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

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

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

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

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

so it must be defined on for all components?

yeah

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

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

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

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

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

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

the vector r'(1) does not exist.

Alright, thanks!

welcome.

what does an undefined derivative tell us?

the x component increases such that it is infinite?

the slope of the line in the x direction is vertical

so does the vector exist?

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

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

I thought it was dx/dt though?

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

that's what we are calculating right?

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

?

k :)

do you recall the domain for the arcCos function?

uh, 0

err, t approaches 1

herp der, I thought you said range >.<

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

Alright - thanks!