differentiation of vector valued functions

- inkyvoyd

differentiation of vector valued functions

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- chestercat

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- inkyvoyd

Find r'(1) if r(t)=

- inkyvoyd

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

- inkyvoyd

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

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## More answers

- anonymous

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

- inkyvoyd

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

- inkyvoyd

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

- anonymous

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

- inkyvoyd

so it must be defined on for all components?

- anonymous

yeah

- anonymous

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

- anonymous

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

- inkyvoyd

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

- anonymous

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

- anonymous

@inkyvoyd yes that's what I used

- anonymous

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

- inkyvoyd

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

- sirm3d

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

- sirm3d

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

- inkyvoyd

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

- sirm3d

the vector r'(1) does not exist.

- inkyvoyd

Alright, thanks!

- sirm3d

welcome.

- amistre64

what does an undefined derivative tell us?

- inkyvoyd

the x component increases such that it is infinite?

- amistre64

the slope of the line in the x direction is vertical

- inkyvoyd

so does the vector exist?

- amistre64

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

- 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...

- amistre64

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

- inkyvoyd

I thought it was dx/dt though?

- amistre64

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

- inkyvoyd

that's what we are calculating right?

- 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?

- amistre64

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

- inkyvoyd

?

- 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!

- amistre64

k :)

- 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.

- 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.

- inkyvoyd

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

- amistre64

do you recall the domain for the arcCos function?

- inkyvoyd

uh, 0

- 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

- amistre64

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

- amistre64

err, t approaches 1

- inkyvoyd

herp der, I thought you said range >.<

- 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.

- inkyvoyd

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

- amistre64

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

- inkyvoyd

Alright - thanks!

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