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inkyvoyd
 3 years ago
differentiation of vector valued functions
inkyvoyd
 3 years ago
differentiation of vector valued functions

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

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0r(t) is a vector valued function, by the way.

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

eseidl
 3 years ago
Best ResponseYou've already chosen the best response.0Take the derivative of each component separately. The input t=1 into the results.

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

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0I mean resubstituting in results in a 0 in the denominator...

eseidl
 3 years ago
Best ResponseYou've already chosen the best response.0yeah good call. then that's your answer. r'(1) is undefined.

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0so it must be defined on for all components?

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

eseidl
 3 years ago
Best ResponseYou've already chosen the best response.0thus r'(1)=<1/4, y'(1),z'(1)>

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

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

eseidl
 3 years ago
Best ResponseYou've already chosen the best response.0@inkyvoyd yes that's what I used

eseidl
 3 years ago
Best ResponseYou've already chosen the best response.0got to go to work...good luck :)

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0alright >.< thanks for helping @amistre64 , helps pwease? :)

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

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.1clearly, when t = 1, the denominator of the first component is zero

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

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.1the vector r'(1) does not exist.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0what does an undefined derivative tell us?

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0the x component increases such that it is infinite?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0the slope of the line in the x direction is vertical

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0so does the vector exist?

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

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0but 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
 3 years ago
Best ResponseYou've already chosen the best response.0x does not change with respect to t that is ....

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0I thought it was dx/dt though?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0dx/dt means the rate of change of x with respect to t

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

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

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0ugh 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!

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.1if 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
 3 years ago
Best ResponseYou've already chosen the best response.1the 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.

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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0do you recall the domain for the arcCos function?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0i 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
 3 years ago
Best ResponseYou've already chosen the best response.0the speed at which the function is changeing in the x direction approaches infinity as t approaches zero

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0herp der, I thought you said range >.<

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

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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0correct, and an undefined derivative is a point of interest, a possible critical point
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