What is the slope of the polar curve #f(theta) = 2theta + tan^2theta - costheta # at #theta = (pi)/8#?

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Answer 1 · 2017-01-06T02:10:46

0.48799, nearly.

In polar form. slope

#(dy)/(dx)#

#=(d/(d theta)(rsintheta))/(d/(d theta)(rcostheta))#

#=(r'sintheta+rcostheta)/(r'costheta-rsintheta)#,

where# r'=(dr)/(d theta)#.

Here,

#r =2 theta+tan^2theta-costheta=pi/4+tan^2(pi/8)-cosn(pi/8)#

at #theta=pi/8 = 22.5^o#

#=0.20466#, nearly.

#r'=2+2tantheta sec^2theta+sinthea#

#=2+2tan(pi/8)sec^2(pi/8)+sin(pi/8)#

#=3.335325, nearly.

Now the slope at #theta = pi/8# is

#=(r'sintheta+rcostheta)/(r'costheta-rsintheta)#,

where # r'=(dr)/(d theta) and theta = pi/8#

Upon substitutions and simplification,

slope = 0.48799, nearly.