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3

## How do I calculate the Fourier Transform of a/(2pi) 1/(a^2 + x^2)?

Steve M
Featured yesterday · Calculus

${\int}_{- \infty}^{\infty} \setminus {e}^{i \omega x} / \left({a}^{2} + {x}^{2}\right) \setminus \mathrm{dx} = \frac{\pi {e}^{- \omega a}}{a}$

#### Explanation:

I never studied Fourier Transforms, but at the end of the day it is just an integral transform similar to Laplace, so:

I won't worry about the $\frac{a}{2 \pi}$ weight of the function or the $\frac{1}{\sqrt{2 \pi}}$ weight of the transform, and just calculate

${I}_{x} = {\int}_{- \infty}^{\infty} \setminus {e}^{i \omega x} / \left({a}^{2} + {x}^{2}\right) \setminus \mathrm{dx}$ where $x \in \mathbb{R}$.

In order to compute this definite integral, consider the following complex variable function over a domain $\mathbb{C}$:

$f \left(z\right) = {e}^{i \omega z} / \left({a}^{2} + {z}^{2}\right)$

And its associated contour integral:

${I}_{z} = {\oint}_{C} \setminus f \left(z\right) \setminus \mathrm{dz}$

Where $C$ is the following semi-circular contour in the complex plane with radius $R > 0$:

We will restrict $R$ to enclose the poles in the upper quadrants once we have analyzed the poles of $f \left(z\right)$. The denominator of the integrand is ${a}^{2} + {z}^{2}$, and so we have simples poles when ${z}^{2} + {a}^{2} = 0 \implies z = \pm a i$.

We need only be concerned with the poles in Q1 and Q2, that (providing we make $R$ large enough) lie within our contour $C$, that is the pole $z = a i$, assuming that $a$ is positive.

So then:

${\oint}_{C} \setminus f \left(z\right) \setminus \mathrm{dz} = {\int}_{- R}^{R} \setminus f \left(x\right) \setminus \mathrm{dx} + {\int}_{{\gamma}_{R}} \setminus f \left(z\right) \setminus \mathrm{dz}$

As $C$ encloses a single pole:

$\alpha = a i$ (assuming $a > 0$)

Then by the residue theorem:

${\oint}_{C} \setminus f \left(z\right) \setminus \mathrm{dz} = 2 \pi i \times \left(\begin{matrix}\text{sum of the residues of the" \\ "poles of " f(z) " within } C\end{matrix}\right)$
 " " = 2pii \ { res_(z=alpha) \ f(z)

And we can calculate the residues as follows:

$r e {s}_{z = \alpha} = {\lim}_{z \rightarrow \alpha} \left(z - \alpha\right) f \left(z\right)$
$\text{ } = {\lim}_{z \rightarrow a i} \left(z - a i\right) \left({e}^{i \omega z} / \left({a}^{2} + {z}^{2}\right)\right)$
$\text{ } = {\lim}_{z \rightarrow a i} \left(z - a i\right) \left({e}^{i \omega z} / \left(\left(z - a i\right) \left(z + a i\right)\right)\right)$
$\text{ } = {\lim}_{z \rightarrow a i} \left({e}^{i \omega z} / \left(z + a i\right)\right)$
$\text{ } = {e}^{i \omega a i} / \left(a i + a i\right)$
$\text{ } = {e}^{- \omega a} / \left(2 a i\right)$

Thus:

${\oint}_{C} \setminus f \left(z\right) \setminus \mathrm{dz} = 2 \pi i {e}^{- \omega a} / \left(2 a i\right) = \frac{\pi {e}^{- \omega a}}{a}$

Now, as we let $R \rightarrow \infty$ we get:

${\oint}_{C} \setminus f \left(z\right) \setminus \mathrm{dz} = {\int}_{- \infty}^{\infty} \setminus f \left(x\right) \setminus \mathrm{dx} + {\int}_{{\gamma}_{R}} \setminus f \left(z\right) \setminus \mathrm{dz}$

As is often the case with contour integrals, we find that:

${\lim}_{R \rightarrow \infty} {\int}_{{\gamma}_{R}} \setminus f \left(z\right) \setminus \mathrm{dz} = 0$

So we will end up with the result:

${\int}_{- \infty}^{\infty} \setminus {e}^{i \omega x} / \left({a}^{2} + {x}^{2}\right) \setminus \mathrm{dx} = {\oint}_{C} \setminus f \left(z\right) \setminus \mathrm{dz}$
$\text{ } = \frac{\pi {e}^{- \omega a}}{a}$

2

## If the volume of the hemisphere is the same as that of the frustum, find the volume of the new solid in terms of pi. Can show me the step plz????

dk_ch
Featured yesterday · Geometry

Given that the volume of the hemisphere of radius $r$ s the same as that of the frustum.

