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Mathematics 29 Online
OpenStudy (anonymous):

Find the curvature κ(t) of the curve X (t) = ( −4 sint )I + ( −4 sint ) j + ( 3 cost ) K

OpenStudy (anonymous):

do you know how to derive the formula for k(t)

OpenStudy (anonymous):

no i don't

OpenStudy (anonymous):

ok derivation of k(t) if you have r(t) then r'(t)/||r'(t)||=T and since ||dT/ds||=k then |dT/dt(ds/dt)|=|dT/ds| so 1/|ds/dt|(dT/dt|=1/|v|(|dT/dt)|=k there you go

OpenStudy (anonymous):

|dT/dt(dt/ds)|=|dT/ds|

OpenStudy (anonymous):

!

OpenStudy (anonymous):

so ill let you do this one, take your velcocity or r'(t) then take the magnitude=1/|r'(t)| then take your r'(t)/|r'(t)|=T then take d/dt(T)=dT/dt and take that magnitude=|dT/dt| 1/|r'(t)|(dT/dt)|=k(t)

OpenStudy (anonymous):

i dont know about this one BUT THESE TURN OUT UGLYYYYYYYY

OpenStudy (anonymous):

it is ugly ! haha

OpenStudy (anonymous):

do u mind giving me a numerical value too !?

OpenStudy (anonymous):

ill do a very easy example r(t)=(cos(t)i+(sin(t))j r'(t)=(-sin(t))i+(cos(t))j |r'(t)|=1 r'(t)/|r'(t)|=(-sin(t))i+(cos(t))j=T d/dt(T)=(-cos(t))i+(-sin(t))j |dT/dt|=1 k(t)=1/1(1)=1

OpenStudy (anonymous):

|dw:1316194279135:dw| N is pointing towards the center of curvature

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