CPM Homework Help : APCALC Problem 12-109

Multiple Choice: Which of the following values is the slope of the line tangent to the polar curve $r ( \theta ) = \frac { 1 } { 1 + \operatorname { sin } ( \theta ) }$ at the point $( \frac { \pi } { 6 } , \frac { 2 } { 3 } )$ ?

$- \frac { \sqrt { 3 } } { 3 }$

$- \frac { 2 \sqrt { 3 } } { 9 }$

$-\frac { \pi } { 3 }$

$\frac { 2 \sqrt { 3 } } { 3 }$

$\frac { \sqrt { 3 } } { 3 }$

Hint:

Remember, the slope is $dy/dx$ , so you will need equations for $x$ and $y$ .

Step:

$y=r\sin(\theta)=\frac{\sin(\theta)}{1+\sin(\theta)}$

Now repeat this process for $x$ .

More Help:

$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$