Find Tangent Slope with Polar coordinates

[JSGhost]

Homework Statement

Find the slope of the tangent line to the given polar curve at the point specified by the value of θ.
r = 9sin(θ)
θ = pi/6

Homework Equations

dy/dx = (dy/dθ) / (dx/dθ)
x=rcosθ
y=rsinθ

(sinx)^2 = (1/2)(1-cos2x)
(cosx)^2 = (1/2)(1+cos2x)
2sinxcosx = sin(2x)

The Attempt at a Solution

r = 9sin(θ)
x=rcos(θ)=9sin(θ)cos(θ)= --> Would this be sin(9θ) or sin(18θ)?
y=rsin(θ)=9sin(θ)sin(θ)= 9sin^2(θ)dy/dx = (dy/dθ) / (dx/dθ) = (9 * 9sinθcosθ) / (cos(18θ) * 9) = sin(18θ)/cos(18θ) = tan(18θ)

θ = pi/6
tan(18*pi/6)=tan(3pi)=0

 

[tiny-tim]

Welcome to PF!

Hi JSGhost! Welcome to PF!

(have a pi: π and try using the X² tag just above the Reply box )

r = 9sin(θ)
x=rcos(θ)=9sin(θ)cos(θ)= --> Would this be sin(9θ) or sin(18θ)?

Nooo! you can't move numbers in and out of a function like sin (you wouldn't do it for √, would you? )

y=rsin(θ)=9sin(θ)sin(θ)= 9sin^2(θ)


dy/dx = (dy/dθ) / (dx/dθ) = (9 * 9sinθcosθ) / (cos(18θ) * 9) = sin(18θ)/cos(18θ) = tan(18θ)

But you're not differentiating.

Start again. ​

 

[JSGhost]

Thanks for replying. I actually figured it out after posting this topic. Saw a similar problem to this one on another site. Thanks anyways.

Find the slope of the tangent line to the given polar curve at the point specified by the value of θ.
r = 9sin(θ)
θ = pi/6

x=rcos(θ)=9sin(θ)cos(θ)= 9(sin(2θ))
y=rsin(θ)=9sin(θ)sin(θ)= 9sin^2(θ)

dx/dθ = 9(cos^2(θ) - sin^2(θ))
dy/dθ = 9(2sinθcosθ)

dy/dx = (dy/dθ)/(dx/dθ)
= 9(2sinθcosθ) / 9(cos^2(θ) - sin^2(θ))
= [2sinθcosθ] / [cos^2(θ)-sin^2(θ)]

when θ = pi/6
dy/dx = [2sin(pi/6)cos(pi/6)] / [cos^2(pi/6) - sin^2(pi/6)]
= [2*(1/2)*(sqrt(3)/2)] / [(sqrt(3)/2)*(sqrt(3)/2)-(1/2)*(1/2)]
= (sqrt(3)/2) / 2/4 = (sqrt(3)/2) / (1/2) = sqrt(3)