Question

compute the velocity vector of the curve (1+cost,sint,2sint/2) for arbitrary t and for t=45,t=90,

Answer 1

just take the derivative of the components

Answer 2

ok!

Answer 3

(-sint,cost,cost)

Answer 4

(1+cost , sint , 2sin t/2)

(-sin t , cos t , cos t/2) is what i get

Answer 5

ofcours!

Answer 6

:)

Answer 7

:) and for unique curve ? how can i find?

Answer 8

unique curve ... ive never heard that phrase before, how is it defined?

Answer 9

and OS is not playing right on my system at the moment. notifs are not refreshing .....

Answer 10

it is,find a unique curve such that alpha(0)=(1,0,5) and alpha'(t)=(t2,t.e^t)

Answer 11

straight line is the simple type of curve in eucledean space

Answer 12

alpha(t)=p+tq q=/0

Answer 13

hmm, im thinking that its wanting you to define the tangent line to the given curve parametrically then?

Answer 14

1+cost=1 ; t = arccos(0)
sint=0 ; t = arcsin(0)
2sint/2 = 5 ; t = arcsin(5/2)

does not seem to be a doable point on the original curve

Answer 15

im thinkin thatwe have to find another curve that Beta(t)

Answer 16

such that bet--->p+tv

Answer 17

alpha, beta .... those do not ring any bells for me. either i havent come across this type of math before, or what i have learned just doesnt use those terms

Answer 18

you r right that i have to find tangent line

Answer 19

i am a stdnt plz help me for solving this qustion

Answer 20

the tangent line at a given point is define parametrically as:\[L=r(t)+n~r'(t)\]

\[x=1+cot(t)-n~sin(t)\\y=sin(t)+n~cos(t)\\z=2sin(\frac{t}{2})+n~cos(\frac{t}{2})\]

Answer 21

im just not sure what point they are wanting you to start on ...

Answer 22

and thats: 1+cos(t) , typoed

Answer 23

ok thnx alot