What is the slope of the tangent line of #r=(2theta+sin2theta)/cos^2theta# at #theta=(-3pi)/8#?

Question

Emily Ammerman · Accepted Answer

#"m"_"tan"=16+8sqrt2+(12+9sqrt2)pi# #(dr)/(d theta)=((cos^2theta)(2+2cos2theta)+(2costhetasintheta)(2theta+sin2theta))/cos^4(theta)# #r'(-3/8pi)=# #((cos^2(-3/8pi))(2+2cos2(-3/8pi))+(2cos(-3/8pi)sin(-3/8pi))(2(-3/8pi)+sin2(-3/8pi)))/cos^4((-3/8pi))# #=16+8sqrt2+(12+9sqrt2)pi#

Christopher Agresta · Answer

undefined To find the slope of the tangent line at a specific point on a polar curve, you can use the formula: $ \frac{dr}{d\theta} = \frac{dy}{dx} = \frac{r' \sin(\theta) + r \cos(\theta)}{r' \cos(\theta) - r \sin(\theta)} $ where $r'$ represents $ \frac{dr}{d\theta} $.

First, calculate $r'$ by finding $ \frac{dr}{d\theta} $. Then plug the given value of $\theta$ into the equation to find the slope of the tangent line.

Given $ r = \frac{2\theta + \sin(2\theta)}{\cos^2(\theta)} $, take the derivative $ r' = \frac{dr}{d\theta} $.

$ r' = \frac{d}{d\theta}\left(\frac{2\theta + \sin(2\theta)}{\cos^2(\theta)}\right) $

After differentiation, evaluate $r'$ at $ \theta = \frac{-3\pi}{8} $.

This will give you the slope of the tangent line at the given point.