differentiate trisecrix, strophoid, rosette

[mppbee]

can someone differentiate trisectrix, strophoid, and rosette in terms of graphs and equation?
sorry but this is the first time ive encountered those terms.
advance thank you for helping


and can someone help me convert this rectangular coordinate (x+3)² + (y+3)² = 18 to polar coordinate?
im trying to do it but i always come up with r² + rsin rcos = 0 . is that right?
im not sure if it is.


ohh i forgot that its said to solve in terms of r.
so i continued r² + rsin rcos = 0
to 1/r [r² = rsin rcos] 1/r divide both sides by r
and became r = sin cos .
is that right?
please help.

 

[stapel]

1. Differentiate trisectrix, strophoid, and rosette in terms of graphs and equation.

sorry but this is the first time ive encountered those terms.

How many graphs of each type have you studied? What where their equations? What did you think about the differences and similarities in their graphs?

2. Convert this rectangular coordinate (x+3)² + (y+3)² = 18 to polar coordinates.

im trying to do it but i always come up with r² + rsin rcos = 0 . is that right?
What were your steps? (And where is the $\displaystyle \theta$ for the angle measure inside the trig functions?)

You know that the rectangular-coordinate equation is for a circle centered at (h, k) = (-3, -3) with a radius of r = sqrt[18]. What did you see when you did the polar-coordinate graph? Did it produce the same circle?

3. Solve in terms of r.
so i continued r² + rsin rcos = 0
to 1/r [r² = rsin rcos] 1/r divide both sides by r

The last line above is somewhat cryptic. Did you mean that you first subtracted the second term to the right-hand side, so you had the following?

. . . . .$\displaystyle r^2\, +\, r\sin\left(\theta\right)\cos\left(\theta\right)\, =\, 0$

. . . . .$\displaystyle r^2\, =\, -r\sin\left(\theta\right)\cos\left(\theta\right)$

Then you divided by sides by r, to get:

. . . . .$\displaystyle r\, =\, -\sin\left(\theta\right)\cos\left(\theta\right)$

(This does not match what you'd posted, is why I ask.) Thank you! ;-)