What is the slope of the polar curve #f(theta) = theta^2 - csctheta # at #theta = (3pi)/4#?

1 Answer
Feb 24, 2018

Answer:

Slope of the curve at
#theta=(3pi)/4#is -1

Explanation:

Given:
#f(theta)=theta^2-csctheta#

Let
#r=f(theta)#

Then,
#r=theta^2-csctheta#

At #theta=(3pi)/4#

#r=((3pi)/4)^2-csc((3pi)/4)#

#((3pi)/4)^2=(9pi^2)/16#

#csc((3pi)/4)=sqrt2#

#r=(9pi^2)/16-sqrt2#

#(dr)/(d(theta))=2theta-(-cscthetacottheta)#

#(dr)/(d(theta))=2theta+cscthetacottheta#

#(dr)/(d(theta))=2((3pi)/4)+csc((3pi)/4)cot((3pi)/4)#

#2((3pi)/4)=(3pi)/2#

#csc((3pi)/4)=sqrt2#

#cot((3pi)/4)=-1#

#(dr)/(d(theta))=(3pi)/2+sqrt2(-1)#

#(dr)/(d(theta))=(3pi)/2-sqrt2#

To find

Slope of the curve at
#theta=(3pi)/4#

#x=rcostheta#

#y=rsintheta#

#dx/(d(theta))=r(-sintheta)+costheta(dr)/(d(theta))#

#dx/(d(theta))=-r(sintheta)+costheta(dr)/(d(theta))#

#r=(9pi^2)/16-1/sqrt2#

#sintheta=sin((3pi)/4)#

#sin((3pi)/4)=1/sqrt2#

#costheta=cos((3pi)/4)#

#cos((3pi)/4)=-1/sqrt2#

#costheta=-1/sqrt2#

#(dr)/(d(theta))=(3pi)/2-sqrt2#

#dx/(d(theta))=-((9pi^2)/16-1/sqrt2)(1/sqrt2)+(-1/sqrt2)((3pi)/2-sqrt2)#

#dx/(d(theta))=1/sqrt2((9pi^2)/16-1/sqrt2)-1/sqrt2((3pi)/2-sqrt2)#

#dx/(d(theta))=1/sqrt2(((9pi^2)/16-1/sqrt2)-((3pi)/2-sqrt2))#

#dy/(d(theta))=rcostheta+sintheta(dr)/(d(theta))#

#r=(9pi^2)/16-1/sqrt2#

#sintheta=sin((3pi)/4)#

#sin((3pi)/4)=1/sqrt2#

#costheta=cos((3pi)/4)#

#cos((3pi)/4)=-1/sqrt2#

#costheta=-1/sqrt2#

#(dr)/(d(theta))=(3pi)/2-sqrt2#

#dy/(d(theta))=((9pi^2)/16-1/sqrt2)(-1/sqrt2)+(1/sqrt2)((3pi)/2-sqrt2)#

#dy/(d(theta))=-1/sqrt2((9pi^2)/16-1/sqrt2)+(1/sqrt2)((3pi)/2-sqrt2)#

#dy/(d(theta))=-1/sqrt2(((9pi^2)/16-1/sqrt2)-((3pi)/2-sqrt2))#

#dy/dx=(dy/(d(theta)))/(dx/(d(theta)))#

#=(-1/sqrt2(((9pi^2)/16-1/sqrt2)-((3pi)/2-sqrt2)))/(1/sqrt2(((9pi^2)/16-1/sqrt2)-((3pi)/2-sqrt2)))#

Cancelling

#dy/dx=-1#

Slope of the curve at
#theta=(3pi)/4#is -1