A particle moves such that its position vector as a function of time is r(t)=4t^2 i+8t^2 j-4t k?

Question

1) Find the unit tangent to the trajectory as a function of time

2) What is the curvature when t=1 (give your answer accurate to 3dp)

any help would be greatly appreciated

Answer 1

r(t)=4t^2i + 8t^2j -4tk

r'(t)=tangent vector = 8Ti+16Tj-4K

magnitude=sqr(8^2 + 16^2 + (-4)^2) = 18.3303028

when t=1

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unit tangent vector:

T= (8t / 18.33)i + (16t/18.33)j + (-4/18.33k)

|T|

magnitude=sqr((8 / 18.33)^2 + (16/18.33)^2 + (-4 /18.33) ^2)

=1 unit

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curvature:

The derivative of the unit tangent is:

T'=[(8t/18.33)i + (16t/18.33)j + (-4/18.33)]/dt

= (8/18.33)i + (16/18.33)j + (0)k

magnitude of unit tangent =|T'|=

sqr((8 / 18.33)^2 + (16 / 18.33)^2 + 0) = 0.975916193

k=|T'| / |r'|

k= 0.975916193/18.3303

k=0.0532