question

- anonymous

question

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

|dw:1433512332090:dw|

- anonymous

@Loser66

- Loser66

R can be understood as direction
\(\mathbb R\) is one direction |dw:1433512991259:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- Loser66

\(\mathbb R^2\) is 2 dimensions

- Loser66

|dw:1433513046896:dw|

- Loser66

\(\mathbb R^3\) is 3 dimensions |dw:1433513079018:dw|

- anonymous

that's a really stupid notation when u could just say we're working in 3 dimensions!!!

- Loser66

\(\vec r(t)=< f(t), g(t) >\) is parametric vector

- anonymous

what's the meaning of those less than greater than symbols?

- Loser66

That is just the notation for YOU to realize that is a parametric vector form

- anonymous

I know \[\vec r(f(t),g(t))=f(t)i+g(t)j\]

- Loser66

For example: you have distance , velocity and time. You can graph a distance w.r.t time by the graph with x-axis represents|dw:1433513283150:dw| time and y-axis represents distance

- Loser66

same as velocity and time |dw:1433513309349:dw|

- Loser66

What if I want you to graph distanc/ velocity and both w.r.t time?

- Loser66

Use parametric equations to jot them out.

- anonymous

I simply know that
\[\vec r(x,y,z)=x i+yj+zk\]
now be it x or x(t)
either way we will have
\[\vec r(x(t),y(t),z(t))=x(t)i+y(t)j+z(t)k\]
I just don't understand those curly braces

- Loser66

Think of this, an accident happens. A car felt down a cliff. |dw:1433513446289:dw|

- Loser66

if it is just an accident, the car's position should be as shown. But it is not, it is here |dw:1433513523649:dw|

- Loser66

Parametric equations answer those question. If you understand why we have to know them, it inspires you a lllllllllllot

- anonymous

tell me one thing,
\[\vec r=<3,4,5>\]
is this the same as
\[3i+4j+5k\]

- Loser66

yup

- Loser66

Have you ever take linear algebra??

- anonymous

I've studied vectors but I've never seen this notation in my school

- Loser66

Do you accept 1 dimension?

- anonymous

what do u mean ?

- Loser66

Do you understand number line? it is 1 dimension |dw:1433513798994:dw|

- anonymous

of course I do

- Loser66

Do you accept 2 dimensions? |dw:1433513841395:dw|

- anonymous

yep...

- Loser66

How about 3 dimensions?

- anonymous

yes, but how is that relevant?I simply wanted to know what those notations were for, I've never seen them before.

- Loser66

1D---yes
2D----yes
3D----yes
why not 4D, 5D.......nD???
Because you don't know them, you reject them? so that you said "that's a really stupid notation when u could just say we're working in 3 dimensions!!!"

- anonymous

Is it also correct to say that if
\[\vec r(t)=\]\[\vec r(t)=f(t)i+g(t)j+h(t)k\]

- Loser66

yes

- anonymous

Alright I understand this notation now

- anonymous

thx

- Loser66

the t in \(r(t),f(t), g(t)\) shows all function are w.r.t t

- anonymous

I'm not rejection higher dimensions but if you're working in 4 dimensions u can simply say in 4 dimensional space

- Loser66

and the comma or i, j, k show they are linearly independent variable.

- Loser66

But they are all relate to each other. In \(\mathbb R^3\) when x =0, it can't become \(\mathbb R^2\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.