Question

Slope of polar equation help needed! r= 2sin(theta)

Answer 1

equation is r= 2sin(theta)
so X = rcos(theta)
Y = rsin(theta)

Answer 2

i take deriv of X to get dx = -rsin(theta) and of Y to get dy = rcos(theta)
put dy/dx to get -cos(theta)/sin(theta)

Answer 3

so far so good?

Answer 4

plug in theta = pi/6 to ultimately get - sqrt3. but its wrong

Answer 5

do i need to convert to cartesian form first before taking derivs?

Answer 6

\[\frac{dy}{dx}=\frac{\frac{d r}{d \theta}\sin(\theta)+rcos(\theta)}{\frac{d r}{d \theta}\cos(\theta)-rsin(\theta)}\]

Answer 7

\[r=2\sin(\theta) \longrightarrow \frac{dr}{d \theta}=2\cos(\theta)\]

Answer 8

\[\frac{dy}{dx}=\frac{2\cos(\theta)\sin(\theta)+2\sin(\theta)\cos(\theta)}{2\cos(\theta)\cos(\theta)-2\sin(\theta)\sin(\theta)}\]

Answer 9

the equation editor is not posting

Answer 10

= 2(2sin(2theta))/2(cos(2theta))=2tan(2theta)

Answer 11

and that is ur slope, if theta equals to a certain value, then u could also find a value for that, if u were to find the equation of a line then u use x=rcos(theta) and y=rsin(theat)...

Answer 12

i hope this helps