What is the slope of the tangent line of #(1-x)(4-y^2)-1/lny = C #, where C is an arbitrary constant, at #(1,2)#?

Question

Scarlett Allison · Accepted Answer

It is not possible to specify a value for C. See graph, with the line y = 2. and the explanation.

C is seemingly single-valued, but really not so. It is so, upon setting x = 1 and y = 2 in the equation. Here, C is shown as -1 /ln 2. With this C for the graph ,the line y = 2 meets it at point(s) given by #(x-1)^2(0)=0. # The solution is x is arbitrary. An attempt to find the equation to the virtual tangent at (1, 2) would result in y = 2. See the (y=2)-inclusive graph. graph{(y-2+10^(-10)x)((x-1)(y^2-4)-1/ln y+1/ln 2)=0 [-10, 10, -5, 5]}

Emilia Aguilar · Answer

The slope is #0#.

Use implicit differentiation (or partial derivatives) to find

#dy/dx = (4-y^2)/(2xy-2y+1/(y(lny)^2))#.

At #(1,2)#, we get

#dy/dx = (4-(2)^2)/(2(1)(2)-2(2)+1/(2(ln2)^2)) = 0#

Given that #(1,2)# is on the curve, we can see that #C = -1/ln2#

Here are some of the curves in this family.

red dots #" "# #c=0#
red dash #" "# #c=-0.9102#
blue dash #" "# #c=-1.4#
black dash #" "# #c=-1.43#
solid black #" "# #c=-1.442695# (that is: #c~~-1/ln2#)

And here are the same near #(1,2)#

The point of at which the two branches meet appears to be #(0.74,2)#

And if we continue with values of #c# less than #-1/ln2#, the branches separate again, but differently.

blue dash #" "# #c=-1.4#
solid black #" "# #c=-1.442695# (that is: #c~~-1/ln2#)
red dash #" "# #c=-1.5#
green dash #" "# #c=-2#

Aurora Alvarado · Answer

undefined To find the slope of the tangent line at a given point, you need to take the derivative of the equation with respect to $ x $, then evaluate it at the given point. Given the equation $ (1-x)(4-y^2)-\frac{1}{\ln y} = C $, and the point $ (1,2) $, the slope of the tangent line can be found by first finding the derivative with respect to $ x $ and then substituting $ x = 1 $ and $ y = 2 $ into the derivative expression.