Polar Coordinates

V

vanbeersj

New member
Joined
Aug 6, 2008
Messages
31
The polar equation of the path of a weather satellite of the earth is r = 7600 / (1+ 0.14cosø), where r is measured in km's. Find the rectangular equation of the path of this satellite. The path is an ellipse, with the earth at one of the foci.

I know that r = (x^2 +Y^2)^1/2 but the difficulty I have right now is simplifying and reducing the 1+0.14cosø. I'm stuck on this could someone point me in the right direction?

Thanks
 
Hello, vanbeers!

The algebra gets really ugly . . .

The polar equation of the path of a weather satellite of the earth is:
. . . .\(\displaystyle r \:=\:\frac{7600}{1+ 0.14\cos\theta}\) . where \(\displaystyle r\) is measured in km's.

Find the rectangular equation of the path of this satellite.
(The path is an ellipse, with the earth at one of the foci.)
Click to expand...

We have: .\(\displaystyle r + 0.14\underbrace{r\cos\theta} \;=\;7600\)
Then: .\(\displaystyle \overbrace{\sqrt{x^2+y^2}} + \;0.14x \:=\:7600 \quad\Rightarrow\quad \sqrt{x^2+y^2}\;=\;7600 - 0.14x\)

Square: .\(\displaystyle x^2 + y^2 \;=\;7600^2 - 2128x + 0.0196x^2 \quad\Rightarrow\quad 0.9804x^2 + 2128x + y^2 \;=\;7600^2\)


Those numbers are too clunky to carry around!

\(\displaystyle \text{Let: }\;\begin{Bmatrix}a &=& 0.9804 \\ b &=& 2128 \\ c &=& 7600^2 \end{Bmatrix}\)

\(\displaystyle \text{So we have: }\;ax^2 + bx + y^2 \;=\;c \quad\Rightarrow\quad a\left(x^2 + \frac{b}{a}x\quad\right) + y^2 \;=\;c\)

\(\displaystyle \text{Complete the square: }\;a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}\right) + y^2 \;=\;c + \frac{b^2}{4a}\)

\(\displaystyle \text{And we have: }\;a\left(x + \frac{b}{2a}\right)^2 + y^2 \;=\;\frac{b^2+4ac}{4a}\)


\(\displaystyle \text{Divide by the constant: }\;\frac{(x+\frac{b}{2a})^2}{\frac{b^2+4ac}{4a^2}} + \frac{y^2}{\frac{b^2+4ac}{4a}} \;=\;1\)


I'll let you back-substitute those constants . . .

 
You must log in or register to reply here.
Top