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

Elijah Abram · Accepted Answer

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#

Paisley Johnson · Answer

undefined To find the slope of the tangent line at a specific point on a curve defined implicitly, we can differentiate the equation with respect to x implicitly, then substitute the given point's coordinates into the resulting expression to find the slope.

Differentiating the given equation implicitly with respect to x:

dy/dx = [2y * dy/dx - (sin(x) - y^2) / x^2] / [2 * (sin(x) - y^2) / x^2 - (sin(x) - y^2) * (-2x) / x^3]

Substitute x = π/3 and y = 1 into the expression for dy/dx to find the slope of the tangent line at the point (π/3, 1).