Question

Find T, N and kappa r(t) = (2t+3)i +(5-t^2)j

Answer 1

Ooo I haven't done these in a while :U a bit rusty.

So ummm r(t) is a position vector.
So the tangent vector is the derivative of positionnnnnnnnn?
And then to get the `unit` tangent vector, we divide by it's magnitude.
\[\Large\bf\sf \hat T\quad=\quad \frac{\vec r\;'}{|\vec r\;'|}\]Does that look right, maybe?

Answer 2

I am only having problems with one part of this problem. that will make it shorter to help

Answer 3

Oh which part XD

Answer 4

when finding T, specifically the j component.

Answer 5

im sorry, not T.. for finding N

Answer 6

specifically dT/dt / |dT/dt| for the j component.

Answer 7

and the problem is taking the derivative of T for the j component. I am stuck on the simplification. I can leave it in a more expanded form.. but that is really nasty to deal with when dividing by the length to get N.

Answer 8

Let's make sure we have the same Tangent Vector first of all.
Did you come up with something like this?\[\Large\bf\sf \hat T\quad=\quad \left<\frac{1}{\sqrt{1+t^2}},\;\frac{-t}{\sqrt{1+t^2}}\right>\]

Answer 9

exactly

Answer 10

Oh boy, quotient rule huh? :( this is gonna be a doozy.

Answer 11

I went through this kinda quick, maybe made a mistake,\[\Large\bf\sf \hat T'\quad=\quad \left<\frac{-t}{(1+t^2)^{3/2}},\;\frac{-1}{(1+t^2)^{3/2}}\right>\]Does this look rightttt?

Answer 12

i will show you where I get to and where I am trying to get
.\[\frac{ \sqrt{1+t^2}-t(\frac{ t }{ \sqrt{1=t^2} }) }{ (\sqrt{1+t^2})^3 }\]

Answer 13

yes.

Answer 14

how do get the above into your j component.

Answer 15

How do you what? :U

Answer 16

You mean, how do you simplify after quotient rule?

Answer 17

yes :)

Answer 18

Yes I think that's the correct simplification :o
Need help getting to that?

Answer 19

yes!! :)

Answer 20

From here,
\[\Large\bf\sf \frac{-\sqrt{1+t^2}-\left(\frac{-t^2}{\sqrt{1+t^2}}\right)}{(1+t^2)}\]Get a common denominator up top.

Answer 21

\[\Large\bf\sf \frac{-\dfrac{(1+t^2)}{\sqrt{1+t^2}}-\left(\dfrac{-t^2}{\sqrt{1+t^2}}\right)}{(1+t^2)}\]

Answer 22

is there supposed to be a root in the denominator?

Answer 23

I multiplied the first term by sqrt(1+t^2), top and bottom,
to get a common denominator with the other fraction.

Answer 24

i see it now!!!

Answer 25

XD

Answer 26

Awesome.. a shade under 2 hours for this one :) well just to figure out that part.. the rest was easy enough.

Answer 27

thanks for your help tremendously!!

Answer 28

ok cool \c:/ good job