## inkyvoyd Group Title differentiation of vector valued functions one year ago one year ago

1. inkyvoyd Group Title

Find r'(1) if r(t)=<cos^-1 (1-2t),6,5^(t^2-1)>

2. inkyvoyd Group Title

r(t) is a vector valued function, by the way.

3. inkyvoyd Group Title

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

4. eseidl Group Title

Take the derivative of each component separately. The input t=1 into the results.

5. inkyvoyd Group Title

the only problem is that r'(1) is undefined for 2/(sqrt(1-(1-2t)^2), @eseidl

6. inkyvoyd Group Title

I mean resubstituting in results in a 0 in the denominator...

7. eseidl Group Title

8. inkyvoyd Group Title

so it must be defined on for all components?

9. eseidl Group Title

yeah

10. eseidl Group Title

but don't you get: $\frac{1}{2t^2+2t}$for the x component of r'(t)?

11. eseidl Group Title

thus r'(1)=<1/4, y'(1),z'(1)>

12. inkyvoyd Group Title

@eseidl , I thought d/dx arccos(u)=1/sqrt(1-u^2)*du/dx?

13. eseidl Group Title

oh, I mean$\frac{1}{2t-2t^2}$missed a neg sign

14. eseidl Group Title

@inkyvoyd yes that's what I used

15. eseidl Group Title

got to go to work...good luck :)

16. inkyvoyd Group Title

alright >.< thanks for helping @amistre64 , helps pwease? :)

17. sirm3d Group Title

$\large \frac {d}{dx} \cos^{-1}(1-2t)=-\frac{ 1 }{ \sqrt{1-(1-2t)^2} }(-2)=\frac{ 2 }{ \sqrt{4t-4t^2} }$

18. sirm3d Group Title

clearly, when t = 1, the denominator of the first component is zero

19. inkyvoyd Group Title

That's my question - what should I do with that 0 in the denominator, @sirm3d ?

20. sirm3d Group Title

the vector r'(1) does not exist.

21. inkyvoyd Group Title

Alright, thanks!

22. sirm3d Group Title

welcome.

23. amistre64 Group Title

what does an undefined derivative tell us?

24. inkyvoyd Group Title

the x component increases such that it is infinite?

25. amistre64 Group Title

the slope of the line in the x direction is vertical

26. inkyvoyd Group Title

so does the vector exist?

27. amistre64 Group Title

|dw:1354543586939:dw| i dont see why not, but id have to do some review to make sure

28. inkyvoyd Group Title

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 Group Title

x does not change with respect to t that is ....

30. inkyvoyd Group Title

I thought it was dx/dt though?

31. amistre64 Group Title

dx/dt means the rate of change of x with respect to t

32. inkyvoyd Group Title

that's what we are calculating right? <dx/dt,dy/dt,dz/dt>

33. amistre64 Group Title

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 Group Title

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

35. inkyvoyd Group Title

?

36. inkyvoyd Group Title

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 Group Title

k :)

38. sirm3d Group Title

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 Group Title

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 Group Title

I'm back @amistre64 and @sirm3d , what I don't understand is that the vector would be infinite in the x direction?

41. amistre64 Group Title

do you recall the domain for the arcCos function?

42. inkyvoyd Group Title

uh, 0<x<pi/2?

43. amistre64 Group Title

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 Group Title

the speed at which the function is changeing in the x direction approaches infinity as t approaches zero

45. amistre64 Group Title

err, t approaches 1

46. inkyvoyd Group Title

herp der, I thought you said range >.<

47. inkyvoyd Group Title

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 Group Title

I'm guessing the function is undefined for r'(1) because the magnitude of vector in the x direction is undefined?

49. amistre64 Group Title

correct, and an undefined derivative is a point of interest, a possible critical point

50. inkyvoyd Group Title

Alright - thanks!