Polar coordinates & Rectangular coordinates

[shivers20]

1. Convert polar coordinates (6, pi/3) to rectangular coordinates

x=6cos pi/3= 3
y=6sin pi/3= 3 sqrt[3]

Answer: (3, 3 sqrt[3])


2. Convert rectangular coordinates (4,2) to polar coordinates.

r=sqrt[4^2+2^2]=sqrt[20]= 2sqrt[5]

then tan O'= 2/4 = 1/2, therefore O' = ? I couldn't find what tan 1/2 converts to? is it undefined?
Answer: (2sqrt[5], 1/2 or whatever tan is)


3. Convert the equation x=y^2 to a polar equation.Answer: r cos (O') = r sin (O')^2

4. Convert r^2= sin(O') cos(O') to an equation in x and y.
substitute: x^2 + y^2 for r^2, x for sin(O') and y for cos (O')

x^2+y^2=xy .................. is y, cos and x for sin or am I right?

5. Sketch the graph of the polar equation r = cos (2O') (0<= O' <= 2pi)

I cant seem to graph this on my calculator. Any hints.



O' represents degree.

 

[shivers20]

I figured out the last problem. Computing the values for r corresponding to some convenient values of θ and sketching the polar curve: I use a chart and plug in example would be: pi/12 = 15 degrees then insert into equation cos (2 (15)) which gives me cos30= 0.86 and continue doing this until I have enough points to sketch a curve.

 

[galactus]

1. Convert polar coordinates (6, pi/3) to rectangular coordinates

x=6cos pi/3= 3
y=6sin pi/3= 3 sqrt[3]

Answer: (3, 3 sqrt[3])

YEP!!. GOOD


2. Convert rectangular coordinates (4,2) to polar coordinates.

r=sqrt[4^2+2^2]=sqrt[20]= 2sqrt[5]

then tan O'= 2/4 = 1/2, therefore O' = ? I couldn't find what tan 1/2 converts to? is it undefined?
Answer: (2sqrt[5], 1/2 or whatever tan is)

\(\displaystyle tan^{-1}(\frac{1}{2})=0.463647609001\;\ radians= \;\ 26.565 \;\ degrees\)


3. Convert the equation x=y^2 to a polar equation.Answer: r cos (O') = r sin (O')^2

\(\displaystyle rcos({\theta})=r^{2}sin^{2}({\theta})\)

\(\displaystyle r=\frac{cos({\theta})}{sin^{2}({\theta})}=cot({\theta})csc({\theta})\)

4. Convert r^2= sin(O') cos(O') to an equation in x and y.
substitute: x^2 + y^2 for r^2, x for sin(O') and y for cos (O')

x^2+y^2=xy .................. is y, cos and x for sin or am I right?

\(\displaystyle \frac{x}{r}=cos({\theta})\;\ and \;\ \frac{y}{r}=sin({\theta})\)

\(\displaystyle x^{2}+y^{2}=(\frac{y}{\sqrt{x^{2}+y^{2}}})(\frac{x}{\sqrt{x^{2}+y^{2}}})\)

5. Sketch the graph of the polar equation r = cos (2O') (0<= O' <= 2pi)

\(\displaystyle \L\\r=cos(2{\theta})\)

I cant seem to graph this on my calculator. Any hints.



O' represents degree.