Finding ω_c Given an Energy Surface E(k) in Constant Magnetic Field

[The Head]

Homework Statement

Consider the energy surface
E(k)=h²((k_x² +k_y² )/m_l+k_z²/m_t
where m_l is the transverse mass parameter and m_l is the longitudinal mass parameter. Use the equation of motion:
h(dk/dt)= -e(vXB) with v=∇_k(E)/h to show that ω_c=eB/(m_l*m_t)^1/2 when the static magnetic field B lies in the x-y plane

Homework Equations

a=(1/h)(d²E(k)/dk²)(dk/dt)
ω=a/v

The Attempt at a Solution

I've tried quite a number of things and spent several hours unsuccessfully, but one route I tried was this:
v= 1/h(∇k(E))=1/h(h²)(k_x(i-hat)/m_l + k_y(j-hat)/m_l + k_z(k-hat)/m_t)
B is in x-y plane, so B= B_x (i-hat) + B_y (j-hat)
so the cross product of v & B = k_z*B_y/(m_l) (i-hat) - k_z*B_x/(m_l) (j-hat) +(k_y*B_x - K_x*B_y)/(m_t) (k-hat)

Now, this is where I am not sure what to do, but I figured I would try the equation:
a=(1/h)(d²E(k)/dk²)(dk/dt) with the idea that this would be centripetal acceleration = ωv, and this way I could solve for ω_c


d²E(k)/dk²= h²(((i-hat)+(j-hat))/m_l +(k-hat)/m_t

dk/dt=-e/h(vXB)=-e/h*k_z*B_y/(m_l) (i-hat) - k_z*B_x/(m_l) (j-hat) +(k_y*B_x - K_x*B_y)/(m_t) (k-hat)

I doted dk/dt with vXB to get rid of the vectors, and divided by h to get:

a=e/m_tm_l(B_x*(k_z-k_y)+B_y(k_x-k_z))=ωv

Although this seems to have somewhat the form I am going for, I have two problems: I don't know how to get rid of all the different k components, and if I divide by the vector v, then I have an answer for ω with components.

Can someone provide some guidance? Thank you!

 

[The Head]

Or perhaps the method I chose isn't even close...if so could someone please just give me a hint on a better method? I don't know if what I'm doing is just creating a bunch of smoke, or if it's on the correct path. I've been struggling with this problem for days!