What is the value of vector a at t=pi given the initial conditions?

[EmmaK]

Homework Statement

The vector a depends on a parameter t, i.e. $a=a(t)=a_x(t)i +a_y(t)j +a_z(t)k$..
it satisfies the equation $da/dt= j$$$\times$$$a$
show that $d^2a_x/dt^2 =-a_x$ , $da_y/dt=0$ and $d^2a_z/dt^2 =-a_z$.

For the vector a, find its value for t=pi if at t=0 $a(0)=i+j$ and $da/dt(0)=0k$

Homework Equations

$a.b = mod(a)mod(b)cos\theta$
$a$ X [itex]b = mod(a)mod(b)sin\theta[/tex] $$\hat{n}$$[/itex]

The Attempt at a Solution

i have absolutely no idea how to start...

 

[Dick]

To start you have to figure out what da/dt=jxa means in terms of the components of a, [ax,ay,az]. Can you find the cross product of the vector j with a?

 

[Altabeh]

Homework Statement

The vector a depends on a parameter t, i.e. $a=a(t)=a_x(t)i +a_y(t)j +a_z(t)k$..
it satisfies the equation $da/dt= j$$$\times$$$a$
show that $d^2a_x/dt^2 =-a_x$ , $da_y/dt=0$ and $d^2a_z/dt^2 =-a_z$.

For the vector a, find its value for t=pi if at t=0 $a(0)=i+j$ and $da/dt(0)=0k$

Homework Equations

$a.b = mod(a)mod(b)cos\theta$
$a$ X [itex]b = mod(a)mod(b)sin\theta[/tex] $$\hat{n}$$[/itex]

The Attempt at a Solution

i have absolutely no idea how to start...

Have you written the second asked-to-be-shown equation correctly? I think it must be modified as follows: $da_y/dt=c$ (c being a constant).

To not get into the trouble of Latex, you can scan a photo of the printed question and put it on the forum.

AB

 

[phsopher]

Have you written the second asked-to-be-shown equation correctly? I think it must be modified as follows: $da_y/dt=c$ (c being a constant).

$da_y/dt=0$ follows from $da/dt= j \times a$

 

[Altabeh]

$da_y/dt=0$ follows from $da/dt= j \times a$

Oh, yes! I straightly put $d^2a_y/{dt^{2}}=0$ without looking at the first derivative. Thanks...

 

[EmmaK]

$j$ X $a$ will be $a_z(t)i+0j-a_x(t).$

so,
$$\stackrel{da_y}{dt}$$=0 , $$\stackrel{da_x}{dt}$$=$a_z (t)$ and $$\stackrel{da_z}{dt}$$= $-a_x(t)$

 

[Dick]

$j$ X $a$ will be $a_z(t)i+0j-a_x(t).$

so,
$$\stackrel{da_y}{dt}$$=0 , $$\stackrel{da_x}{dt}$$=$a_z (t)$ and $$\stackrel{da_z}{dt}$$= $-a_x(t)$

That's a good start. Can you continue from there?

 

[EmmaK]

I need to differentiate $a_z (t)$ with respect to t... can i just say it's $-a_x$ ?

 

[Dick]

I need to differentiate $a_z (t)$ with respect to t... can i just say it's $a_z$ ?

Well, no. da_z/dt isn't just a_z. Your last result says it's -a_x. Try looking at the second derivative part. You want to show e.g. d/dt(da_z/dt))=(-a_z). How would that work?

 

[EmmaK]

ahh i meant to write 'is it just -a_x'

can you integrate both sides? so da_z/dt is -a_x t , which is a function of t??

 

[Dick]

ahh i meant to write 'is it just -a_x'

can you integrate both sides? so da_z/dt is -a_x t , which is a function of t??

a_x isn't a constant. You can't integrate it by multiplying it by t. Just differentiate da_z/dt, that will give you the second derivative, right?

 

[EmmaK]

ok...but how do i differentiate -a_x(t) ?

ohhh, it's just -(the x -component of j x a)?

 

[Dick]

ok...but how do i differentiate -a_x(t) ?

ohhh, it's just -(the x -component of j x a)?

Right. da/dt=jxa tells you how to differentiate the components of a.