# vectors and differentiation

• Jan 8th 2010, 05:29 AM
Emma L
vectors and differentiation
The vector a depends on a parameter t, i.e. $\displaystyle a=a(t)=a_x(t)i +a_y(t)j +a_z(t)k$.. it satisfies the equation $\displaystyle da/dt= j (Vector product) a$
show that $\displaystyle d^2a_x/dt^2 =-a_x$ , $\displaystyle da_y/dt=0$ and $\displaystyle d^2a_z/dt^2 =-a_z$.

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

i have absolutely no idea how to start...
• Jan 8th 2010, 07:14 AM
Shanks
can you use latex to restate your problem?
• Jan 9th 2010, 02:16 PM
Emma L
Hope that's a bit clearer :) sorry, wasnt sure how to do the vector product in latex(Doh)
• Jan 9th 2010, 02:20 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.

$\displaystyle \times$ is given by \times
• Jan 9th 2010, 03:21 PM
Emma L
ye sorry, it's the cross product
• Jan 10th 2010, 06:00 AM
HallsofIvy
Quote:

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

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

i have absolutely no idea how to start...

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

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

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

For the last part you need to solve those equations. $\displaystyle \frac{da_y}{dt}= 0$ is easy: $\displaystyle a_y$ 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 $\displaystyle \frac{d^2a_x}{dt^2}= -a_x$ and $\displaystyle \frac{d^2a_z}{dt^2}= -a_z$. Solve those differential equations using the initial values $\displaystyle a_x(0)= 1$, $\displaystyle a_x'(0)= 0$, $\displaystyle a_z(0)= 0$, and $\displaystyle a_z'(0)= 0$.