Question

Please help...I have been doing homework for 4+ hours, and my brain is fried...Can someone do the steps and show me how to do this? Award At End.

Given r(t), find the following: r(t) = ... Write a(t) in terms of its tangential and normal components.

Answer 1

http://www.wolframalpha.com/input/?i=+%3Ct%5E2%2C+sin%28t%29+-+tcos%28t%29%2C+cos%28t%29+%2B+tsin%28t%29%3E%27

Answer 2

divide that by its magnitude to get the unit tangent vector

Answer 3

Divide what? <2t, tsin(t), tcos(t)>

Answer 4

How do I solve for the magnitude
?

Answer 5

\[\large \hat{T(t)} = \dfrac{1}{t\sqrt{5}}\langle 2t, ~t\sin t, ~t\cos t\rangle\]

Answer 6

magnitude = \(\large \sqrt{(2t)^2 + (t\sin t)^2 + (t\cos t)^2} = t\sqrt{5}\)

Answer 7

Oh well, u can cancel \(t\) as well

Answer 8

Unit tangent vector :
\[\large \hat{T(t)} = \dfrac{1}{\sqrt{5}}\langle 2, ~\sin t, ~\cos t\rangle\]

Answer 9

So its < 2/Sqr(5) , sin(t)/Squr(5) , cos(5)/Sqr(5) >

Answer 10

yes but thats just the unit tangent vector, not the component of acceleration

Answer 11

Haha yeah I saw that extra t and canceled it out. :)

Answer 12

Okay and to get acceleration whats the formula?

Answer 13

I have it written down somewhere in my notes...Gotta rummage through them /.\

Answer 14

Acceleration :

http://www.wolframalpha.com/input/?i=+%3Ct%5E2%2C+sin%28t%29+-+tcos%28t%29%2C+cos%28t%29+%2B+tsin%28t%29%3E%27%27

Answer 15

velocity = first derivative
acceleration = second derivative

Simple.

Answer 16

So basically we have to solve for velocity first. So we need to take the derivative of r(t) = <t^2, sin(t) - tcos(t), cos(t) + tsin(t)>

Answer 17

yes

Answer 18

Then we take the derivative of that! Hah!

Answer 19

Okay hold on...Let me try this..

Answer 20

So < 2 , sin(t) + tcos(t) , cos(t) - tsin(t) > ??

Answer 21

@ganeshie8 Do I just remove the 2 because this is the second derivative?

Answer 22

nope, thats the acceleration vector.

Answer 23

next, find the tangent and normal components

Answer 24

for tangent component, do below :
1) find the magnitude of acceleration
2) multiply it by the unit tangent vector

Answer 25

\[\overrightarrow{A(t)} = \langle 2 , \sin(t) + t\cos(t) , \cos(t) - t\sin(t) \rangle \]

\[|\overrightarrow{A(t)} | = ?\]

Answer 26

Wait I though..Magnitude of acceleration = Change of velocity / Time interval

Answer 27

thought*

Answer 28

and ur definition of Magnitude of velocity is Change of position / Time interval

?

Answer 29

what u have there are average acceleration and average velocity -
for instantaneous acceleration u need to use derivatives.

Answer 30

\[\overrightarrow{A(t)} = \langle 2 , \sin(t) + t\cos(t) , \cos(t) - t\sin(t) \rangle\]

can u find its magnitude ?

Answer 31

Ohhh okay...I'm kinda lost now. :/

Answer 32

well, you haven't answered my question on whether u can find the magnitude of given Acceleration vector :/

Answer 33

Yeah I am trying to...Im getting lost...:/

Answer 34

To find the magnitude of A(t) what do we do? What formula?

Answer 35

magnitude of <x, y, z> = sqrt(x^2 + y^2 + z^2)

Answer 36

http://www.learningaboutelectronics.com/images/Magnitude-formulas.png

Answer 37

So how would I find the magnitude of A(t). Do I just square < 2 , sin(t) + tcos(t) , cos(t) - tsin(t) > ?

Answer 38

After I times x,y, and z by ^2 of course. @ganeshie8

Answer 39

\(\large \overrightarrow{A(t)} = \langle \color{green}{2} , \color{blue}{\sin(t) + t\cos(t)} , \color{red}{\cos(t) - t\sin(t)} \rangle \)


Magnitude = \(\large \sqrt{( \color{green}{2})^2 + ( \color{blue }{\sin(t)+t\cos(t)})^2 + ( \color{red}{\cos(t) - t\sin(t)})^2}\)

Answer 40

simplify^

Answer 41

Uhhh.... 4 + t^2 cos^2(t)+2 t cos(t) sin(t)+sin^2(t) + cos^2(t)-2 t cos(t) sin(t)+t^2 sin^2(t)

Answer 42

I can simplify that more I think..

Answer 43

it simplifies to \(\large \sqrt{t^2+5}\)

http://www.wolframalpha.com/input/?i=simplify+sqrt%282%5E2+%2B+%28sint+%2B+tcost%29%5E2+%2B+%28cost-tsint%29%5E2%29

Answer 44

Ohh...I expanded it out.

Answer 45

So, the tangent component of \(\large \overrightarrow{A(t)} \) is :

\(|\large \overrightarrow{A(t)}| *\hat{T(t)} = \sqrt{t^2+5} * \dfrac{1}{\sqrt{5}}\langle 2, ~\sin t, ~\cos t\rangle \)

Answer 46

you may find the normal component of \(\large \overrightarrow{A(t)}\) same way :

u just need to find the unit normal vector and multiply it by the magnitude of \(\large \overrightarrow{A(t)}\)

good luck :)