Compute the pushforward of the vector field $\frac{\partial}{\partial x^1}$ via a given function $\phi$ We define $\phi:=(\phi^1,\phi^2):\Omega\subset\mathbb{R}^2\to\phi(\Omega)$ with $\Omega$ such that $\phi$ is a diffeomorphism by
$$x^1:=\phi^1(r,\theta)=r\cos\theta\qquad\text{and}\qquad x^2:=\phi^2(r,\theta)=r\sin\theta$$
for $r,\theta\in\mathbb{R}$.
Compute the pushforward $(\phi^{-1})_*V$ of the vector field $V(x^1,x^2) = \frac{\partial }{\partial x^1}$
 A: The OP clearly does not know where to start, so I will try to give a full answer.
Forget all of the fancy notation. Your function $\phi$ changes from polar coordinates to Cartesian coordinates. If you give it an $r$ and a $\theta$ then $\phi$ will give you an $x$ and a $y$. We have:
$$\phi : (r,\theta) \longmapsto (r\cos\theta,r\sin\theta).$$
The Jacobian matrix of $\phi$ is the matrix of partial derivatives:
$$J_{\phi} = \left[\begin{array}{cc} \partial\phi_1/\partial r & \partial\phi_1/\partial \theta \\ \partial\phi_2/\partial r & \partial\phi_2/\partial \theta\end{array}\right] = \left[\begin{array}{cc} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{array}\right]$$
Notice that $\det(J_{\phi}) = r\cos^2\theta + r\sin^2\theta \equiv r$, and so $\phi$ is a diffeomorphism if and only if $r \neq 0$.
Given a vector field in polar form, say $v=a(r,\theta)\partial_r + b(r,\theta)\partial_{\theta}$, we can find $(\phi_*)(v)$:
$$(\phi_*)(v) \sim \left[\begin{array}{cc} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{array}\right]\left[\begin{array}{c} a \\ b \end{array}\right]=\left[\begin{array}{c} a\cos\theta-rb\sin\theta \\ a\sin\theta + br\cos\theta \end{array}\right]$$
Hence $(\phi_*)(v) = (a\cos\theta-rb\sin\theta)\partial_x+(a\sin\theta+br\cos\theta)\partial_y$.
Similarily, given the vector field $\partial_x$, we might want to know which vector field in polar form $v$, gets sent to $\partial_x$. For this you want $(\phi_*)(v)=\partial_x$. In other words, $v=(\phi^{-1}_*)(\partial_x)$. Let $v=a\partial_r+b\partial_\theta$, then $(\phi_*)(a\partial_r+b\partial_{\theta}) = \partial_x$ becomes the matrix equation
$$\left[\begin{array}{cc} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{array}\right]\left[\begin{array}{c} a \\ b \end{array}\right]=\left[\begin{array}{c} 1 \\ 0 \end{array}\right]$$
The matrix of $\phi$ is $J_{\phi}$ and the matrix of $\phi^{-1}$ is $(J_{\phi})^{-1}$. If we multiply on the left by the inverse of $J_{\phi}$ we get $a = \cos\theta$ and $b=-\frac{1}{r}\sin\theta$. Hence:
$$(\phi_*)(\cos\theta\partial_r-\tfrac{1}{r}\sin\theta\partial_{\theta}) = \partial_x$$
Equivalently, this may be written as
$$(\phi^{-1}_*)(\partial_x) = \cos\theta\partial_r-\tfrac{1}{r}\sin\theta\partial_{\theta}$$
However, we need to write $r$ and $\theta$ in terms of $x$ and $y$. Notice that since $x=r\cos\theta$ and $y=r\sin\theta$ we have $r=\sqrt{x^2+y^2}$ and so:
$$(\phi^{-1}_*)(\partial_x) = \frac{x}{\sqrt{x^2+y^2}}\partial_r-\frac{y}{\sqrt{x^2+y^2}}\partial_{\theta}$$
The vector $\partial_x$ based at $(x,y)$ gets sent the what is written above.
A: To elaborate upon Fly by Night's response, if $F: M \to N$ is a smooth map and $p\in M$ and $f: M \to \mathbb R$, then in a coordinate chart $(x^i)$ about $p$ and $(y^i)$ about $F(p)$ we have
$$\left(F_*\left. \frac{\partial}{\partial x^i}\right|_p\right)f = \left( \frac{\partial F^j}{\partial x^i}(p) \frac{\partial}{\partial y^j}\right) f$$
So the Jacobian of your transformation describes how to compute pushforwards in coordinates. 
A: HINT: 
If you have fixed coordinates then the Jacobian matrix is the matrix of the push-forward. 
Calculate the Jacobian and then multiply it by the appropriate vector.
