- Thread starter
- #1

$\alpha = 2\arcsin\left(\sqrt{\frac{s}{2a}}\right)$

$\beta = 2\arcsin\left(\sqrt{\frac{s-c}{2a}}\right)$

$$

\lim_{a\to\infty}\left[a^{3/2}(\alpha - \beta -(\sin(\alpha) - \sin(\beta))\right]

$$

- Thread starter dwsmith
- Start date

- Thread starter
- #1

$\alpha = 2\arcsin\left(\sqrt{\frac{s}{2a}}\right)$

$\beta = 2\arcsin\left(\sqrt{\frac{s-c}{2a}}\right)$

$$

\lim_{a\to\infty}\left[a^{3/2}(\alpha - \beta -(\sin(\alpha) - \sin(\beta))\right]

$$

- Admin
- #2

- Mar 5, 2012

- 9,165

First I would rewrite the expressions for $\alpha, \beta, \sin \alpha, \sin \beta$.

$\alpha = 2\arcsin\left(\sqrt{\frac{s}{2a}}\right)$

$\beta = 2\arcsin\left(\sqrt{\frac{s-c}{2a}}\right)$

$$

\lim_{a\to\infty}\left[a^{3/2}(\alpha - \beta -(\sin(\alpha) - \sin(\beta))\right]

$$

The method to do so, is to draw a right triangle with sides that match your angles.

Combine that with the double-angle-formulas.

You'll get for instance:

$$\quad \alpha = \arccos\left(1-\frac s a\right) \\

\quad \alpha = \arcsin\left(\sqrt{\frac {2as -s^2}{a^2}}\right) \\

\quad \sin \alpha = \sqrt{\frac {2as -s^2}{a^2}}$$

The next step that springs to mind is a Taylor expansion, which is basically a more advanced form of l'Hôpital's rule.

Let's define $A=\sqrt{\dfrac {2as -s^2}{a^2}}$, then for instance, since $\dfrac s a \ll 1$ (near the limit), we have:

$$\quad \alpha = \arcsin(A) = A + \frac {A^3}{6} + \frac{3A^5}{40} + ... \\

\quad \sin \alpha = \sqrt{\frac {2s}{a}}\sqrt{1 - \frac{s}{2a}} = \sqrt{\frac {2s}{a}} \left(1 - \frac 1 2 \cdot \frac{s}{2a} + ...\right)$$

Last edited:

- Thread starter
- #3

I dont think that will arrive at the solution of $\frac{\sqrt{2}}{3}(s^{3/2} - (s-c)^{3/2})$ though.First I would rewrite the expressions for $\alpha, \beta, \sin \alpha, \sin \beta$.

The method to do so, is to draw a right triangle with sides that match your angles.

Combine that with the double-angle-formulas.

You'll get for instance:

$$\quad \alpha = \arccos\left(1-\frac s a\right) \\

\quad \alpha = \arcsin\left(\sqrt{\frac {2as -s^2}{a^2}}\right) \\

\quad \sin \alpha = \sqrt{\frac {2as -s^2}{a^2}}$$

The next step that springs to mind is a Taylor expansion, which is basically a more advanced form of l'Hôpital's rule.

Let's define $A=\sqrt{\dfrac {2as -s^2}{a^2}}$, then for instance, since $\dfrac s a \ll 1$ (near the limit), we have:

$$\quad \alpha = \arcsin(A) = A + \frac {A^3}{6} + \frac{3A^5}{40} + ... \\

\quad \sin \alpha = \sqrt{\frac {2s}{a}}\sqrt{1 - \frac{s}{2a}} = \sqrt{\frac {2s}{a}} \left(1 - \frac 1 2 \cdot \frac{s}{2a} + ...\right)$$

- Admin
- #4

- Mar 5, 2012

- 9,165

The method I described is guaranteed to work.I dont think that will arrive at the solution of $\frac{\sqrt{2}}{3}(s^{3/2} - (s-c)^{3/2})$ though.

Moreover, it is more general than l'Hôpital.