How do you find the slope of the polar curve $r = 3 sec (2 θ)$ $r=3sec(2theta)$ at $θ = \frac{π}{6}$ $theta=pi/6$ ?

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Answer 1 · 2014-09-21T11:13:22

The slope of the curve $r = 3 sec (2 θ)$ $r=3sec(2theta)$ at $θ = \frac{π}{6}$ $theta=pi/6$ is $\frac{3 \sqrt{3}}{5}$ ${3sqrt{3}}/5$ .

Let us look at some details.

We can write

${(x=rcos theta=3sec(2theta)cos theta),(y=rsin theta=3sec(2theta)sin theta):}$

${dx}/{d theta}=3[2sec(2theta)tan(2theta)cdot cos theta+sec(2 theta)cdot(-sin theta)]$

$=3sec(2theta)[2tan(2theta)cos theta-sin theta]$

${dy}/{d theta}=3[2sec(2theta)tan(2theta)cdot sin theta+sec(2theta)cdot cos theta]$

$=3sec(2theta)[2tan(2theta)sin theta+cos theta]$

So,

${dy}/{dx}={{dy}/{d theta}}/{{dx}/{d theta}}={3sec(2theta)[2tan(2theta)sin theta+cos theta]}/{3sec(2theta)[2tan(2theta)cos theta-sin theta]}$

by cancelling out $3sec(2theta)$ ,

$={2tan(2theta)sin theta+cos theta}/{2tan(2theta)cos theta-sin theta}$

Now, we can find the slope $m$ by pluggin in $theta=pi/6$ .

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

How do you find the slope of the polar curve r=3sec(2theta)r=3sec(2θ)r=3sec(2theta) at theta=pi/6θ=π6theta=pi/6 ?