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)r=cos(2θ) at theta=pi/2θ=π2 ?

    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