What is the slope of the polar curve #f(theta) = sectheta - csctheta # at #theta = (5pi)/4#?

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Answer 1 · 2017-09-03T04:15:45

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Answer 2 · 2017-09-04T16:38:24

#1#

#r=f(theta)=sectheta-csctheta#

We want #dy/dx# of this curve at #theta=(5pi)/4#.

To find the slope in terms of #x# and #y#, we have to use the identities #{(x=rcostheta),(y=rsintheta):}#

So here,

#{(x=(sectheta-csctheta)costheta=1-cottheta),(y=(sectheta-csctheta)sintheta=tantheta-1):}#

Note that #dy/dx=(dy/(d theta))/(dx/(d theta))#.

#{(dx/(d theta)=d/(d theta)(1-cottheta)=csc^2theta),(dy/(d theta)=d/(d theta)(tantheta-1)=sec^2theta):}#

So:

#dy/dx=(dy/(d theta))/(dx/(d theta))=sec^2theta/csc^2theta=tan^2theta#

So the slope at #theta=(5pi)/4# is:

#m=tan^2((5pi)/4)=(-1)^2=1#