How do I solve: The normal to the curve $\ xe^{-y} + e^{y} = 1 + x$ at the point $(c, \ln c)$ has a $y$-intercept $\ c^2+ 1$. Determine c.

Question

How do I solve: The normal to the curve $\ xe^{-y} + e^{y} = 1 + x$ at the point $(c, \ln c)$ has a $y$-intercept $\ c^2+ 1$. Determine c.

I tried substituting $\ x=0$ to find out the y-intercept, but it came out to $\ e^y = 1$, which is 0.

I don't understand what I should do. How do I differentiate the left and right sides of the equation?

Answer 1

Setting $x = 0$ in $xe^{-y}+ e^y= x+ 1$ gives the $y$-intercept of $xe^{-y}+ e^y= x+ 1$. That is NOT what is asked for! You are asked to find the $y$-intercept of the normal line to that curve at $(c, \ln(c))$. So first you need to find that line!

Differentiating $xe^{-y}+ e^y= x+ 1$ with respect to $x$,

$$e^{-y}- xe^{-y}y'+ e^y y'= 1$$

At $(c, \ln(c))$ that is

$$e^{-\ln(c)}- ce^{-\ln(c)}y'+ e^{\ln(c)}y'= 1$$

$$1/c- c(1/c)y'+ (1/c)y'= (1/c- 1)y'+ 1/c= 1$$

so

$$(1/c- 1)y'= 1- 1/c$$

$$y'= (1-1/c)/(1/c- 1)= -1$$

The tangent line to this curve at $(c, \ln(c))$ has slope $-1$ so the normal line there has slope $1$ and has equation $y= x- c+ \ln(c)$. The $y$-intercept of that line is $(0, \ln(c)- c)$.

Answer 2

You may write the factor $e^{-y}$ as $\frac{1}{e^{y}}$. Then your equation is a quadratic equation in $e^{y}$ with solutions $y = 0$ and $y = \ln(x)$ (which are obvious by inspection). From the derivative of $y = \ln(x)$, the slope at $x = c$ is thus $1/c$ and the tangent equation at this point is $y = x/c + \ln(c) - 1$. At this same point, the normal line has the equation $y = -x/c + \ln(c) - 1$. Therefore the y-intercept of the normal line is obtained by setting $x = 0$ and $y = 1 + c^2$, yielding the solution for "c" as $\ln(c) = 2 + c^2$ which can be solved by the Lambert W-function.