What is the slope of the tangent line of #secx-cscy= C #, where C is an arbitrary constant, at #(pi/3,pi/3)#?

Question

William Abdalla · Accepted Answer

#- 3 sqrt 3#

As a point of the curve is #T ( pi/3, pi/3 )#,

#C = sec (1/3pi) - csc (1/3pi) = 2 - 2/sqrt3#, So,

#sec x - csc y = 2 ( 1- 1/3sqrt3 )#. And so,

#secx tanx - (- csc y cot y )y' = 0#, giving

y' at T #= - ( sec (1/3pi ) tan ( 1/3pi))/( csc ( 1/3pi )(cot ( 1/3pi ))#

#= - (( 2 ) ( sqrt 3 ))/(( 2/sqrt3 )(1/sqrt3 )#

#= - 3 sqrt 3#

Graph, with tangent: graph{(1/cos x- 1/ sin y - 2 ( 1- 1/3 sqrt3 ))(y-1/3pi+3sqrt3(x-1/3pi))(y-1/3pi-3sqrt3(x+1/3pi))((x-1/3pi)^2+(y-1/3pi)^2-0.025)((x+1/3pi)^2+(y-1/3pi)^2-0.025)=0}

Indeed, vivid.

Did you observe that I have shown also

the tangent at #T'( -1/3pi, 1/3pi )#.

Charles Arocha · Answer

undefined To find the slope of the tangent line at the point $(\frac{\pi}{3}, \frac{\pi}{3})$, we need to find the derivative of the given function with respect to $x$ and $y$, and then evaluate it at that point.

Given equation: $\sec(x) - \csc(y) = C$

Taking partial derivatives with respect to $x$ and $y$, we get:

$\frac{\partial}{\partial x}(\sec(x) - \csc(y)) = \frac{\partial}{\partial x}(C)$

$\frac{\partial}{\partial y}(\sec(x) - \csc(y)) = \frac{\partial}{\partial y}(C)$

To compute these partial derivatives, we use the following identities:

$\frac{d}{dx} \sec(x) = \sec(x) \tan(x)$

$\frac{d}{dy} \csc(y) = -\csc(y) \cot(y)$

Substituting these derivatives into the partial derivatives, we have:

$\sec(x) \tan(x) = 0$

$-\csc(y) \cot(y) = 0$

Solving these equations gives us $x = \frac{\pi}{2}$ and $y = \frac{\pi}{2}$.

So, at the point $(\frac{\pi}{3}, \frac{\pi}{3})$, the slope of the tangent line is undefined.