2016.3.6 Question 6

• In the case where $B>A>0$ or $-B<-A<0$, notice that

  $$R\text{cosh}<!--\; FUNCTION\; APPLICATION\; -->(x+\gamma )=R\text{cosh}<!--\; FUNCTION\; APPLICATION\; -->x\text{cosh}<!--\; FUNCTION\; APPLICATION\; -->\gamma +R\text{sinh}<!--\; FUNCTION\; APPLICATION\; -->x\text{sinh}<!--\; FUNCTION\; APPLICATION\; -->\gamma .$$

  Therefore, we would like $R\text{sinh}<!--\; FUNCTION\; APPLICATION\; -->\gamma =A$ and $R\text{cosh}<!--\; FUNCTION\; APPLICATION\; -->\gamma =B$.

  Since $\text{cosh}<!--\; FUNCTION\; APPLICATION\; -->{\gamma }^2-\text{sinh}<!--\; FUNCTION\; APPLICATION\; -->{\gamma }^2=1$, we have $R^2=B^2-A^2$.

  We also have $\text{tanh}<!--\; FUNCTION\; APPLICATION\; -->\gamma =\frac{A}{B}$, and therefore $\gamma =\text{artanh}<!--\; FUNCTION\; APPLICATION\; -->\frac{A}{B}$.

  Notice that $\text{cosh}<!--\; FUNCTION\; APPLICATION\; -->\gamma >0$, so $R$ must have the same sign as $B$.

  • If $B>A>0$, $R=\sqrt{B^2-A^2}$.
  • If $B<-A<0$, $R=-\sqrt{B^2-A^2}$.

• In the case where $-A<B<A$, notice that

  $$R\text{sinh}<!--\; FUNCTION\; APPLICATION\; -->(x+\gamma )=R\text{sinh}<!--\; FUNCTION\; APPLICATION\; -->\gamma \text{cosh}<!--\; FUNCTION\; APPLICATION\; -->x+R\text{cosh}<!--\; FUNCTION\; APPLICATION\; -->\gamma \text{sinh}<!--\; FUNCTION\; APPLICATION\; -->x.$$

  Therefore, we would like $R\text{cosh}<!--\; FUNCTION\; APPLICATION\; -->\gamma =A$ and $R\text{sinh}<!--\; FUNCTION\; APPLICATION\; -->\gamma =B$.

  Since $\text{cosh}<!--\; FUNCTION\; APPLICATION\; -->{\gamma }^2-\text{sinh}<!--\; FUNCTION\; APPLICATION\; -->{\gamma }^2=1$, we have $R^2=B^2-A^2$.

  We also have $\text{tanh}<!--\; FUNCTION\; APPLICATION\; -->\gamma =\frac{B}{A}$, and therefore $\gamma =\text{artanh}<!--\; FUNCTION\; APPLICATION\; -->\frac{B}{A}$.

  Notice that $\text{cosh}<!--\; FUNCTION\; APPLICATION\; -->\gamma >0$, so $R$ will have the same sign as $A$, and hence $R=\sqrt{A^2-B^2}$.

• When $B=A$, we have

  $$\begin{array}{llll}\hfill A\text{sinh}<!--\; FUNCTION\; APPLICATION\; -->x+B\text{cosh}<!--\; FUNCTION\; APPLICATION\; -->x& =A\frac{e^x-e^{-x}}{2}+A\frac{e^x+e^{-x}}{2}& \hfill & \\ \hfill & =A{e}^x.& \hfill & \end{array}$$

• When $B=-A$, we have

  $$\begin{array}{llll}\hfill A\text{sinh}<!--\; FUNCTION\; APPLICATION\; -->x+B\text{cosh}<!--\; FUNCTION\; APPLICATION\; -->x& =A\frac{e^x-e^{-x}}{2}-A\frac{e^x+e^{-x}}{2}& \hfill & \\ \hfill & =A{e}^{-x}.& \hfill & \end{array}$$

Therefore, in conclusion,

1. We have $\text{sech}<!--\; FUNCTION\; APPLICATION\; -->x=a\text{tanh}<!--\; FUNCTION\; APPLICATION\; -->x+b$, and hence $1=a\text{sinh}<!--\; FUNCTION\; APPLICATION\; -->x+b\text{cosh}<!--\; FUNCTION\; APPLICATION\; -->x$. If $b>a>0$, we have

   $$\sqrt{b^2-a^2}\text{cosh}<!--\; FUNCTION\; APPLICATION\; -->\left(x+\text{artanh}<!--\; FUNCTION\; APPLICATION\; -->\frac{a}{b}\right)=1.$$

   Therefore,

   $$\begin{array}{llll}\hfill \text{cosh}<!--\; FUNCTION\; APPLICATION\; -->\left(x+\text{artanh}<!--\; FUNCTION\; APPLICATION\; -->\frac{a}{b}\right)& =\frac{1}{\sqrt{b^2-a^2}}& \hfill & \\ \hfill x+\text{artanh}<!--\; FUNCTION\; APPLICATION\; -->\frac{a}{b}& =\pm \text{arcosh}<!--\; FUNCTION\; APPLICATION\; -->\frac{1}{\sqrt{b^2-a^2}}& \hfill & \\ \hfill x& =\pm \text{arcosh}<!--\; FUNCTION\; APPLICATION\; -->\frac{1}{\sqrt{b^2-a^2}}-\text{artanh}<!--\; FUNCTION\; APPLICATION\; -->\frac{a}{b},& \hfill & \end{array}$$

   as desired.

