Determining the Slope and Tangent Lines for a Polar Curve

Key Questions

  • How do you find the slope of the polar curve #r=cos(2theta)# at #theta=pi/2# ?

    By converting into parametric equations,

    #{(x(theta)=r(theta)cos theta=cos2theta cos theta), (y(theta)=r(theta)sin theta=cos2theta sin theta):}#

    By Product Rule,

    #x'(theta)=-sin2theta cos theta-cos2theta sin theta#

    #x'(pi/2)=-sin(pi)cos(pi/2)-cos(pi)sin(pi/2)=1#

    #y'(theta)=-sin2thetasin theta+cos2theta cos theta#

    #y'(pi/2)=-sin(pi)sin(pi/2)+cos(pi)cos(pi/2)=0#

    So, the slope #m# of the curve can be found by

    #m={dy}/{dx}|_{theta=pi/2}={y'(pi/2)}/{x'(pi/2)}=0/1=0#

    I hope that this was helpful.

    Wataru · 20 · Oct 16 2014
  • How do you find the equation of the tangent line to the polar curve #r=3+8sin(theta)# at #theta=pi/6# ?

    Answer:

    The equation of the tangent line is

    #y-7/2={11sqrt{3}}/5(x-{7sqrt{3}}/2)#

    Explanation:

    In order to find the equation of a line, we need two pieces of information:

    #{(1. "Point: " (x_1,y_1)),(2. "Slope: " m):}#

    Let us find #(x_1,y_1)#.

    Since

    #{(x(theta)=rcos theta=(3+8sin theta)cos theta),(y(theta)=rsin theta=(3+8sin theta)sin theta):}#,

    #x_1=x(pi/6)=[3+8sin(pi/6)]cos(pi/6)={7sqrt{3}}/2#

    #y_1=y(pi/6)=[3+8sin(pi/6)]sin(pi/6)=7/2#

    Now, let us find #m#.

    By differentiating with respect to theta#,

    #{dx}/{d theta}=8cos theta cdot cos theta+(3+8sin theta)cdot(-sin theta)#

    #=8(cos^2theta-sin^2theta)-3sin theta#

    #=8cos(2theta)-3sin theta#

    #{dy}/{d theta}=8cos theta cdot sin theta+(3 + 8sin theta)cdot cos theta#

    #=8(2sin theta cos theta)+3cos theta#

    #=8sin(2theta)+3cos theta#

    So,

    #{dy}/{dx}={{dy}/{d theta}}/{{dx}/{d theta}}={8sin(2theta)+3cos theta}/{8cos(2theta)-3sin theta}#

    Now, we can find #m#.

    #m={dy}/{dx}|_{theta=pi/6} ={8({sqrt{3}}/2)+3({sqrt{3}}/2)}/{8(1/2)-3(1/2)}={11sqrt{3}}/5#

    By Point-Slope Form: #y-y_1=m(x-x_1)#,

    #y-7/2={11sqrt{3}}/5(x-{7sqrt{3}}/2)#

    Wataru · 13 · Sep 21 2014
  • How do you find the slope of the tangent line to a polar curve?

    A polar equation of the form #r=r(theta)# can be converted into a pair of parametric equations

    #{(x(theta)=r(theta)cos theta),(y(theta)=r(theta)sin theta):}#.

    The slope #m# of the tangent line at #theta=theta_0# can be expressed as

    #m={dy}/{dx}|_{theta=theta_0}={{dy}/{d theta}|_{theta=theta_0}}/{{dx}/{d theta}|_{theta=theta_0}}={y'(theta_0)}/{x'(theta_0)}#.

    I hope that this was helpful.

    Wataru · · Oct 9 2014

Questions

Polar Curves