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

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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 $3/(2pi)+(pi)/12$

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