Question

find the tangential and normal components of the acceleration vector: r(t) = (cosT)i +(sinT)j + (T)K

Answer 1

looks familiar

Answer 2

the tangent and normal are unit lngths tho right?

Answer 3

r' = tangent
r'' = normal

Answer 4

ummm i know the formula for tangent is (v dot a)/magnitude v

Answer 5

that just gives you a scalar, not a vector

Answer 6

v.a is a number and |v| is a number; so they dont produce a vector

Answer 7

the answer is 0 for tangential and 1 for normal, i just don't know how they got it

Answer 8

can you type up the generic form of an acceleration vector? I know its got something like dT/dt + kNdT/dt or some such

Answer 9

umm i think its a=v'T = k(v^2)N?

Answer 10

where T and N are units tan and norm; gonna have to check the google on that tho :)

Answer 11

i meant a + in-between, not =

Answer 12

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

Answer 13

you use this formula to solve it: but i just can't get the right answer, can you see what you get when you plug it in

Answer 14

\[a_T=\frac{r'.r''}{|r'|}\]
\[a_N=\frac{|r'xr''|}{|r'|}\]

Answer 15

\[a _{T} = r'(T) * r''(T)/\left|r'(t)\right|\]

Answer 16

yes with that formula

Answer 17

and you're supposed to get 0 and 1 but I'm not getting either of those

Answer 18

r(t) = < cosT, sinT, T >

r'(t) = < -sinT, cosT, 1 >

r''(t) = < -cosT, -sinT, 0 >

we agree on that?

Answer 19

yup

Answer 20

r'(t) = < -sinT, cosT, 1 >
.r''(t) = < -cosT, -sinT, 0 >
-------------------------
sincos - sincos + 0 = 0

so At = 0 regardless of |r'|

Answer 21

ohhhhh ohhhh i had plus 1k instead of 0

Answer 22

r'(t) = < -sinT, cosT, 1 >
x r''(t) = < -cosT, -sinT, 0 >
--------------------------
0 + sinT = sinT
-(0+cos(t)) = -cosT
sin^2 + cos^2 = 1

we agree with the cross?

Answer 23

since the cross is the same components as r'; its |cross|/|r'| = 1

Answer 24

yeah

Answer 25

sin^2+cos^2+1^2
---------------- = 1
sin^2+cos^2+1^2

Answer 26

you got 1 for for magnitude r'(t)?

Answer 27

doesnt matter what |r'| actually is; since its larger than 0

Answer 28

we have the same results top to bottom so anyting/itself = 1

Answer 29

i got |r'(t)| = 2 though

Answer 30

actually i got root2

Answer 31

then |r'xr''| will be 2 and 2/2 = 1

Answer 32

in this instance:

|r'xr''| = |r'| = a

a/a = 1

Answer 33

why does |r'xr''| = |r'|

Answer 34

what are your compenets for r'xr'' ?

Answer 35

components ..

Answer 36

(-sin t, cos t, 1) x (-cos t, -sin t, 0)

Answer 37

thats not the end result; that is what its made from

Answer 38

hang on let me do it really quick

Answer 39

when you actually cross them, i did this above, what do we get as components?

Answer 40

sin, -cos, sin^2 + cos^2

Answer 41

good, and s^2+c^2 = 1 right?

Answer 42

yup

Answer 43

now, look at the components of r' and tell me if there is any difference

Answer 44

just the negative signs

Answer 45

but you square it!

Answer 46

so it won't matter!

Answer 47

cross = sin, -cos, 1
r' = sin, cos, 1

when we take the magnitude of each we square the components, add them and sqrt them


cross r'
sin^2 = sin^2
.....

correct

Answer 48

ahhhh genius! thank you so much!

Answer 49

youre welcome :)