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inkyvoyd
differentiation of vector valued functions
Find r'(1) if r(t)=<cos^-1 (1-2t),6,5^(t^2-1)>
r(t) is a vector valued function, by the way.
I have differentiated individual components, getting r'(t)=<2/(sqrt(1-(1-2t)^2),0,5*ln(5)*2t*5^(t^2)>
Take the derivative of each component separately. The input t=1 into the results.
the only problem is that r'(1) is undefined for 2/(sqrt(1-(1-2t)^2), @eseidl
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?
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)>
@eseidl , I thought d/dx arccos(u)=1/sqrt(1-u^2)*du/dx?
oh, I mean\[\frac{1}{2t-2t^2}\]missed a neg sign
@inkyvoyd yes that's what I used
got to go to work...good luck :)
alright >.< thanks for helping @amistre64 , helps pwease? :)
\[\large \frac {d}{dx} \cos^{-1}(1-2t)=-\frac{ 1 }{ \sqrt{1-(1-2t)^2} }(-2)=\frac{ 2 }{ \sqrt{4t-4t^2} }\]
clearly, when t = 1, the denominator of the first component is zero
That's my question - what should I do with that 0 in the denominator, @sirm3d ?
the vector r'(1) does not exist.
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
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...
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? <dx/dt,dy/dt,dz/dt>
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?
and, yes that is what we are trying to determine can we write y=x^(1/3) in vector form ?
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!
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.
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.
I'm back @amistre64 and @sirm3d , what I don't understand is that the vector would be infinite in the x direction?
do you recall the domain for the arcCos function?
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
the speed at which the function is changeing in the x direction approaches infinity as t approaches zero
herp der, I thought you said range >.<
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.
I'm guessing the function is undefined for r'(1) because the magnitude of vector in the x direction is undefined?
correct, and an undefined derivative is a point of interest, a possible critical point