## inkyvoyd 2 years ago differentiation of vector valued functions

1. inkyvoyd

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

2. inkyvoyd

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

3. inkyvoyd

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

4. eseidl

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

5. inkyvoyd

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

6. inkyvoyd

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

7. eseidl

8. inkyvoyd

so it must be defined on for all components?

9. eseidl

yeah

10. eseidl

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

11. eseidl

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

12. inkyvoyd

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

13. eseidl

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

14. eseidl

@inkyvoyd yes that's what I used

15. eseidl

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

16. inkyvoyd

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

17. sirm3d

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

18. sirm3d

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

19. inkyvoyd

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

20. sirm3d

the vector r'(1) does not exist.

21. inkyvoyd

Alright, thanks!

22. sirm3d

welcome.

23. amistre64

what does an undefined derivative tell us?

24. inkyvoyd

the x component increases such that it is infinite?

25. amistre64

the slope of the line in the x direction is vertical

26. inkyvoyd

so does the vector exist?

27. amistre64

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

28. inkyvoyd

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

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

30. inkyvoyd

I thought it was dx/dt though?

31. amistre64

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

32. inkyvoyd

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

33. amistre64

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

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

35. inkyvoyd

?

36. inkyvoyd

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

k :)

38. sirm3d

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

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

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

41. amistre64

do you recall the domain for the arcCos function?

42. inkyvoyd

uh, 0<x<pi/2?

43. amistre64

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

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

45. amistre64

err, t approaches 1

46. inkyvoyd

herp der, I thought you said range >.<

47. inkyvoyd

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

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

49. amistre64

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

50. inkyvoyd

Alright - thanks!