Begining Complex Analysis Q's

[Guest]

:?: These problems seem easy but I have no idea how to do them.

For \(\displaystyle x=cos, y=sint, -\pi \leq t \leq 0\)please

a. draw the smooth curve, mark its initialand terminal point and determine its orientation.

For this part I know the initial point of a path y:[a,b] to C are the points y(a) and y(b) respectively...but what does that mean?

b. eliminate the parameter and express te curve in its rectangular form.

c. evaluate the line integral \(\displaystyle $\int{(\frac{y}{x^2+y^2}dx-\frac{x}{x^2+y^2}dy})$\)

and

Evaluate both line integrals of the function \(\displaystyle $M(x,y)=xy-y^2$\) along the path \(\displaystyle $x=t^2,y=t,1\leq t\leq 3.$\)

Thank You for your time, consideration and assistance.

 

[soroban]

Hello, BlueFalcon!

I can help you with the first two parts . . .

\(\displaystyle x\,=\,\cos\theta,\;\;y\,=\,\sin\theta,\;\;\; -\pi\,\leq\,\theta\,\leq\,0\)

a) Draw the curve, mark its initial and terminal point and determine its orientation.

It starts at: \(\displaystyle \,\begin{array}{cc}x\:=\:\cos(-\pi)\:=\:-1\\ y\:=\:\sin(-\pi)\:=\:0\end{array}\;\;\Rightarrow\;\;(-1,\,0)\)

And ends at: \(\displaystyle \:\begin{array}{cc}x\:=\:\cos(0)\:=\:1 \\ y\:=\:\sin(0)\:=\:0\end{array}\;\;\Rightarrow\;\;(1,0)\)

If we plot a few intermetdiate points,
\(\displaystyle \;\;\)we have the lower half of a unit circle at the origin.

The points are generated from the 9:00 position, counterclockwise to 3:00.

b) Eliminate the parameter and express te curve in its rectangular form.

We have: \(\displaystyle \:\begin{array}{cc}x\:=\:\cos x\\ y\:=\:\sin x\end{array}\)

Square: \(\displaystyle \:\begin{array}{cc}x^2\:=\:\cos^2x\\ y^2\:=\:\sin^2x\end{array}\)

Add: \(\displaystyle \:x^2\,+\,y^2\:=\:\cos^2x\,+\,\sin^2x\)

We have: \(\displaystyle \,x^2\,+\,y^2\:=\:1\;\;\Rightarrow\;\;y\:=\:\pm\sqrt{1\,-\,x^2}\)

Since we want the lower semicircle: \(\displaystyle \,y\:= \:-\sqrt{1\,-\,x^2}\)

 

[pka]

Because \(\displaystyle \cos ^2 (t) + \sin ^2 (t) = 1\), the denominators in part c are 1.
Thus \(\displaystyle \L
\int\limits_{ - 1}^0 {\left[ {\sin (t)( - \sin (t)) - \cos (t)(\cos (t))} \right]dt}\).

For the last line integral
\(\displaystyle \L
\begin{array}{l}
M(x,y) = xy - y^2 \\
x(t) = t^2 ,\quad y(t) = t\quad \Rightarrow \quad M(t) = t^3 - t^2 ,\quad dx = 2tdt,\quad dy = dy \\
\int\limits_C {M(x,y)dx} = \int\limits_1^3 {\left( {t^3 - t^2 } \right)(2tdt)} \\
\end{array}\).
You can set the other integral the same way.