So volume of frustum = volume of hemisphere
$= \frac{1}{2} \cdot \frac{4}{3} \pi {r}^{3} = \frac{2}{3} \pi {r}^{3}$

So volume of new solid = volume of the sphere $= \frac{4}{3} \pi {r}^{3}$

Now height of the cone

$V O = \sqrt{V {A}^{2} - O {A}^{2}}$

$\implies V O = \sqrt{{50}^{2} - {30}^{2}} = 40 c m$

$\Delta s V O A \mathmr{and} V C B$ are similar

So $\frac{V C}{V O} = \frac{B C}{O A}$

$\implies V C = \frac{B C}{O A} \times V O = \frac{r}{30} \times 40 = \frac{4 r}{3}$

So volume of frustum

$\frac{1}{3} \pi O {A}^{2} \times V O - \frac{1}{3} \pi {r}^{2} \times V C$

$= \frac{1}{3} \pi \left({30}^{2} \times 40 - {r}^{2} \times \frac{4 r}{3}\right)$

As the volume of frustum = volume of hemisphere

$\frac{1}{3} \pi \left({30}^{2} \times 40 - {r}^{2} \times \frac{4 r}{3}\right) = \frac{2}{3} \pi {r}^{3}$

$\implies \left({30}^{2} \times 40 - \frac{4}{3} {r}^{3}\right) = 2 {r}^{3}$

$\implies \left(\frac{4}{3} + 2\right) {r}^{3} = 36000$

$\implies {r}^{3} = 10800$

So volume of new solid = volume of the sphere

$= \frac{4}{3} \pi {r}^{3} = \frac{4}{3} \pi \times 10800 = 14400 \pi c {m}^{3}$

3

## Heisenberg's uncertainty principle and its significance?

Urip
Featured yesterday · Chemistry

DeltaxDeltap>=ħ/2

#### Explanation:

The Heisenberg's Uncerttainty principle stablishes the limit to the precision in which a pair of measurable variables, such as position, $\Delta x$, and linear momentum, $\Delta p$, can be determined. And is always represented by an inequality such as

DeltaxDeltap>=ħ/2

Where:

• $\Delta x$ is the uncertainty of the position of the particle;
• $\Delta p$ is the uncertainty of the momentum of the particle.

This means that for a quantum object, such as an electron or a photon, the more precision we get on it's position uppon a measure, the less information we will have about it's momentum, and vice versa. From the expression, we see that:

Deltax>=ħ/(2Deltap)

So if we take a measure of the particle's momentum and obtain it's exact momentum, this means that $\Delta p \to 0$, so

Deltax>=ħ/0=>Deltax->oo

Although this is the most known uncertainty equation, there are others.

Notice that this is a natural physical phenomena and it does not depend on the measure techniques used in the experiment, the Heisenberg's Uncertainty Principle holds and since it's discovery, has never been violated.

Now let's try to shed some light on what actually happens on an example. Suppose we have an electron moving on a tube in our lab. We know where it's coming from (somewhere on the left), since we arranged that. But since it is moving, we don't know its exact position (other than it is on the tube) and we don't know its exact velocity. If we let it go across the whole tube, it will hit a wall.

If we do nothing, the uncertainty of position along the tube axis of our electron is the whole tube's length. If the length is L, then $\Delta x = L$

And this gives us the uncertainty of the momentum Deltap>=ħ/(2L).

But now, we want to determine the exact momentum of the electron while it is travelling across the tube. To do so, we must throw something at him. And what scientists usually do is hit it with a photon, which has known energy and momentum. As in the picture.

Our "test photon" hits the electron, being scattered on the process, having its energy and momentum changed.
The electron, on other hand, has its momentum changed and its initial path due to the collision.

So if our photon is emmited transversaly from one side of the tube to the other, in a way that we know when it arrives on the other side or not, and we know that if it doesn't, that means it hit the electron on that part. Check the image below, that is the view as if you were looking from the end of the tube, just before the end wall.

