Question

Determine the tangent vector, unit normal,and bionormal for r(t)= (t^2 )+(2t+1)(t^3)

Answer 1

is r(t) in component form? you know, with the ijk s?

Answer 2

if so; then

r' = tangent
r'' = normal
r'xr'' = binormal

Answer 3

and any vector divided by its own magnitude goes unit

Answer 4

can you show me

Answer 5

what if it ask me to calculate at the point on the curve where t=1

Answer 6

then take the derivatives; input t=1, and voila!

Answer 7

id try to type up a better response, but at the moment trying so on this site is pointless

Answer 8

im still confuse with the binormal

Answer 9

with any luck that fixed some problems

Answer 10

its good

Answer 11

ill assumes this is in vector component form, since thats about the only way itll make sense:

r(t)= < t^2, (2t+1)(t^3) >

r'(t) = <2t, 2t^3 + 3t^2(2t+1)>

r''(t) = <2, 12t^2 + 6t(2t+1)>

Answer 12

if t=1; and you want units; plug in t=1 to determine the vectors; and since a cross works best with 3 components; lets make z=0


r(1)= < 1, 3, 0 >

r'(1) = <2, 11, 0>

r''(1) = <2, 30, 0>

Answer 13

i got some thing different.

Answer 14

unitT = r'/|r'| = <2,11,0>/sqrt{125}

unitN = r''/|r''| = <2,30,0>/sqrt{904}

the Binormal is just the TxN, or r'xr''

Answer 15

well, since im no good at mindreading; feel free to present what youve done :)

Answer 16

its question number 4

Answer 17

vectors are not single digits; unless you wanna consider the number line itself as the vector space

Answer 18

make sure we got the right r(t)

Answer 19

(2t+1)(t^3) = 6t^4 + t^3
derived = 24t^3 + 3t^2

which is not what youve got on your paper

Answer 20

opps, typo

Answer 21

2t^4 + t^3

8t^3 + 3t^2

Answer 22

what you posted was this

r(t)= (t^2)i+(2t+1)(t^3)j

or is it spose to be:

r(t)= (t^2)i +(2t+1)j +(t^3)k ???

Answer 23

r(t)= (t^2)i +(2t+1)j +(t^3)k

Answer 24

ahh, well that does change things then :)

Answer 25

r' = 2t i + 2 j + 3t^2 k
|r'| = sqrt(4t^2 + 4 + 9t^4)

T(1) = <2,2,3>/5


r'' = 2 i + 0 j + 6t k
|r''| = sqrt{4+0+36t^2}

N(1) = <1,0,6>/sqrt(40)

the wolfram is being retarded with this and cant verify if ive made a mistake ....

Answer 26

ive been using wolfham all day

Answer 27

unit T is spose to be: \(\large \frac{r'}{|r'|}\), which in this case is coming up hideous

unit N is defined as; \(\large \frac{T'}{|T'|}\), which is every bit as much of a pain

Answer 28

the agnitudes are not playing nice and making a monster of trying to go that route

Answer 29

i also read up that tangent and normals are spose to be perp, so they dot to zero; but i cant seem to get this thing to dot up regradless

Answer 30

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

this might be useful ...