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

1 Answer
Nov 22, 2015

#(pi^2+18sin(pi^2/4))/(6pi^2sin(pi^2/4))#

Explanation:

The constant #C#
If we know that #(pi/3,pi/2)# is a point on the graph, then #C = sin(pi/3)-sin((pi/2)^2)/(pi/3) = sqrt3/2-(3sin((pi/2)^2))/pi#.
So #C# is determined by the values of #x# and #y#. It is not arbitrary. (It is not independent of the values of the variables.)

Slope of Tangent
Regardless of that, we can find the slope of the tangent line by implicit differentiation.

#d/dx(sinx-sin(y^2)/x)=d/dx( C) #

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

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

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

The slope at #(pi/3,pi/2)#, is therefore

#dy/dx = ((pi/3)^2cos(pi/3)+sin((pi/2)^2))/(2(pi/3)(pi/2)sin((pi/2)^2))#

# = (pi^2+18sin(pi^2/4))/(6pi^2sin(pi^2/4))#