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what is the curvatu… - QuestionCove
OpenStudy (anonymous):

what is the curvature of r(t)=(sint-tcos t )i+ (cost+tsin t)j+ (t^2)k

5 years ago
OpenStudy (turingtest):

can you get the unit tangent vector?

5 years ago
OpenStudy (anonymous):

i have the derivative of r(t) as tsint i + t cost j + 2t k

5 years ago
OpenStudy (turingtest):

\[\kappa=\frac{\|\vec T'(t)\|}{\|\vec r'(t)\|}\]is probably the easiest formula to use here what you have written so far is r'(t)

5 years ago
OpenStudy (anonymous):

i am having a togh time figuring out the unit tangent

5 years ago
OpenStudy (turingtest):

and I have not checked that r'(t) yet once you have it though, divide it by the magnitude |r'(t)| that is the unit tangent T(t)

5 years ago
OpenStudy (anonymous):

magnitude of r'(t)=t(sqrt5)

5 years ago
OpenStudy (turingtest):

yeah I agree with your r'(t) so now just divide by the magnitude

5 years ago
OpenStudy (turingtest):

divide, that is your unit tangent

5 years ago
OpenStudy (turingtest):

\[\vec T(t)=\frac{\vec r'(t)}{\|\vec r'(t)\|}=?\]

5 years ago
OpenStudy (anonymous):

T=[(sint - tcost)i +(cost + tsint)j + 2t]/(tsqrt5) this is where i get confused. I dont know if i need to use the second derivative of r(t)???

5 years ago
OpenStudy (turingtest):

\[\kappa=\frac{\|\vec T'(t)\|}{\|\vec r'(t)\|}\]

5 years ago
OpenStudy (turingtest):

no you divided r(t) by the magnitude of r'(t)

5 years ago
OpenStudy (anonymous):

ok.let me try to solve that!

5 years ago
OpenStudy (anonymous):

k=1/5t

5 years ago
OpenStudy (turingtest):

I got k=t, but I could have made a mistake

5 years ago
OpenStudy (turingtest):

oh wait\[\kappa=\frac1{5t}\]?yeah that's what I get

5 years ago
OpenStudy (turingtest):

is that what you meant with t on the bottom?

5 years ago
OpenStudy (anonymous):

yes

5 years ago
OpenStudy (turingtest):

sweet, then we're good :)

5 years ago
OpenStudy (anonymous):

thank you...first time and it may not be the last one

5 years ago
OpenStudy (turingtest):

hope not, see ya!

5 years ago
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