What is the equation of the tangent line to the polar curve #f(theta)=theta- sin((3theta)/2-pi/2)+tan((2theta)/3-pi/2) # at #theta = pi#?

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Answer 1 · 2016-11-09T21:28:36

#y = -18/61(pi + sqrt(3)/3)( x + pi + sqrt(3)/3)#

Let's begin by determining the #(x,y)# point through which the line must pass. Because #theta = pi#, we know that #y = rsin(pi) = 0#

#x = rcos(pi)#

#x = -r#

#r = f(pi) = pi - sin((3pi)/2 - pi/2) + tan((2pi)/3 - pi/2)#

#r = pi + sqrt(3)/3#

The point through which the line must pass is:

#(-pi - sqrt(3)/3, 0)#

This reference Tangents to polar curves gives us an equation for #dy/dx#:

#dy/dx = (((dr)/(d theta))sin(theta) + r cos(theta))/(((dr)/(d theta))cos(theta) - r sin(theta))#.

To obtain the slope of the tangent line, m, we evaluate the above at #theta = pi# so the above equation becomes:

#m = r/(((dr)/(d theta))#

Therefore, we need to evaluate #(dr)/(d theta)# at #pi#. I am going to use WolframAlpha to do the computation:

#(dr)/(d theta) = 61/18#

#m = -18/61(pi + sqrt(3)/3)#

Using the point-slope form of the equation of a line:

#y = -18/61(pi + sqrt(3)/3)( x + pi + sqrt(3)/3)#