MHB High school inequality find b in √[(x−1)^2+(y−2)^2]<b⟹|xy2−4|<a

solakis1
Messages
407
Reaction score
0
given a>0 find b>0 such that:
$$\sqrt{(x-1)^2+(y-2)^2}<b\Longrightarrow |xy^2-4|<a$$
 
Mathematics news on Phys.org
My attempt:
The $x$ and $y$ values are confined to the area of an open disc of radius $b$ with center at $(1,2)$.

The largest allowed $y$-variation ($y_{max}-y_{min} = 2b$) happens for $x=1$.

The largest allowed $x$-variation happens for $y=2$.

Hence, we have the relations:

$x=1$:

$|y-2| < b\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$(1).

and $|y^2-4| < a\;\;\;\;\;\;\;$(2).$y=2$:

$|x-1| < b\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$(3).

and $|4x-4| < a$ or $|x-1| < \frac{a}{4}\;\;$(4).(2). Implies: $\sqrt{4-a}<y <\sqrt{4+a}$. Note, that $a \le 4$. (There is a symmetric case for $y < 0$, which is omitted).

(1). Implies: $2-b < y < 2+b$.

Now, $a >0$ is given, therefore:

$2+b \le \sqrt{4+a}$ and $2-b \ge \sqrt{4-a} $

$\Rightarrow b \le 2- \sqrt{4-a}$ and $b \leq \sqrt{4+a}-2$.

We need to choose the smaller of the two, which is the latter: $\sqrt{4+a}-2$. This can be seen e.g. from their Taylor expansion:

\[2- \sqrt{4-a} \approx \frac{a}{4}+\frac{a^2}{64}+O(a^3) \\\\ \sqrt{4+a}-2 \approx \frac{a}{4}-\frac{a^2}{64}+O(a^3)\]

From (3) and (4) we immediately get: $b \le \frac{a}{4} $.

The difference between the two possible boundaries - according to the Taylor expansion - is:

\[\sqrt{4+a}-2 -\frac{a}{4} \approx -\frac{a^2}{64}+O(a^3) < 0\]Thus, $b =\sqrt{4+a}-2$ is the only possible choice. The largest possible $b$-value is obtained when $a = 4$:
$b = \sqrt{8}-2 \approx 0.8284$.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Back
Top