Heisenberg's Uncertainty Principle Question

[stephen8686]

Homework Statement

Show that the smallest possible uncertainty in the position of an whose speed is given by $\beta=\frac{v}{c}$ is: $$\Delta x_{min}=\frac{h}{4\pi m_o c}\sqrt{1-\beta^2}$$

Relevant Equations

$$\Delta x \Delta p=\frac{h}{4\pi}$$

So with the \gamma=\frac{1}{\sqrt{1-\beta^2}} it seems obvious that relativistic momentum, p=\gamma m_o v is supposed to be used.
Then \frac{ dp}{dv}=m_o(1-\beta^2)^{-1/2}+m_o v (\frac{-1}{2}(1-\beta^2)^{-3/2}(\frac{-2v}{c^2}))=m_o(\frac{1}{\sqrt{1-B^2}}+\frac{\beta}{(1-\beta^2)^{3/2}})=\frac{m_o}{(1-B^2)^{3/2}}
so \Delta p=\frac{m_o\Delta v}{(1-B^2)^{3/2}}

But this doesn't exactly fit the expression that I'm supposed to show. I don't know what do do with the \Delta v. In addition to this, I don't think I conceptually understand the question. Why should there be a \Delta x_{min}? Why can't the uncertainty in momentum be huge and the uncertainty in x be reduced to a dirac delta function?

 

[vela]

Hint: You know the speed, but you don't know the direction the object is moving.