imporant question about geodesic plane I have question

Q
  Show that geodesics in the plane are straight lines. Use $X(u,v)=(u\cos(v),u\sin(v))$
  as the parametrization of the plane.

I hope someone can answer.
Thanks
please at least give to me hint to solve this question thanks
Proposition 9.1.4
Any (part of a) straight line on a surface is a geodesic.
By this, we mean that every straight line can be parametrized so that it is
a geodesic. A similar remark applies to other geodesics we consider and whose
parametrization is not specified (see Exercise 9.1.2).
Proof
This is obvious, for any straight line has a (constant speed) parametrization of
the form
γ(t) = a + bt,
where a and b are constant vectors, and clearly ¨γ = 0.
Example 9.1.5
All straight lines in the plane are geodesics, as are the rulings of any ruled
surface, such as those of a (generalized) cylinder or a (generalized) cone, or the
straight lines on a hyperboloid of one sheet.

You have a curve in a surface. This curve has a "velocity" vector which is tangent to the surface. It also has an "acceleration" vector. The curve is called a geodesic if there is no tangential component to this acceleration. Let's consider your idea: ${\bf X}(u,v) = (u \cos v, u \sin v , 0)$, where $u$ and $v$ can be thought of as functions of a parameter $t$. To find the velocity vector we differentiate with respect to $t$:
$$\dot{\bf X} = (\dot{u}\cos v - u\dot{v}\sin v, \dot{u}\sin v + u\dot{v}\cos v, 0)$$
To find the acceleration vector, we need to differentiate a second time:
$$\ddot{\bf X} = ((\ddot{u}-u\dot{v}^2)\cos v - (2\dot{u}\dot{v}+u \ddot{v})\sin v, (\ddot{u}-u\dot{v}^2)\sin v + (2\dot{u}\dot{v}+u\ddot{v})\cos v,0)$$
For ${\bf X}(t)$ to be a geodesic, we need there to be no tangential component to the acceleration. Since ${\bf X}$ is the $xy$-plane, that means we need no $x$- and no $y$-components to $\ddot{\bf X}$. Hence:
\begin{array}{ccc}
(\ddot{u}-u\dot{v}^2)\cos v - (2\dot{u}\dot{v}+u \ddot{v})\sin v &=& 0 \\
(\ddot{u}-u\dot{v}^2)\sin v + (2\dot{u}\dot{v}+u\ddot{v})\cos v &=& 0
\end{array}
Since sine and cosine are linearly independent functions, we need $\ddot{u}-u\dot{v}^2=0$ and $2\dot{u}\dot{v}+u \ddot{v}=0$.
