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Differentiation in history, Pierre de Fermat and René Descartes.
- Thread starter NinjaKiwi
- Start date
Then what is required is a general formula for the tangent of \(\displaystyle \sqrt{ax}\) at some point.
Per the reference, we can let \(\displaystyle y=\sqrt{ax}\).
This gives \(\displaystyle x^{2}+(a-2c)x+2c-2=0\)
The discriminant must be 0:
\(\displaystyle (a-2c)^{2}-4(1)(2c-2)=0\)
Solving for c gives \(\displaystyle c=\frac{\pm 2\sqrt{a-1}+a+2}{2}\)
Per the reference, we can let \(\displaystyle y=\sqrt{ax}\).
This gives \(\displaystyle x^{2}+(a-2c)x+2c-2=0\)
The discriminant must be 0:
\(\displaystyle (a-2c)^{2}-4(1)(2c-2)=0\)
Solving for c gives \(\displaystyle c=\frac{\pm 2\sqrt{a-1}+a+2}{2}\)
D
Deleted member 4993
Guest
Thank you!
One question though, does it matter if I use (a-x)^2 instead of (a-y)^2 in the circle equation?
Do the results differ?
That'll depend on your problem -
which way the center of the circle is shifted (relative to the origin)?
ok, so it is either positive or negative? therefore the +- in the equation that galactus solved?
http://math.kennesaw.edu/~jdoto/13.pdf looking at the suggestion given of Fermat(scroll down 5 pages), should i use E*sqrt(ax) instead of E^2?
http://math.kennesaw.edu/~jdoto/13.pdf looking at the suggestion given of Fermat(scroll down 5 pages), should i use E*sqrt(ax) instead of E^2?