$e^y = \frac{x}{a} + \sqrt{( 1 + (\frac{x}{a})^2}$
The problem asks to solve this equation to x. My problem stands still by the square root.
Asked
Active
Viewed 27 times
0
-
2Do you mean $$e^{y}=\frac{x}{a}+\sqrt{1+\left(\frac{x}{a}\right)^2}$$? – Dr. Sonnhard Graubner Sep 21 '19 at 16:17
-
1Hint: $(\sqrt{1+u^2} + u)(\sqrt{1 + u^2} - u) = 1$ – achille hui Sep 21 '19 at 16:19
-
Well in general if you have $M = cx + \sqrt{a + dx^2}$ you can do $M-cx =\sqrt {a + dx^2}$ and $(M-cx)^2 = a+dx^2$ and now it's just a quadratic. – fleablood Sep 21 '19 at 16:40
2 Answers
1
If so we write $$e^y-\frac{x}{a}=\sqrt{1+\left(\frac{x}{a}\right)^2}$$ squaring we get $$e^{2y}+\left(\frac{x}{a}\right)^2-2e^y\times \frac{x}{a}=1+\left(\frac{x}{a}\right)^2$$ Combining like terms and isolating $x$ we get $$x=a\times \left(\frac{e^{2y}-1}{2e^y}\right)$$
Dr. Sonnhard Graubner
- 95,283
1
Note that
- $\operatorname{arsinh} t = \ln\left(t+\sqrt{1+t^2}\right)$
Setting $t = \frac{x}{a}$, your equation becomes $$e^y = t+\sqrt{1+t^2} \Leftrightarrow y = \operatorname{arsinh} t$$
Hence,
$$t = \frac{x}{a} = \sinh y \Leftrightarrow \boxed{x= a\sinh y = a\frac{e^y-e^{-y}}{2}}$$
trancelocation
- 32,243
-
Thank you for precious your time and the arcsine transformation notation. – Ilgaz Kuscu Sep 21 '19 at 17:19