Question

give the corresponding equation in rectangular coordinates

Answer 1

\[\huge \color {blue} { r=e ^{a \theta}}\]

Answer 2

well, i recall that x = r cos(t), and y = r sin(t)

Answer 3

that might suggest solving t in terms of x, and subbing it into y

Answer 4

\[\huge \ln(r)=a \theta \text{ i wanna sketch this}\]

Answer 5

\[r=e^{a \theta}\implies \ln(r)=a \theta\]

Answer 6

no symmetry around any axis!!!

Answer 7

Looks to me like you could just substitute away... after some clever manipulations, probably...
\[\huge r = \sqrt{x^2+y^2}\]\[\huge \tan\theta = \frac{y}x\]

Answer 8

\[\frac{1}{2}\ln(x^2+y^2)=a \tan^{-1}(\frac{ x }{ y })\]

Answer 9

not looking good

Answer 10

don't make the equation feel bad about itself :D

Answer 11

And if that's the case, I'm out of ideas :)

Answer 12

my idea was that since parametrics define the rectangular coords seperately, and since the original curve is not a "function" ....

Answer 13

theres not problem in plotting

Answer 14

@amistre64 cud we use that o sketch the polar cordinates

Answer 15

you could yes; not to sure what the a is in e^(a theta) im thinking its just an arbitrary constant at the moment.