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

Question

1464 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-20T16:52:44

Slope is #1#

As we are seeking slope at #(pi/3,pi/3)#, this point lies on the curve given by #sin^2y-cos^2x=C#.

Putting values from #(pi/3,pi/3)#, we get #3/4-1/4=C#

or #C=1/2# and function is #sin^2y-cos^2x=1/2#

Now differentiating the implicit function #sin^2y-cos^2x=1/2#

#2sinycosy(dy)/(dx)+2cosxsinx=0#

or #(dy)/(dx)=-(2sinxcosx)/(2sinycosy)=-sin(2x)/sin(2y)#

and at #(pi/3,pi/3)#, slope is #(1/2)/(1/2)=1#