Question

HELP PLEASE!!!!!
Find the vectors T, N, and B at the given point... r(t) = 7cost, 7sint, 7lncost
(7,0,0)

Answer 1

For the unit tangent vector I got \[T(t)= <sintcost, \cos^2 t,sint>\]

Answer 2

Then evaluating for T(0) I got <0,1,0> and webassign said that answer was correct. I evaluated T(0) because I set the i-th component equal to the x value of the point. Then I set the j-th component equal to the y value of the point and both indicate that the value of t=0-degrees.

Answer 3

For the derivative I got \[T'(t)= <\cos^2 t-\sin^2 t/-2costsint, cost>\]

Answer 4

If I vealuate T'(0)= <1,0,1> so I use that to find the magnitude which is just the sqaure root of two. When I divide the vector through by that I get the unit normal vetoer "V(t)"\[V'(t)= < \frac{ 1 }{ \sqrt{2} },0, \frac{ 1 }{ \sqrt{2} } >\]

Answer 5

My online homework liked the first answer for T(t) but not the second for V(t)...... and that last posting should not have been V'(t) but V(t)

Answer 6

@Psymon

Answer 7

what is V'(t)? unit vector of T?

Answer 8

no it is the unit normal vector...... found by taking the derivative of the unit tangent vector and dividing it by the magnitude of the derivative of the unit tangent vector

Answer 9

so, you should use N as its symbol

Answer 10

I meant to but I accidentally put that and then I retyped it twice making errors on both tries and finally I got everything in the equation accept the N and I write v' instead but yes I agree it should be N

Answer 11

I think you make mistake at N, you cannot take T' =<1,0,1> and |T'| = 1/sqrt 2.....
because it's not that . you should do something like
\[N=\frac{T'}{|T'|}=\frac{<cos 2t, sin2t, cost>}{\sqrt{cos^2(2t)+ sin^2(2t)+ cos^2t}}\]

Answer 12

and then plug t to calculate N.

Answer 13

I actually agree. But remember we found t=0 or 180 degrees. If you plug that in and calculate it I seem to get that same result. Do you?

Answer 14

oh , yea, you are right

Answer 15

However the homework assignment disagrees with you on that point

Answer 16

It says that this is not the answer

Answer 17

@theEric

Answer 18

Any Idea?

Answer 19

No, sorry... I tried a little bit, but I got a different \(\vec T(t)\) from yours, and you said yours was correct.


\(\vec r(t) = \left<7\cos t,\ 7\sin t,\ 7\ln(\cos t)\right>
\\\implies \vec r'(t)=\left< -7\sin t,\ 7\cos t,\ 7\tan t\right>
\\\implies \dfrac{\vec r'(t)}{\left|\left|\vec r'(t)\right|\right|}=\vec T(t)=\dfrac{\left< -7\sin t,\ 7\cos t,\ 7\tan t\right>}{\sqrt{49\sin^2 t+\ 49\cos^2 t+\ 49\tan^2 t}}
\\~\\\qquad\qquad=\dfrac{\left< -7\sin t,\ 7\cos t,\ 7\tan t\right>}{7\sqrt{1+\tan t}}=\dfrac{\left< -\sin t,\ \cos t,\ \tan t\right>}{\sqrt{1+\tan t}}
\)

Answer 20

Good luck!

Answer 21

@theEric the only thing wrong with what you put in was that you started with tan^2 in the denominator then after you factored the 49 out you only wrote tan when it should have been tan^2 my paper looks identical to yours but with tan^2

Answer 22

The 1+tan^2 would just be sec^2 and that comes out of the radical as just 7sec

Answer 23

@ybarrap ?

Answer 24

yes, I agree @chrisplusian that it should be just 7sec t after taking square root in denom, but otherwise @theEric has the right solution for the unit tangent

Answer 25

Ok so when I simplify that I get <-sintcost, cos^2(t), sint> do you agree?

Answer 26

yes!

Answer 27

Ok so then the unit normal vector is the derivative of the unit tangent vector over the magnitude of the derivative of the unite tangent vector right?

Answer 28

yes

Answer 29

And you can check by taking dot product, should be zero

Answer 30

I see my mistake, thank you both!
@chrisplusian yep!

Answer 31

For the derivative of the unit tangent vector I got \[<\sin^2 t -\cos^2 t,-2sintcost, cost >\]

Answer 32

do you agree with that?

Answer 33

yes, I get : <-cos(2 t), -2 sin(t) cos(t), cos(t)>

Answer 34

\(
T(t)= \left<\sin t\cos t,\ \cos^2 t,\ \sin t\right>
\)
Looks good to me, too!

Answer 35

Ok @ybarrap do you agree that this should be evaluated at t=0? from the original question?

Answer 36

yes

Answer 37

because cos(0)=1, sin(0)=0

Answer 38

\(
\left<-cos(2 t), -2 sin(t) cos(t), cos(t)\right>
\)

@chrisplusian You do as we have, finding all of the vectors in terms of \(t\). Then you let \(t=0\). That looks good! :)

Answer 39

Ok at this point can't I evaluate this and get T'(t)= < -1,0,1> ??

Answer 40

excuse me T'(0)= <-1,0,1>

Answer 41

@ybarrap ?

