I Detailed solution of an envelopes example in a math book

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I am looking for steps for the solution of an envelopes example in Mathematical Methods for Physics and Engineering book
Can someone please show me the steps from circled equation 1 to 2?
Envelopes.JPG

Thank you.
 
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If you replace ##x \leftrightarrow y## and ##a \leftrightarrow b## in the previous equation, you get the last equation.
 
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I believe the author of your text book? @Orodruin
 
erobz said:
I believe the author of your text book? @Orodruin
No. There are several textbooks with that or similar names. This is from Riley et al. I believe.

@vgarg This is why you should never just quote the name of the book, but also the author(s).
 
Hill said:
If you replace ##x \leftrightarrow y## and ##a \leftrightarrow b## in the previous equation, you get the last equation.
To clarify: By the symmetry of the problem, you could have solved it for ##b## to get
$$
b = \frac{L y^{1/3}}{(x^{2/3} + y^{2/3})^{1/2}}
$$
instead of
$$
a = \frac{L x^{1/3}}{(x^{2/3} + y^{2/3})^{1/2}}
$$

Then simply substitute ##b## in the first equation using ##a^2 + b^2 = L^2##.
 
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Thank you very much for your help. I will sure list the author of the book in any future post.
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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