

  37. amistre64
    • 3 years ago
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    k :)

  38. sirm3d
    • 3 years ago
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    if the vector r'(1) does exist, then (1) it points in which direction; and (2) it has a magnitude. i'd be interested to see the magnitude of the this vector.

  39. sirm3d
    • 3 years ago
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    the physical interpretation of this problem is that as t approches 1, the tangent vector r'(t) points to the direction of the x-axis, and the length of this vector stretches to infinity.

  40. inkyvoyd
    • 3 years ago
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    I'm back @amistre64 and @sirm3d , what I don't understand is that the vector would be infinite in the x direction?

  41. amistre64
    • 3 years ago
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    do you recall the domain for the arcCos function?

  42. inkyvoyd
    • 3 years ago
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    uh, 0<x<pi/2?

  43. amistre64
    • 3 years ago
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    i think arcCos is from -1 to 1; and at t=1, we have arccos(-1); this space curve us just a short little section across the reals

  44. amistre64
    • 3 years ago
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    the speed at which the function is changeing in the x direction approaches infinity as t approaches zero

  45. amistre64
    • 3 years ago
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    err, t approaches 1

  46. inkyvoyd
    • 3 years ago
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    herp der, I thought you said range >.<

  47. inkyvoyd
    • 3 years ago
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    so you can't draw the vector because in the x-direction the magnitude is infinity, but in the y and z it's finite.

  48. inkyvoyd
    • 3 years ago
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    I'm guessing the function is undefined for r'(1) because the magnitude of vector in the x direction is undefined?

  49. amistre64
    • 3 years ago
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    correct, and an undefined derivative is a point of interest, a possible critical point

  50. inkyvoyd
    • 3 years ago
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    Alright - thanks!

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