What is the slope of the tangent line of #sinx-y^2/x= C #, where C is an arbitrary constant, at #(pi/3,1)#?

Question

1333 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-29T08:32:46

Slope of the tangent at #(pi/3,1)# is #3/(2pi)+(pi)/12#

At #(pi/3,1)# we have #sin(pi/3)-1^2/(pi/3)=C# or #C=sqrt3/2-(1xx3)/pi=(pisqrt3-3)/(2pi)#

Slope of the function is given by its first derivative and differentiating the function #sinx-y^2/x=(pisqrt3-3)/(2pi)#, we get

#cosx-(x*2y*(dy/(dx))-y^2*1)/x^2=0# or

#cosx-2y/x*(dy)/(dx)+y^2/x^2=0# or

#2y/x*(dy)/(dx)=y^2/x^2+cosx# or

#(dy)/(dx)=y^2/x^2*x/(2y)+x/(2y)cosx# or

#(dy)/(dx)=y/(2x)+x/(2y)cosx#

and at #(pi/3,1)#, #(dy)/(dx)=1/(2(pi)/3)+((pi)/3)/(2*1)cos(pi/3)#

or #(dy)/(dx)=3/(2pi)+(pi)/6*1/2=3/(2pi)+(pi)/12#

Hence slope at #(pi/3,1)# is #3/(2pi)+(pi)/12#