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.
 A: The length of the arc can be defined as
$$ds=\sqrt{dx^2+dy^2}$$
Consider the family of curves lying wholly on a given surface and passing through two given points both on the same plane. Among these curves there will be one for which the length of arc is a minimum. Such curves are known geodesics. Because of that we are required to minimize
$$L=\int ds=\int \sqrt{dx^2+dy^2}$$
where because of parameterization
$$x=u\cos(v)\rightarrow dx=du\cos (v)-u\sin(v)dv$$
$$y=u\sin(v)\rightarrow dy=du\sin (v)+u\cos(v)dv$$
By rewriting the integral
$$L=\int \sqrt{dx^2+dy^2}=\int \sqrt{du^2+u^2dv^2}=\int dv\sqrt{\bigg(\frac{du}{dv}\bigg)^2+u^2}$$
The Euler Lagrange equation (for sake of simplicity $u'=\frac{du}{dv}$)
$$\frac{\partial L}{\partial u}-\frac{d}{dv}\bigg(\frac{\partial L}{\partial u'}\bigg)=0$$
$$\frac{u}{\sqrt{u^2+u'^2}}-\frac{d}{dv}\bigg(\frac{u'}{\sqrt{u^2+u'^2}}\bigg)=0$$
$$\frac{u}{\sqrt{u^2+u'^2}}-u\frac{u\,u''-u'^2}{(u^2+u'^2)^{3/2}}=0$$
$$u\frac{u^2+u'^2-u\,u''}{(u^2+u'^2)^{3/2}}=0$$
One solution is $u=0$ and the other one
$$u^2+u'^2-u\,u''=0\Rightarrow u=A\sec(v+B)$$
where $A$ and $B$ are cosntants. In cartesian coordinates parametric equations for the geodesics are
$$x=u\cos(v)=A\sec(v+B)\cos(v)=A\frac{\cos(v)}{\cos(v+B)}$$
$$y=u\sin(v)=A\sec(v+B)\sin(v)=A\frac{\sin(v)}{\cos(v+B)}$$
To eliminate $v$ we can use trigonometric identities and change of variable such that $w=v+B$ without loss of generality
$$x=A\frac{\cos(w-B)}{\cos(w)}=A\cos(B)-A\sin(B)\frac{\sin(w)}{\cos(w)}\rightarrow \frac{\sin(w)}{\cos(w)}=\cot(B)-\frac{x}{A\sin(B)}$$
and
$$y=A\frac{\sin(w-B)}{\cos(w)}=A\cos(B)\frac{\sin(w)}{\cos(w)}-A\sin(B)$$
by replacing $\frac{\sin(w)}{\cos(w)}$
$$y=A\cos(B)\bigg(\cot(B)-\frac{x}{A\sin(B)}\bigg)-A\sin(B)$$
$$y=-\cot(B)\,x+A\big(\cos(B)\cot(B)-\sin(B)\big)$$
which is a straight line.
A: 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 $2\dot{u}\dot{v}+u \ddot{v}=0$.
