Construction of Curve With Angle of Tangent Vector With x Axis Equal to Three Times Angle of Radial Vector With x Axis

A curve is drawn such that for each point  
  on the curve, the angle between the tangent vector at the point and the  
  axis is three times the angle between the radial vector and the  
  axis. Find the equation of the curve.
  is the angle between the radial vector and the  
  axis, then  
\[\frac{y}{x} = tan \theta\]
\[tan 3 \theta = \frac{y}{x}\]
\[tan(A+B)= \frac{tanA+tanB}{1-tanAtanB}\]
/ With  
\[A=B= \theta\]
\[tan 2 \theta = \frac{2 tan \theta }{1- tan^2 \theta}\]
  and with 
\[A= \theta , \; B= 2 \theta\]
\[\begin{equation} \begin{aligned} tan(\theta + 2 \theta ) &= \frac{tan \theta tan2 \theta}{1- tan \theta tan 2 \theta } \\ &= \frac{tan \theta +\frac{2 tan \theta }{1- tan^2 \theta}}{1- tan \theta \frac{2 tan \theta }{1- tan^2 \theta}}= \tan \theta \frac{3-tan^2 \theta}{1-3 tan^2 \theta} \end{aligned} \end{equation}\]
\[\frac{dy}{dx}= \frac{y}{x} \frac{3-y^2/x^2}{1-3y^2/x^2}= \frac{y}{x} \frac{3x^2-y^2}{x^2-3y^2}\]
This equation is homogeneous so let  
\[\frac{dy}{dx}= v + x \frac{dv}{dx}\]
  and the equation becomes  
\[v + x \frac{dv}{dx}=v \frac{3-v^2}{1-3v^2} \rightarrow x \frac{dv}{dx}= v \frac{3-v^2}{1-3v^2}-v = \frac{2v(1+v^2)}{1-3v^2}\]
Separating variables gives  
\[\frac{1-3v^2}{v(1+v^2)} dv= \frac{2}{x} dx\]
We can write this as  
\[(\frac{1}{v} - \frac{4v}{1+v^2})dv = \frac{2}{x} dx\]
Integration gives  
\[ln(v)- 2ln(1+v^2)=2 ln(x)+C\]
\[ln(\frac{v}{(1+v^2)^2})=ln(Kx^2) \rightarrow \frac{v}{(1+v^2)^2}=Kx^2\]
Now substitute  
\[v= y/x\]
  to get  

You have no rights to post comments