Lie vs. covariant derivative: Visual motivation I'm currently teaching a course that "applies" differential geometry to computational problems but doesn't have time to go through theorems/proofs in detail.  We're taking a visual approach to help people see from a high level the differential geometry toolbox. I'd like to cover derivatives of vector fields on surfaces.  Both the Lie and covariant derivatives come up in such a lecture.
Is there a clear/concrete example of a pair of vector fields $(X,Y)$ on the plane that illustrates (1) why the Lie derivative $\mathcal L_X Y$ is different from the covariant derivative $\nabla_X Y$ and (2) why both derivatives might be useful in different contexts?
I'm looking for a succinct, plot-able visualization to help explain what's going on.
 A: At the encouragement of a commenter, I'll post my attempt at a simple example.  My apologies if this is completely incorrect.
Take $V(x,y):=(1,-y)$ and $W(x,y):=(-y,x)$, the circular vector field.
To compute the Lie derivative, note that the field $V$ implies a diffeomorphism of the plane $\psi_t(x,y)=(t+x,ye^{-t})$.  This effectively warps the $y$ axis.
The Jacobian of $\psi_t$ is:
$$
J_t(x,y)=\left(
\begin{array}{cc}
1 & 0\\0 &e^{-t}
\end{array}
\right)
$$
Now, let's compute a Lie derivative:
$$
\begin{array}{rl}
\mathcal L_V W(x,y)&=
\lim_{t\rightarrow0}\frac{1}{t}
[
(d\psi_{-t})_{\psi_t(p)}(W_{\psi_t(p)})-W_p
]\\
&=
\lim_{t\rightarrow0}\frac{1}{t}
\left[
\left(
\begin{array}{cc}
1&0\\0&e^{t}
\end{array}
\right)
\begin{pmatrix}
-ye^{-t}\\t+x
\end{pmatrix}
-
\begin{pmatrix}
-y\\x
\end{pmatrix}
\right]
\\
&=\lim_{t\rightarrow0}\frac{1}{t}\begin{pmatrix}
-y(e^{-t}-1)\\
e^{t}(t+x)-x
\end{pmatrix}
\\
&=\begin{pmatrix}
y\\1+x
\end{pmatrix}
\end{array}
$$
Because we're in the plane, the covariant derivative coincides with the directional derivative.  Hence:
$$
\begin{array}{rl}
\nabla_VW(x,y)
&=
\lim_{t\rightarrow0}\frac{1}{t}(W(x+tV_x(x,y),y+tV_y(x,y))-W(x,y))
\\
&=
\lim_{t\rightarrow0}\frac{1}{t}(W(x+t,(1-t)y)-W(x,y))
\\
&=
\lim_{t\rightarrow0}\frac{1}{t}\begin{pmatrix}
-(y-ty)+y\\
(t+x)-x
\end{pmatrix}
\\
&=\begin{pmatrix}
y\\1
\end{pmatrix}
\end{array}
$$
Notice the $y$ components of these two derivatives differs. 
A: Here's a qualitative illustration I have lying around, showing the difference between a vector field having zero Lie derivative and zero covariant derivative. 

Certainly doesn't capture the whole idea, but might help. I think the main idea I was trying to communicate is that covariant derivatives are all about a single flow line of $X$, while Lie derivatives care about all of $X$.
