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

Question

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

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

Mr. Lisp · Answer

For b)  Note that Firs value is $f(x,0)=(e^x,0).$
Since $xin [0,1]$ this is the line segment between the points $(1,0)$ and $(e,0)$.
Second value is $f(x,pi)=(-e^x,0).$
Since $xin [0,1]$ this is the line segment between the points $(1,0)$ and $(-e,0)$.
Third value is $f(0,theta)=(cos theta, sin theta).$
Since $thetain [0,pi],$  that gives the upper unit semicircle.
Final value is $f(1,theta)=(e cos theta, e sin theta).$
Since $theta in [0, pi]$, that gives the upper unit semicircle which its center at the origin and with radius $e.$
The graph is

Ennar · Answer

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

amd · Answer

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

Answered by amd on December 20, 2021

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

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