In this case, you know with exact precision the position of the electron with respect to the tube's axis on that particular moment. But, for that, you lost all information about the electron's momentum, since you have no idea of how much energy and momentum the scattered photon exchanged with the electron, you can't apply conservation laws to determine it's energy and momentum after the collision.

Hope this helps you, but bare in mind that quantum mechanics can be very confusing and usually takes some very hardcore study time to understand its basics.

3

## Resistance?

dk_ch
Featured yesterday · Physics

Considering the side view of the given solid showing trapezium section.The mid point of rectangular surface $A$ is considered as origin O and $O X$ ,the X-axis is perpendicular to surface $A$ at O.
We also consider a rectangular slice of length $y$ and width $w$ parallel to the surface $A$ at a distance $x$ from origin $O$. The slice has infinitesimally small thickness $\mathrm{dx}$.

Now length of the rectangular cross section increases from A to B at a definite rate along X-axis and the rate is$= \frac{{y}_{2} - {y}_{1}}{L}$

So $y = {y}_{1} + x \left(\frac{{y}_{2} - {y}_{1}}{L}\right)$

Hence area of the selected slice

$A = w \times y$

If $\mathrm{dR}$ be the resistance of the slice then we can write

$\mathrm{dR} = \rho \frac{\mathrm{dx}}{A} = \rho \frac{\mathrm{dx}}{w y} = \frac{\rho}{w} \left(\frac{\mathrm{dx}}{{y}_{1} + x \left(\frac{{y}_{2} - {y}_{1}}{L}\right)}\right)$

Taking integration between $\left(0 , L\right)$ we get the resistance of the solid across AB

$R = {\int}_{0}^{L} \frac{\rho}{w} \left(\frac{\mathrm{dx}}{{y}_{1} + x \left(\frac{{y}_{2} - {y}_{1}}{L}\right)}\right)$

$\implies R = \frac{\rho}{w} {\int}_{0}^{L} \left(\frac{\mathrm{dx}}{{y}_{1} + x \left(\frac{{y}_{2} - {y}_{1}}{L}\right)}\right)$

$\implies R = \frac{\rho}{w} \times \frac{L}{{y}_{2} - {y}_{1}} {\left[\ln \left({y}_{1} + x \left(\frac{{y}_{2} - {y}_{1}}{L}\right)\right)\right]}_{0}^{L}$

$\implies R = \frac{\rho}{w} \times \frac{L}{{y}_{2} - {y}_{1}} \left[\ln \left({y}_{1} + L \left(\frac{{y}_{2} - {y}_{1}}{L}\right)\right) - \ln \left({y}_{1} + 0 \times \left(\frac{{y}_{2} - {y}_{1}}{L}\right)\right)\right]$

$\implies R = \rho \frac{L}{w \left({y}_{2} - {y}_{1}\right)} \ln \left({y}_{2} / {y}_{1}\right)$

3

## How do you do this question?

George C.
Featured yesterday · Calculus

${\sum}_{k = 1}^{n} 3 k \left({k}^{2} + 4\right) = \frac{3}{4} n \left(n + 1\right) \left({n}^{2} + n + 8\right)$

#### Explanation:

We want to find:

${s}_{n} = {\sum}_{k = 1}^{n} 3 k \left({k}^{2} + 4\right)$

The first few terms of this sum are:

$15 , 48 , 117 , 240 , 435 , 720$

So the sequence we want a formula for is the sums:

$\textcolor{b l u e}{15} , 63 , 180 , 420 , 855 , 1575$

The sequence of differences of that sequence is:

$\textcolor{b l u e}{48} , 117 , 240 , 435 , 720$

The sequence of differences of those differences is:

$\textcolor{b l u e}{69} , 123 , 195 , 285$

The sequence of differences of those differences is:

$\textcolor{b l u e}{54} , 72 , 90$

The sequence of differences of those differences is:

$\textcolor{b l u e}{18} , 18$

Having reached a constant sequence, we can use the first term of each of these sequences as coefficients to give us a formula for the sum to $n$ terms...

s_n = color(blue)(15)/(0!)+color(blue)(48)/(1!)(n-1)+color(blue)(69)/(2!)(n-1)(n-2)+color(blue)(54)/(3!)(n-1)(n-2)(n-3)+color(blue)(18)/(4!)(n-1)(n-2)(n-3)(n-4)

