Question

What is the standard form of the equation with rho=18, phi=(5pi)/4?

Please no direct answers

I have gotten to:
x cos phi + y sin phi - p=0
x cos (5 pi/4) + y sin (5 pi/4) - 18 = 0

But I have no idea on where to go from here.

Answer 1

@Nnesha

Answer 2

@Directrix

Answer 3

you rectangular form?

Answer 4

the very fact that you talking about rho=18
i think you were essentially looking for a rectangular equation

Answer 5

converting from spherical to rectangular
x^2+y2+z^2=rph^2

Answer 6

rho^2

Answer 7

hey talk to me! haha so i know what exactly you want to find

Answer 8

Sorry haha was afk

Answer 9

i see this is not conversion anyway

Answer 10

I have no idea...the information ^^ up there was all I was given.

Answer 11

so the standard form is given by xcos(phi)+ysin(phi)-p=0

Answer 12

which you already did
xcos(5pi/4)+ysin(5pi/4)-18=0

Answer 13

yeah...

Answer 14

now you just need to evaluate cos(5pi/4) and sin(5pi/4)

Answer 15

you should be familiar with trigonometry, so what the problem here?

Answer 16

One second..

Answer 17

cos(5pi/4)=?

Answer 18

.998

Answer 19

sin(5pi/4)=.068

Answer 20

don't use calculator!
if you know trig you know that \[\cos (\pi/4) =\sqrt{2}/2\]

Answer 21

-_-

Answer 22

but we are looking for 5pi/4 so you need to know what quadrant is the angle 5pi/4 in

Answer 23

that angle is in the 3th quadrant so cos there is negative
then cos(5pi/4)=-sqrt(2)/2

Answer 24

with the same strategy sin(5pi/4)=-sqrt(2)/2
so the equation becomes
\[-\frac{\sqrt{2}}{2 }x-\frac{\sqrt{2}}{2}y=18\]

Answer 25

Sorry