Let $f: \mathbb{R}^2\to \mathbb{R}^2$ given by $f(x, y) = (e^x \cos y, e^x \sin y)$.

Question

Let $f: \mathbb{R}^2\to \mathbb{R}^2$ given by $f(x, y) = (e^x \cos y, e^x \sin y)$.

Take $S$ to be the set $S = [0, 1]\times [0, \pi]$.

(a) Calculate $Df$ and $\det Df$.

(b) Sketch the image under $f$ of the set $S$.

We remark that if one identifies $\mathbb{C}$ with $\mathbb{R}^2$ as usual, then $f$ is the function $f(z) = e^z$.

For (a), $Df(x,y)=\begin{bmatrix}e^x \cos y & -e^x \sin y\\e^x \sin y & e^x \cos y\end{bmatrix}$ and $\det \begin{bmatrix}e^x \cos y & -e^x \sin y\\e^x \sin y & e^x \cos y\end{bmatrix}=e^{2x}\cos^2 y+e^{2x}\sin^2 y=e^{2x}$

I do not understand what I have to do in (b), could someone help me please? Thank you

Answer 1

Hint: The domain is a rectangular region. Start by sketching the images of its sides. E.g., one side of the rectangle is $\{(t,0)\mid t\in[0,1]\}$. Plug this into $f$, producing $(e^t,0)$, and draw the resulting curve.

Answer 2

Hint: Fix an $x$. Can you draw the image of $\{x\}\times [0,\pi]$ under $f$? What happens when you start varying $x$?