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roadrunner
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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]
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
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
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