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\timesa
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 b = mod(a)mod(b)sin\theta[/tex] \hat{n}

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\timesa
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 b = mod(a)mod(b)sin\theta[/tex] \hat{n}

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.