What is the equation of the tangent line of #r=tan^2theta - costheta# at #theta=pi/4#?

Question

Luke Alvarez · Accepted Answer

slope= #5/(3+2/sqrt2)# = # 5/(3+sqrt2)# It is known that Cartesian to Polar is x= r#cos theta# and y= r #sin theta# Hence dx=# (dr)/(d theta) cos theta -r sin theta# and dy= #(dr)/(d theta) sin theta +r cos theta# Therefore slope #dy/dx =((dr)/(d theta) sin theta + r cos theta)/((dr)/(d theta) cos theta - r sin theta# When #theta= pi/4# , slope would be =# ((dr)/(d theta) +r)/((dr)/(d theta) -r )# , because for # theta= pi/4#, #sin theta = cos theta # Now given r= #tan^2 theta -cos theta#, #(dr)/(d theta ) =2 tan theta sec^2 theta + sin theta #. For #theta = pi/4#, #(dr)/(d theta)= 2*1*2 + 1/sqrt2# = #4+ 1/sqrt2# and r= 1- #1/sqrt2# Hence slope= #5/(3+2/sqrt2)# = # 5/(3+sqrt2)#

Emily Ammerman · Answer

undefined To find the equation of the tangent line to the polar curve $ r = \tan^2(\theta) - \cos(\theta) $ at $ \theta = \frac{\pi}{4} $, first, we need to find the derivative of $ r $ with respect to $ \theta $, which gives us $ \frac{dr}{d\theta} $. Then, we evaluate $ r $ and $ \frac{dr}{d\theta} $ at $ \theta = \frac{\pi}{4} $ to get the slope of the tangent line. Finally, we use the point-slope form of a line to find the equation of the tangent line.

$$ r = \tan^2(\theta) - \cos(\theta) $$

$$ \frac{dr}{d\theta} = \frac{d}{d\theta}(\tan^2(\theta) - \cos(\theta)) $$

$$ \frac{dr}{d\theta} = 2\tan(\theta)\sec^2(\theta) + \sin(\theta) $$

At $ \theta = \frac{\pi}{4} $:

$$ r(\frac{\pi}{4}) = \tan^2(\frac{\pi}{4}) - \cos(\frac{\pi}{4}) = 1 - \frac{\sqrt{2}}{2} $$

$$ \frac{dr}{d\theta}(\frac{\pi}{4}) = 2\tan(\frac{\pi}{4})\sec^2(\frac{\pi}{4}) + \sin(\frac{\pi}{4}) = 2\sqrt{2} + \frac{\sqrt{2}}{2} $$

So, the slope of the tangent line at $ \theta = \frac{\pi}{4} $ is $ m = 2\sqrt{2} + \frac{\sqrt{2}}{2} $.

Using the point-slope form of a line:

$$ y - y_1 = m(x - x_1) $$

$$ y - (1 - \frac{\sqrt{2}}{2}) = (2\sqrt{2} + \frac{\sqrt{2}}{2})(x - \frac{\pi}{4}) $$

$$ y = (2\sqrt{2} + \frac{\sqrt{2}}{2})x + (1 - \frac{\sqrt{2}}{2}) - (2\sqrt{2} + \frac{\sqrt{2}}{2})\frac{\pi}{4} $$

$$ y = (2\sqrt{2} + \frac{\sqrt{2}}{2})x + (1 - \frac{\sqrt{2}}{2}) - \frac{\pi}{2} $$