What is the slope of the tangent line of xarctan(y/pi)= C xarctan(yπ)=C, where C is an arbitrary constant, at (pi/3,pi/3)(π3,π3)?

1 Answer
Mar 19, 2017

dy/dx=-10/9arctan(1//3)dydx=109arctan(1/3)

Explanation:

Taking the derivative, we first use the product rule:

(d/dxx)arctan(y/pi)+x(d/dxarctan(y/pi))=0(ddxx)arctan(yπ)+x(ddxarctan(yπ))=0

The typical arctangent derivative is d/dxarctan(x)=1/(1+x^2)ddxarctan(x)=11+x2, so here we apply the chain rule:

arctan(y/pi)+x(1/(1+(y/pi)^2))(d/dxy/pi)=0arctan(yπ)+x11+(yπ)2(ddxyπ)=0

The derivative with respect to xx of y/piyπ, which is the linear equation 1/piy1πy, so its derivative is 1/pidy/dx1πdydx:

arctan(y/pi)+x/(1+y^2/pi^2)(1/y)dy/dx=0arctan(yπ)+x1+y2π2(1y)dydx=0

Solving for dy/dxdydx at (pi/3,pi/3)(π3,π3):

arctan(1//3)+(pi//3)/(1+(pi^2//9)//pi^2)(1/(pi//3))dy/dx=0arctan(1/3)+π/31+(π2/9)/π2(1π/3)dydx=0

arctan(1//3)+(1/(1+1//9))dy/dx=0arctan(1/3)+(11+1/9)dydx=0

arctan(1//3)+9/10(dy/dx)=0arctan(1/3)+910(dydx)=0

dy/dx=-10/9arctan(1//3)dydx=109arctan(1/3)