How do you find the equation of the tangent line to the polar curve $r = 3 + 8 sin (θ)$ $r=3+8sin(theta)$ at $θ = \frac{π}{6}$ $theta=pi/6$ ?

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Answer 1 · 2014-09-21T23:04:13

The equation of the tangent line is

$y - \frac{7}{2} = \frac{11 \sqrt{3}}{5} (x - \frac{7 \sqrt{3}}{2})$ $y-7/2={11sqrt{3}}/5(x-{7sqrt{3}}/2)$

In order to find the equation of a line, we need two pieces of information:

${(1. "Point: " (x_1,y_1)),(2. "Slope: " m):}$

Let us find $(x_1,y_1)$ .

Since

${(x(theta)=rcos theta=(3+8sin theta)cos theta),(y(theta)=rsin theta=(3+8sin theta)sin theta):}$ ,

$x_1=x(pi/6)=[3+8sin(pi/6)]cos(pi/6)={7sqrt{3}}/2$

$y_1=y(pi/6)=[3+8sin(pi/6)]sin(pi/6)=7/2$

Now, let us find $m$ .

By differentiating with respect to theta#,

${dx}/{d theta}=8cos theta cdot cos theta+(3+8sin theta)cdot(-sin theta)$

$=8(cos^2theta-sin^2theta)-3sin theta$

$=8cos(2theta)-3sin theta$

${dy}/{d theta}=8cos theta cdot sin theta+(3 + 8sin theta)cdot cos theta$

$=8(2sin theta cos theta)+3cos theta$

$=8sin(2theta)+3cos theta$

So,

${dy}/{dx}={{dy}/{d theta}}/{{dx}/{d theta}}={8sin(2theta)+3cos theta}/{8cos(2theta)-3sin theta}$

Now, we can find $m$ .

$m={dy}/{dx}|_{theta=pi/6} ={8({sqrt{3}}/2)+3({sqrt{3}}/2)}/{8(1/2)-3(1/2)}={11sqrt{3}}/5$

By Point-Slope Form: $y-y_1=m(x-x_1)$ ,

$y-7/2={11sqrt{3}}/5(x-{7sqrt{3}}/2)$

How do you find the equation of the tangent line to the polar curve r=3+8sin(theta)r=3+8sin(θ)r=3+8sin(theta) at theta=pi/6θ=π6theta=pi/6 ?