# vectors and differentiation

• Jan 8th 2010, 06:29 AM
Emma L
vectors and differentiation
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 $da/dt= j (Vector product) a$
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 $da/dt(0)=0k$

i have absolutely no idea how to start...
• Jan 8th 2010, 08:14 AM
Shanks
can you use latex to restate your problem?
• Jan 9th 2010, 03:16 PM
Emma L
Hope that's a bit clearer :) sorry, wasnt sure how to do the vector product in latex(Doh)
• Jan 9th 2010, 03:20 PM
Drexel28
Quote:

Originally Posted by Emma L
Hope that's a bit clearer :) sorry, wasnt sure how to do the vector product in latex(Doh)

I'm sorry. This may be a stupid question, but what is "vector product"? Is it the cross product or the dot product? Both involve vectors.

EDIT: It's obvious by context that it is cross product.

$\times$ is given by \times
• Jan 9th 2010, 04:21 PM
Emma L
ye sorry, it's the cross product
• Jan 10th 2010, 07:00 AM
HallsofIvy
Quote:

Originally Posted by Emma L
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 $da/dt= j (Vector product) a$
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 $da/dt(0)=0k$

i have absolutely no idea how to start...

So the problem is that you are given $\vec{a}$ such that $\frac{d\vec{a}}{dt}= \vec{j}\times \vec{a}$ and you are asked to show that, in that case, $\frac{d^2a_x}{dt^2}= -a_x$, $\frac{da_y}{dt}= 0$, and $\frac{d^2a_z}{dt^2}= -a_z$.

Okay, go ahead and calculate $\vec{j}\times \vec{a}$. That's easy, it is just $a_z\vec{i}- a_x\vec{k}$. Setting $\frac{d\vec{a}}{dt}= \frac{da_x}{dx}\vec{i}+ \frac{da_y}{dt}\vec{j}+ \frac{da_z}{dt}\vec{k}$ equal to that gives you three equations: $\frac{da_x}{dt}= a_z$, $\frac{da_y}{dt}= 0$, and $\frac{da_z}{dt}= -a_x$.

differentiating the first of those, with respect to t, gives $\frac{d^2 a_x}{dt^2}= \frac{d^a_z}{dt}= -a_x$. Get the point? I'll leave the others to you now.

For the last part you need to solve those equations. $\frac{da_y}{dt}= 0$ is easy: $a_y$ is a constant and the last part tells us that that constant is 1.

To solve the other two, use the fact, that you have now shown, that $\frac{d^2a_x}{dt^2}= -a_x$ and $\frac{d^2a_z}{dt^2}= -a_z$. Solve those differential equations using the initial values $a_x(0)= 1$, $a_x'(0)= 0$, $a_z(0)= 0$, and $a_z'(0)= 0$.