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

Question

Emilia Aguilar · Accepted Answer

0.1197y-0.2143x-0.0103=0.
I have given my best answer. For this purpose, I had made self-corrections,

The period is #2pi#. What is seen in the graph is redrawn periodically. For #theta in [0, pi/2), (pi/2, pi], [pi, 3/2pi) and (3/2pi, 2pi]#, it is in #Q_1,Q_2, Q_3 and Q_4#, respectively, Pole is not a node. The two infinitely long vertical loops touch here. #theta = pi/2 uarr and 3/2pi darr # represent asymptotes. graph{(x^2(x^2+y^2)-y^2sqrt(x^2+y^2)-x^2y)((x-.207)^2+(y-.207)^2-.02)=0 [-10, 10, -5, 5]} The equation to the tangent at #P( a, alpha )# is #rsin(theta-psi)=asin(alpha-psi)#, where #psi# is the inclination of the tangent to #theta = 0#, and is given by, #m=tanpsi=(r'sintheta+rcostheta)/(r'costheta-rsintheta)#, at #(a, alpha) and r'=(dr)/(d theta)#. Here, the point of contact of the tangent P is #(a, alpha)=(1-1/sqrt2, pi/4)=(0.2929, 0.7854) #, nearly.. #r'=2tantheta sec^2theta-costheta=4-1/sqrt2=3.293#, nearly. #tanpsi=((r'sintheta+rcostheta)/(r'costheta-rsintheta))# #= ((4-1/sqrt2)(1/sqrt2)+(1-1/sqrt2)(1/sqrt2))/((4-1/sqrt2)(1/sqrt2)-(1-1/sqrt2)(1/sqrt2))# #=(5sqrt2-2)/(3sqrt2)=1.195#, nearly. #psi=tan^(-1)(1.195)=50.08^o=0.8740# radian. So, the equation to the tangent is rsin(theta-0.8740)=0.2929sin(0.7854-0.8740)# #=-0.2929sin(0.0886 radian)=-0.2929sin5.076^o)# #=-0.0259#, nearly, giving #rsin(theta-0.8740)=-0.0259#, with #theta# in radian. or rsin(theta^o-50.08^o)=-0.0259# Expanding and converting to Cartesian form, this is 0.6417y-0.7889x+0.0259=0. The Socratic twin graph is for the Cartesian frame. See the graph below. graph{(x^2(x^2+y^2)-y^2sqrt(x^2+y^2)-x^2y)(0.6417y-0.7889x+0.0259)((x--.207)^2+(y-.207)^2-.001)=0 [-139.4, 139.1, -66.7, 72.5]}

Christopher Agresta · Answer

undefined To find the equation of the tangent line of $ r = \tan^2(\theta) - \sin(\theta) $ at $ \theta = \frac{\pi}{4} $, we first find the derivative of $ r $ with respect to $ \theta $, then evaluate it at $ \theta = \frac{\pi}{4} $.

The derivative $ \frac{dr}{d\theta} $ is given by $ \frac{d}{d\theta}(\tan^2(\theta) - \sin(\theta)) $.

Differentiating $ \tan^2(\theta) $ with respect to $ \theta $ using the chain rule yields $ 2\tan(\theta) \sec^2(\theta) $.

Differentiating $ -\sin(\theta) $ with respect to $ \theta $ gives $ -\cos(\theta) $.

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

Evaluate $ \frac{dr}{d\theta} $ at $ \theta = \frac{\pi}{4} $ to get $ \frac{dr}{d\theta} = 2\tan(\frac{\pi}{4})\sec^2(\frac{\pi}{4}) - \cos(\frac{\pi}{4}) = 2(1)(2) - \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2} $.

Now, we need the slope of the tangent line, which is $ \frac{dr}{d\theta} $ evaluated at $ \theta = \frac{\pi}{4} $, so $ \frac{3\sqrt{2}}{2} $.

To find the point on the curve where the tangent line touches, plug $ \theta = \frac{\pi}{4} $ into $ r = \tan^2(\theta) - \sin(\theta) $ to get $ r = 1 - \frac{\sqrt{2}}{2} $.

Using the point-slope form of a line $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) $ is a point on the line and $ m $ is the slope, and substituting $ (x_1, y_1) = (\frac{\pi}{4}, 1 - \frac{\sqrt{2}}{2}) $ and $ m = \frac{3\sqrt{2}}{2} $, the equation of the tangent line is $ r - (1 - \frac{\sqrt{2}}{2}) = \frac{3\sqrt{2}}{2}(\theta - \frac{\pi}{4}) $.