How do you find the polar equation to the tangent, at $θ = \frac{π}{12}$ $theta = pi/12$ , on $r = sin 3 θ$ $r = sin 3theta$ ?

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How do you find the polar equation to the tangent, at $θ = \frac{π}{12}$ $theta = pi/12$ , on $r = sin 3 θ$ $r = sin 3theta$ ?

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Answer 1 · 2017-01-24T10:19:40

So, $r = \frac{1}{4 cos θ - 2 sin θ}$ $r =1/(4cos theta - 2sin theta )$ . In cartesian form, this is $4 x - 2 y - 1 = 0$ $4x-2y-1=0$

Explanation:

The given equation represents a three-petal rose.

The equation to the tangent

at $P (f (α), α)$ $P (f(alpha), alpha)$ , on $r = f (θ)$ $r = f(theta)$ is

$r (sin θ - tan ψ cos θ) = f (α) (sin α - tan ψ cos α)$ $r(sin theta-tan psi cos theta) =f(alpha)(sin alpha-tan psi cos alpha)$ ,

where

$tan ψ = m =$ $tan psi = m=$

$(\frac{d r}{d θ} sin α + r cos α)$ $((dr)/(d theta )sinalpha+ r cos alpha)$ $/ (\frac{d r}{d θ} cos θ - r sin θ)$ $/((dr)/(d theta)cos theta- r sin theta)$ , at P.

Briefly,

$r sin (θ - ψ) = f (α) sin (α - ψ)$ $r sin (theta-psi)= f(alpha)sin(alpha-psi)$

Here, $r = sin 3 θ and \frac{d r}{d θ} = 3 cos 3 θ$ $r = sin 3 theta and (dr)/(d theta)=3 cos 3 theta$ .

P is $(\frac{1}{\sqrt{2}}, \frac{π}{12})$ $(1/sqrt2, pi/12)$ ,

m at P = $\frac{\frac{3}{2} + \frac{1}{2}}{\frac{3}{2} - \frac{1}{2}} = 2$ $(3/2+1/2)/(3/2-1/2)= 2$ .

Now, the equation to the tangent at P is

$r (sin θ - 2 cos θ) = \frac{1}{\sqrt{2}} (\frac{1}{\sqrt{2}} - \frac{2}{\sqrt{2}}) = - \frac{1}{2}$ $r(sin theta- 2 cos theta) = 1/sqrt2(1/sqrt2-2/sqrt2)=-1/2$

So, $r = 1 / (4 cos θ - 2 sin θ)$ $r =1//(4cos theta - 2sin theta )$

In cartesian form, this is 4x-2y-1=0#

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How do you find the polar equation to the tangent, at $θ = \frac{π}{12}$ $theta = pi/12$ , on $r = sin 3 θ$ $r = sin 3theta$ ?

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Explanation:

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How do you find the polar equation to the tangent, at $θ = \frac{π}{12}$ $theta = pi/12$ , on $r = sin 3 θ$ $r = sin 3theta$ ?