Question

For the vector function r(t) = < cos t, sin t, ln(cos t)>
a) Find T(t) at the point (1,0,0) and explain what this tells you about the motion of a particle on the curve at this point.
b) Find an equation of the line tangent to r(t) at this point.

Answer 1

I have worked out so far that T(t) = r' / | r' | and that is:

<-sin t, cos t, -tan t > / sec t

I now need evaluate this but I dont know what value of t to use. Do I need to use the point given to find a value of t that will work?

Answer 2

Ok, using the original function solving for a value that works for each gives t=0...getting father

Answer 3

I think I worked out the first part. and got <0, 1, 0> but now I need to find N(t) and am hitting a brick wall. I think Im doing it right, but I cant get a proper vector as a result. Can you please help me. I really need to understand this

Answer 4

i believe that N = T'/|T'|

Answer 5

Yes, I got T' = <-cos(2t) , -2sin(t)cos(t), -cos(t)> and for

Answer 6

i was right :)

http://tutorial.math.lamar.edu/Classes/CalcIII/TangentNormalVectors.aspx

Answer 7

|T'| I get sin^2(t) (4 cos(t) +3) But when I plug in 0 to t, I dont get real results

Answer 8

r = < cos t, sin t, ln(cos t)>

r'/|r'| = T = <-sin(t) , cos(t), -tan(t)>/sqrt(1+tan^2)

T = cos(t) <-sin(t) , cos(t), -tan(t)>

T = <-sin(t)cos(t) , cos^2(t), -sin(t)>
.............................................................

T' = <-cos^2(t)+sin^2(t) , -2sin(t)cos(t), -cos(t)>

right so far?

Answer 9

yes

Answer 10

im thinking that: -sin(t)cos(t) can prolly be written a different way to make the setup easier

Answer 11

sin(2a) = 2sin(a)cos(a); so double the innards and half the outer

-sin(t)cos(t) = -sin(2t)/2

Answer 12

I also wrote -cos^2(t)+sin(t) as -cos(2t) but im not sure that makes it easier

Answer 13

-sin(2t)/2 to -cos(2t)

-2sin(t)cos(t) = -sin(2t)

T' = <-cos(2t), -sin(2t), -cos(t)>
|T'| = sqrt(1+cos^2(t))

Answer 14

@snaef999 I think ur T(t) is wrong
should it be <sint,cost,-tant>/square root of 1-(sec) ^2

Answer 15

nah, its right

Answer 16

tan^2 + 1 = sec^2

sqrt(tan^2+1)=sec

Answer 17

N=<-cos(2t), -sin(2t), -cos(t)>/sqrt(1+cos^2(t)) ; at t=0

N=<-1, 0, -1>/sqrt(2)

N=<-1/sqrt2, 0, -1/sqrt2>

Answer 18

oh oppss sorry i got the derivative of tan !!
@amistre64 thnkx !!
m working on the same problem n was stuck !!

Answer 19

ok, that makes things work out better

Answer 20

Not sure where I went wrong though...I think I tried to simplify too much and made a mistake

Answer 21

prolly, algebra/trig can be a bother ;)

Answer 22

Thank you so much!

Answer 23

Living_Dreams, you working on this for a class?