2. When $a>b>0$,

   $$\sqrt{a^2-b^2}\text{sinh}<!--\; FUNCTION\; APPLICATION\; -->\left(x+\text{artanh}<!--\; FUNCTION\; APPLICATION\; -->\frac{b}{a}\right)=1.$$

   Therefore,

   $$\begin{array}{llll}\hfill \text{sinh}<!--\; FUNCTION\; APPLICATION\; -->\left(x+\text{artanh}<!--\; FUNCTION\; APPLICATION\; -->\frac{b}{a}\right)& =\frac{1}{\sqrt{a^2-b^2}}& \hfill & \\ \hfill x+\text{artanh}<!--\; FUNCTION\; APPLICATION\; -->\frac{b}{a}& =\text{arsinh}<!--\; FUNCTION\; APPLICATION\; -->\frac{1}{\sqrt{a^2-b^2}}& \hfill & \\ \hfill x& =\text{arsinh}<!--\; FUNCTION\; APPLICATION\; -->\frac{1}{\sqrt{a^2-b^2}}-\text{artanh}<!--\; FUNCTION\; APPLICATION\; -->\frac{b}{a}.& \hfill & \end{array}$$

3. We would like to have two solutions to the equation $1=a\text{sinh}<!--\; FUNCTION\; APPLICATION\; -->x+b\text{cosh}<!--\; FUNCTION\; APPLICATION\; -->x$.

   • $0<a<b$, this gives

     $$x=\pm \text{arcosh}<!--\; FUNCTION\; APPLICATION\; -->\frac{1}{\sqrt{b^2-a^2}}-\text{artanh}<!--\; FUNCTION\; APPLICATION\; -->\frac{a}{b},$$

     For this to make sense, we must have $\frac{1}{\sqrt{b^2-a^2}}\ge 1$, and therefore $0<\sqrt{b^2-a^2}\le 1$, which is $0<b^2-a^2\le 1$.

     For this to have two distinct points, we would like to have $\text{arcosh}<!--\; FUNCTION\; APPLICATION\; -->\frac{1}{\sqrt{b^2-a^2}}\ne 0$ as well. This means $b^2-a^2\ne 1$.

     Therefore, in this case, this means that $a<b<\sqrt{a^2+1}$.

   • $b=a$, this gives $a{e}^x=1$, which gives a unique solution $x=-\text{ln}<!--\; FUNCTION\; APPLICATION\; -->a$.

   • $-a<b<a$, this gives

     $$\sqrt{A^2-B^2}\text{sinh}<!--\; FUNCTION\; APPLICATION\; -->\left(x+\text{artanh}<!--\; FUNCTION\; APPLICATION\; -->\frac{B}{A}\right)=1,$$

     which can only give the solution $x=\text{arsinh}<!--\; FUNCTION\; APPLICATION\; -->\frac{1}{\sqrt{A^2-B^2}}-\text{artanh}<!--\; FUNCTION\; APPLICATION\; -->\frac{B}{A}$.

   • $b=-a$, this gives $-a{e}^{-x}=1$, which does not have a solution.

   • $-b<-a<0$, this gives

     $$-\sqrt{b^2-a^2}\text{cosh}<!--\; FUNCTION\; APPLICATION\; -->\left(x+\text{artanh}<!--\; FUNCTION\; APPLICATION\; -->\frac{a}{b}\right)=1,$$

     but this is impossible, since both square root and $\text{cosh}<!--\; FUNCTION\; APPLICATION\; -->$ are always positive.

   Therefore, the only possibility is when $a<b<\sqrt{a^2+1}$.

4. When they touch at a point, this will mean at this value, the number of solutions will change on both sides. This is only possible when $b=\sqrt{a^2+1}$.

   Therefore,

   $$x=-\text{artanh}<!--\; FUNCTION\; APPLICATION\; -->\frac{a}{\sqrt{a^2+1}}.$$

   Hence,

   $$\begin{array}{llll}\hfill y& =a\text{tanh}<!--\; FUNCTION\; APPLICATION\; -->x+b& \hfill & \\ \hfill & =-a\cdot \frac{a}{\sqrt{a^2+1}}+\sqrt{a^2+1}& \hfill & \\ \hfill & =\frac{-a^2+a^2+1}{\sqrt{a^2+1}}& \hfill & \\ \hfill & =\frac{1}{\sqrt{a^2+1}}.& \hfill & \end{array}$$