What Are the Kinematics of a Particle Moving Along a Curve?

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Homework Help Overview

The problem involves the kinematics of a particle moving along a curve defined by the position vector r(t) = . Participants are tasked with finding the velocity, speed, acceleration, curvature, arc length, and various vector quantities related to the motion of the particle.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to compute the velocity, acceleration, curvature, and arc length, while also seeking clarification on the correctness of their calculations.
  • Some participants question the correctness of the cross product used in the curvature calculation.
  • Others express uncertainty about the results for the unit tangent and the subsequent calculations needed for the principal normal and binormal vectors.

Discussion Status

The discussion is ongoing, with participants providing feedback on specific calculations and seeking further clarification on various aspects of the problem. There is no explicit consensus on the correctness of the original poster's attempts, and multiple interpretations of the calculations are being explored.

Contextual Notes

Participants note the urgency of the original poster's situation, as they are looking for timely assistance due to time constraints.

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Homework Statement



The position of a particle at time t is given by
r(t) =<t,t^2,(2/3)t^3>

(a) Find the velocity v, speed |v| and acceleration a of the particle, and the
curvature k of its path, as a function of t.


(b) Find the arc length function s(t), and the arc length of the curve as t increases
from t = 0 to t = 3.

(c) At the point <1,1,2/3>, find the unit tangent, principal normal and binormal
vectors T, N and B, and find the curvature k and radius


Homework Equations


*NOTE* | | arround somethign mean the magnitude of the vector not absolute value
IE |2i,3j,4k| = sqrt(4+9+16)
Code:
T(t)= r'(t)
       |r'(t)|

k=|r'(t)Xr"(t)|
       |r'(t)|^3

N(t)=T'(t)
      |T'(t)|

T(t)=r'(t)
     |r'(t)|

B(t)=T(t)XN(t)

r'(t)=v
r"(t)=v'(t)=a

The Attempt at a Solution





a) r(t) =<t,t^2,(2/3)t^3 (given)
v= r'(t)=1 i, 2t j, 2t^2 k
a= r"(t)=v'(t)=0 i, 2 j, 4t k


|v|=|r'(t)|=sqrt(1+4t^2+4t^4)=(2t^2+1)=|V|
Code:
k=|r'(t)xr"(t)|
     |r'(t)|^3

r'(t) X r"(t)= | i    j    k  |
             |1  2t   2t^2  |  =[B](4t^2)i, -4t j, 2 k[/B]
             |0   2   4t    | 

|r'(t) X r"(t)|= sqrt(16t^4+16t^2+4)
=2sqrt(4t^4+4t^2+1)
=2sqrt(2t^2+1)^2 
= [B]2(2t^2+1)[/B]

k= 2(2t^2+1)
   (2t^2+1)^3

[B]= 2
(2t^2+1)^2[/B]


did i do this right?

s(t)=(integral form a to b of) |r'(t)|
s(t)=(int of)2t^2+1

integral form 0 to 3 is (2/3)t^3+t = (2/3)3^3 +3 = 18+3=21?


c)At the point <1,1,2/3>, find the unit tangent, principal normal and binormal
vectors T, N and B, and find the curvature k and radius


my attempt...
Code:
T(t)=r'(t)             r'(t)=1 i, 2t j 2t^2 k
     |r'(t)|             |r'(t)|=2t^2+1

T(t)=(1/(2t^2+1)) i + ((2t)/(2t^2+1)) j + ((t^2)/(2t^2+1)) k

the point 1,1,2/3 means t=1
[B]T(1)=1/3 i + 2/3 j + 1/3 k[/B]
CORRECT??!



IM STUCK i can't find T'(t)

but i found |T'(t)| with
Code:
|T'(t)|=|r'(t) X r"(t)|      =       2(2t^2+1)          =   2
             |r'(t)|^2              (2t^2+1)            2t^2+1
also i found k with the solution from part a


k= 2
(2t^2+1)^2


k=2/9?

and once i get that T'(t) i can easy use B(t)=T(t) X N(t) to finish the problem

so please point out any errors and help me figure out how to get T'(t) :D
 
Last edited:
Physics news on Phys.org
Your v x a is incorrect
 
look better now?
 
Last edited:
ok i fixed it, does everythign else look k correct? (b also)? 'cuse I am running onto problems on C
 
dont mean to be pushy but i need to sleep soon, can any1 shed some light please?
 
bed in 20 minutes...any1?
 
bump?
 

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