MHB Slope of polar curve at indicated point

[ineedhelpnow]

$r=5$ and $\theta=\pi/6$$\frac{dy}{dx}=\frac{\frac{dy}{d \theta}}{\frac{dx}{d \theta}}=\frac{\frac{dr}{d \theta}sin(\theta)+rcos(\theta)}{\frac{dr}{d \theta}cos(\theta)-rsin(\theta)}$

$\frac{0*sin(\pi/3)+5cos(\pi/3)}{0*cos(\pi/3)-5sin(\pi/3)}=-\frac{\sqrt{3}}{3}$is that right?

 

[lfdahl]

Hi, ineedhelpnow - greetings from Denmark :)

The formula for the slope seems to be OK (please cf. Pauls Online Notes : Calculus II - Tangents with Polar Coordinates)

Are you sure, that $r \ne r(\theta)$? If so, then your calculation is all right except that $\theta$
should be $\frac{\pi}{6}$

 

[ineedhelpnow]

i think i did everything right. that's the slope equation in my book so i just plugged everything in.