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inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Find r'(1) if r(t)=<cos^1 (12t),6,5^(t^21)>
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
r(t) is a vector valued function, by the way.
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
I have differentiated individual components, getting r'(t)=<2/(sqrt(1(12t)^2),0,5*ln(5)*2t*5^(t^2)>
 one year ago

eseidl Group TitleBest ResponseYou've already chosen the best response.0
Take the derivative of each component separately. The input t=1 into the results.
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
the only problem is that r'(1) is undefined for 2/(sqrt(1(12t)^2), @eseidl
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
I mean resubstituting in results in a 0 in the denominator...
 one year ago

eseidl Group TitleBest ResponseYou've already chosen the best response.0
yeah good call. then that's your answer. r'(1) is undefined.
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
so it must be defined on for all components?
 one year ago

eseidl Group TitleBest ResponseYou've already chosen the best response.0
but don't you get: \[\frac{1}{2t^2+2t}\]for the x component of r'(t)?
 one year ago

eseidl Group TitleBest ResponseYou've already chosen the best response.0
thus r'(1)=<1/4, y'(1),z'(1)>
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
@eseidl , I thought d/dx arccos(u)=1/sqrt(1u^2)*du/dx?
 one year ago

eseidl Group TitleBest ResponseYou've already chosen the best response.0
oh, I mean\[\frac{1}{2t2t^2}\]missed a neg sign
 one year ago

eseidl Group TitleBest ResponseYou've already chosen the best response.0
@inkyvoyd yes that's what I used
 one year ago

eseidl Group TitleBest ResponseYou've already chosen the best response.0
got to go to work...good luck :)
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
alright >.< thanks for helping @amistre64 , helps pwease? :)
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
\[\large \frac {d}{dx} \cos^{1}(12t)=\frac{ 1 }{ \sqrt{1(12t)^2} }(2)=\frac{ 2 }{ \sqrt{4t4t^2} }\]
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
clearly, when t = 1, the denominator of the first component is zero
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
That's my question  what should I do with that 0 in the denominator, @sirm3d ?
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
the vector r'(1) does not exist.
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Alright, thanks!
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
what does an undefined derivative tell us?
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
the x component increases such that it is infinite?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
the slope of the line in the x direction is vertical
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
so does the vector exist?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354543586939:dw i dont see why not, but id have to do some review to make sure
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
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...
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
x does not change with respect to t that is ....
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
I thought it was dx/dt though?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
dx/dt means the rate of change of x with respect to t
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
that's what we are calculating right? <dx/dt,dy/dt,dz/dt>
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
and, yes that is what we are trying to determine can we write y=x^(1/3) in vector form ?
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
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!
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
the physical interpretation of this problem is that as t approches 1, the tangent vector r'(t) points to the direction of the xaxis, and the length of this vector stretches to infinity.
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
I'm back @amistre64 and @sirm3d , what I don't understand is that the vector would be infinite in the x direction?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
do you recall the domain for the arcCos function?
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
uh, 0<x<pi/2?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
the speed at which the function is changeing in the x direction approaches infinity as t approaches zero
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
err, t approaches 1
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
herp der, I thought you said range >.<
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
so you can't draw the vector because in the xdirection the magnitude is infinity, but in the y and z it's finite.
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
I'm guessing the function is undefined for r'(1) because the magnitude of vector in the x direction is undefined?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
correct, and an undefined derivative is a point of interest, a possible critical point
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Alright  thanks!
 one year ago
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