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Question

r(t)=<2cost, 3sint,t^2> at t=0 and t=pi/4

I've tried solving this but it keeps getting more and more complicated, and i don't see how anything can drop out.

For the normal vector i got <0,1,0> and <-sqrt(2)/3, sqrt(2)/3, pi/2> respectively.

Is this right?

Indica · Accepted Answer

r = ( 2cost, 3sint, t² )

r’ = ( −2sint, 3cost, 2t )

We can write the normalisation of this as T√(4+4t²+5cos²t) = ( −2sint, 3cost, 2t )
(it’s just a bit easier to differentiate again in this form)

viz. T’√(4+4t²+5cos²t) + T(4t−5costsint)/√(4+4t²+5cos²t) = ( −2cost, −3sint, 2 )

For N find v = ( −2cost, −3sint, 2 ) − T(4t−5costsint)/√(4+4t²+5cos²t) and normalise

………………….

t=0  :  r’ = ( 0, 3, 0 )  :  T = r’/||r’|| = ( 0, 1, 0 )

Using above, v = ( −2, 0, 2) ) – ( 0, 1, 0 )(0) = ( −2, 0, 2 )

N = v/||v|| = ( −1/√2, 0, 1/√2 )

………………….

t=π/4  :  r’ = ( −√2, 3/√2, π/2 )  :  T = r’/||r’|| = ( −0.472261, 0.708391, 0.52455 )

v = ( −√2, −3/√2, 2 ) –  ( −0.472261, 0.708391, 0.52455 )(π−5/2)/√(4+π²/4+5/2)  

v = ( −1.31303, −2.27309, 1.88761 )

N = v/||v|| = ( −0.406099, −0.703031, 0.583808 )

There’s no simple closed form for this N

………………….

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