Answer 42

Don't you need to divide N by mag of N to make it a unit vector?

Answer 43

I thought that\[N(t)=\frac{ T'(t) }{ \left| T'(t) \right| }\]

Answer 44

There might be a difference in ways of doing it. I do as @chrisplusian does, where \(\vec N\) is always a unit vector.

Answer 45

Then if I evaluate T'(0) I come up with <-1,0,1> and the magnitude of the derivative of the unit tangent vector can be found by \[\sqrt{(-1)^2 +(0)^2 +1^2}=\sqrt{2}\]

Answer 46

I think you can do \(N(t)=\dfrac{ T'(0) }{ \left| T'(0) \right| }\) also.

Answer 47

I agree with this, @chrisplusian .

Answer 48

I agree as well.

Answer 49

:D

Answer 50

So for the unit normal vector I came up with \[N(t)=<\frac{ -1 }{ \sqrt{2} },0,\frac{ 1 }{ \sqrt{2} }>\]

Answer 51

Looks good to me. :)

What about the fun part, \(\vec B=\vec T\times\vec N\)

Answer 52

oh, yeah!

Answer 53

should be a snap now!

Answer 54

Well the probem is that I have to put the answers in for each part before I can proceed to the next and web assign (an online homework) says that we are all three wrong and that is not the unit normal vector

Answer 55

I did the dot product between the two, and it is zero.

Answer 56

@ybarrap will you explain what you are doing the dot product of?

Answer 57

T and N should be perpendicular

Answer 58

the unit tangent vector and the original vector? And then the unit tangent vector and the unit normal vector?

Answer 59

the unit tangent and the unit normal should be perpendicular and their dot products must be zero. I checked, it is zero.

Answer 60

Huh!

Answer 61

What about the unit tangent vector and the original vector? The should not be perpendicular should they?

Answer 62

Is it a formatting issue? Online questions are not smart..

Answer 63

Smart enough to count as a grade :)

Answer 64

The original vector is neither tangent nor normal, generally.

Answer 65

It can vary. They're not strictly related, like \(\vec T\) and \(\vec N\).

The \(\vec T\), \(\vec N\), and \(\vec B\) are all orthogonal. They make up a little coordiante system! :)
http://www.math.tamu.edu/~tom.kiffe/calc3/tnb/tnb.html

Answer 66

The unit normal can be pointing in the wrong direction. Remember that the unit normal can be the same vector but pointing 180 degrees in the other direction

Answer 67

@ybarrap I am not getting zero as the dot product, to be sure I am doing it correctly..... when I dot a vector with another vector I add the corresponding elements correct?

Answer 68

So try the negative of what you have

Answer 69

I don't remember that... But they are both normal, and it's worth a shot if it might be correct.

Answer 70

@ybarrap how would you check to be sure the normal is pointing in the correct direction?

Answer 71

I know I can use the right hand rule for a cross product but how do you know for a dot product?

Answer 72

here is how I checked the dot product:

http://www.wolframalpha.com/input/?i=%28d%2Fdt+%28-sin%28t%29cos%28t%29%2C+cos^2%28t%29%2C+sin%28t%29%29%29dot%28-sin%28t%29cos%28t%29%2C+cos^2%28t%29%2C+sin%28t%29%29

Answer 73

Call me old fashioned.... I just took \[<0,1,0> <-\frac{ 1 }{ \sqrt{2} },0,\frac{ 1 }{ \sqrt{2} }>\]

Answer 74

I couldn't find the dot operator

Answer 75

I type them, and I just use `\bullet`, like \(\vec A\bullet\vec B\)

Answer 76

Where did you get <0,1,0> is that the tangent at (7,0,0)?

Answer 77

That is the unit tangent vector evaluated at t=0

Answer 78

Also, I use `\left< stuff \right>` for the \(\left< stuff \right>\)

Answer 79

\cdot works also: \(\Huge \cdot \)

Answer 80

So you get zero, we agree. I would try negative.

Answer 81

\(\vec A\cdot\vec B\)
Awesome, thanks! :)

Answer 82

I don't see how I got zero because by my calulations I did not

Answer 83

http://www.wolframalpha.com/input/?i=%3C0%2C1%2C0%3Edot%3C%E2%88%922^%28-.5%29%2C0%2C2^%28-.5%29%3E

Answer 84

Just look at where the zeroes are, actually!

Answer 85

\(\left<0,a,0\right>\cdot\left<b,0,d\right>\)
Doesn't matter what your numbers are! Each component has a \(0\).

Answer 86

\(\left<0,a,0\right>\cdot\left<b,0,d\right>=0b+a0+0d=0\)

Answer 87

I see why I didn't get zero, I messed up but how would you decide if the normal is pointing in the right direction?

Answer 88

well that was my last attempt, I tried to change the signs, but that was wrong too!!Arghhhhhh Thanks for the help you guys, I wonder if we are doing something wrong, or if it has the wrong answer programmed in

Answer 89

It could just be that it has the formatting done incorrectly. I'm pretty confident in our solution, mostly because there are three of us.

Answer 90

Also, it didn't ask for unit vectors, just vectors. Maybe?

Answer 91

So, \(\vec T'(t)\) but not \(\dfrac{\vec T'(t)}{\left|\left|\vec T'(t)\right|\right|}\)?