Question

differentiation of vector valued functions

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

yeah good call. then that's your answer. r'(1) is undefined.

Answer 8

so it must be defined on for all components?

Answer 9

yeah

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

@inkyvoyd yes that's what I used

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

the vector r'(1) does not exist.

Answer 21

Alright, thanks!

Answer 22

welcome.

Answer 23

what does an undefined derivative tell us?

Answer 24

the x component increases such that it is infinite?

Answer 25

the slope of the line in the x direction is vertical

Answer 26

so does the vector exist?

Answer 27

i dont see why not, but id have to do some review to make sure

Answer 28

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...

Answer 29

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

Answer 30

I thought it was dx/dt though?

Answer 31

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

Answer 32

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

Answer 33

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?

Answer 34

and, yes that is what we are trying to determine

can we write y=x^(1/3) in vector form ?

Answer 35

?

Answer 36

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!

Answer 37

k :)

Answer 38

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.

Answer 39

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.

Answer 40

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

Answer 41

do you recall the domain for the arcCos function?

Answer 42

uh, 0<x<pi/2?

Answer 43

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

Answer 44

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

Answer 45

err, t approaches 1

Answer 46

herp der, I thought you said range >.<

Answer 47

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.

Answer 48

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

Answer 49

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

Answer 50

Alright - thanks!