What is the slope of the tangent line of #r=(sin^2theta)/(-thetacos^2theta)# at #theta=(pi)/4#?

Question

Samantha Adkison · Accepted Answer

The slope is #m = (4 - 5pi)/(4 - 3pi)#

Here is a reference to Tangents with polar coordinates

From the reference, we obtain the following equation:

#dy/dx = ((dr)/(d theta)sin(theta) + rcos(theta))/((dr)/(d theta)cos(theta) - rsin(theta))#

We need to compute #(dr)/(d theta)# but please observe that #r(theta)# can be simplified by using the identity #sin(x)/cos(x) = tan(x)#:

#r = -tan^2(theta)/theta#

#(dr)/(d theta) =(g(theta)/(h(theta)))' = (g'(theta)h(theta) - h'(theta)g(theta))/(h(theta))^2#

#g(theta) = -tan^2(theta)# #g'(theta) = -2tan(theta)sec^2(theta)#

#h(theta) = theta# #h'(theta) = 1#

#(dr)/(d theta) = (-2thetatan(theta)sec^2(theta) + tan^2(theta))/(theta)^2#

Let's evaluate the above at #pi/4#

#sec^2(pi/4) = 2# #tan(pi/4) = 1#

#r'(pi/4) = (-2(pi/4)(1)(2) + 1)/(pi/4)^2#

#r'(pi/4) = (-2(pi/4)(1)(2) + 1)(16/(pi^2))#

#r'(pi/4) = (16- 16pi)/(pi^2)#

Evaluate r at #pi/4#:

#r(pi/4) = -4/pi = -(4pi)/pi^2#

Note: I made the above denominator #pi^2# so that it was common with the denominator of #r'# and would, therefore, cancel when we put them in the following equation:

#dy/dx = ((dr)/(d theta)sin(theta) + rcos(theta))/((dr)/(d theta)cos(theta) - rsin(theta))#

At #pi/4# the sines and cosines are equal, therefore, they will cancel.

We are ready to write an equation for the slope, m:

#m = (16 - 16pi + -4pi)/(16 - 16pi - -4pi)#

#m = (4 - 5pi)/(4 - 3pi)#

Ezekiel Alexander · Answer

undefined The slope of the tangent line of $ r = \frac{\sin^2(\theta)}{-\theta \cos^2(\theta)} $ at $ \theta = \frac{\pi}{4} $ is $ -1 $.