Differential displacement: dr= (dr/dt)dt
for the parametric curve x=t y=t z=t^2

Question

Differential displacement: dr= (dr/dt)dt
for the parametric curve x=t y=t z=t^2

turingtest · Answer

um... unless I am misreading the Q all you need to do is write out the sum of the differentiate $$\vec r(t)=\langle t,t,t^2\rangle$$add all the components, and then mulitply by dt

turingtest · Answer

wow I wrote that terribly, but I hope you understand what I mean
differentiate the vector function, add the components, then multiply by dt

anonymous · Answer

Well from what im reading I need to start with $$\sqrt{16t ^{2}+16t+17}dt$$ and in normally I would end up with something simple like 4dt but also I am reading some examples where I need to take the integral

turingtest · Answer

hm... I'm confused
I probably just don't understand the problem

anonymous · Answer

It's from Calculus three. I honestly think that all I need to do is solve that square root

turingtest · Answer

I'm seeing stuff like this http://www.ittc.ku.edu/~jstiles/220/handouts/The%20Differential%20Line%20Vector%20for%20Coordinate%20Systems.pdf though this is not for parametrics still, if it is just a differential you are looking for I can't see why you would need to integrate ...what do you mean "solve the sqrt?" and where did 16 and 17 come from ?

anonymous · Answer

sorry I posted a sample question and forgot the question I have is x=2t y=3t z=2(t+t^2) so for this equation the first step is $$ds= \sqrt{(dx/dt)^2+ (dy/dt)^2+ (dz/dt)^2}$$