$\textcolor{w h i t e}{{s}_{n}} = 15 + 48 n - 48 + \frac{69}{2} {n}^{2} - \frac{207}{2} n + 69 + 9 {n}^{3} - 54 {n}^{2} + 99 n - 54 + \frac{3}{4} {n}^{4} - \frac{15}{2} {n}^{3} + \frac{105}{4} {n}^{2} - \frac{75}{2} n + 18$

$\textcolor{w h i t e}{{s}_{n}} = \frac{3 {n}^{4}}{4} + \frac{3 {n}^{3}}{2} + \frac{27 {n}^{2}}{4} + 6 n$

$\textcolor{w h i t e}{{s}_{n}} = \frac{3}{4} n \left({n}^{3} + 2 {n}^{2} + 9 n + 8\right)$

$\textcolor{w h i t e}{{s}_{n}} = \frac{3}{4} n \left(n + 1\right) \left({n}^{2} + n + 8\right)$

That is $A = 1$, $B = 1$, $C = 8$

Then:

${s}_{100} = \frac{3}{4} \left(\textcolor{b l u e}{100}\right) \left(\textcolor{b l u e}{100} + 1\right) \left({\left(\textcolor{b l u e}{100}\right)}^{2} + \left(\textcolor{b l u e}{100}\right) + 8\right)$

$\textcolor{w h i t e}{{s}_{100}} = 75 \left(101\right) \left(10000 + 100 + 8\right) = 7575 \cdot 10108 = 76568100$

${s}_{60} = \frac{3}{4} \left(\textcolor{b l u e}{60}\right) \left(\textcolor{b l u e}{60} + 1\right) \left({\left(\textcolor{b l u e}{60}\right)}^{2} + \left(\textcolor{b l u e}{60}\right) + 8\right)$

$\textcolor{w h i t e}{{s}_{60}} = 45 \cdot 61 \cdot 3668 = 10068660$

So:

${\sum}_{k = 61}^{100} 3 k \left({k}^{2} + 4\right) = {s}_{100} - {s}_{60} = 76568100 - 10068660 = 66499440$

2

## How do the pituitary and hypothalamus interact to regulate thyroxine levels?

Mandira P.
Featured yesterday · Anatomy & Physiology

Thyroid hormone thyroxine (T3, T4, etc) is secreted in response to a trophic hormone released by anterior pituitary. It is named thyrotropin or thyroid stimulating hormone (=TSH).

Anterior pituitary in turn acts under the influence of thyrotropin releasing hormone (=TRH): a neurohormone released by hypothalamus in blood of hypothalamic hypophyseal portal system.

This is summarised in the following diagram:

So when there is secretion of TRH, there is subsequent release of TSH and Thyroxine level in blood will rise.

Higher level of thyroxine, in turn will send negative feedback to both anterior pituitary and hypothalamus. The negative feedback will ensure a decreased production of TRH and TSH, which means level of thyroxine will not increase beyond a normal threshold in normal healthy individuals.

All these hormones are released in highly controlled manner to maintain a steady homoeostasis.

3

## How to prove tan x = sin x / cosx rather mathematically?

Tom M.
Featured yesterday · Trigonometry

Yes, there is a way. You're on the right tracks, but there's also another way which takes us back to where the trigonometric functions come from.

#### Explanation:

We can prove this the way you are doing this.

$\sin x = \frac{O}{H}$
$\cos x = \frac{A}{H}$

$\sin \frac{x}{\cos} x = \frac{O}{H} \div \frac{A}{H}$
$= \frac{O}{\cancel{H}} \times \frac{\cancel{H}}{A}$
$= \frac{O}{A} = \tan x$ as required.

The other way to prove this is to go back to where the trigonometric functions come from in the first place.

Take a unit circle. This is a circle, radius 1.

(all angles are measured in degrees from this point on).

Let us make a right-angled triangle from the origin, with the hypotenuse being the radius. This triangle has an angle $\theta$ (theta) to the x-axis.

We will give this triangle a height of $\sin \theta$. This is the side opposite to the angle. Since the hypotenuse is 1 (as is the radius), $O = \sin \theta$

This triangle has another non-right-angle, however. This angle is $90 - \theta$. This angle is known as complementary to $\theta$, which means that they sum to ${90}^{o}$.

The line adjacent to our original angle $\theta$ is the line opposite to our complementary angle $90 - \theta$. So $A = \sin \left(90 - \theta\right)$. Since this the sin of the complementary angle, we shall call this the complementary sine, cosine. $A = \cos \theta$

The other thing we can do is to draw a tangent to the unit circle from the "top" of the triangle to the x-axis. We will call this length $\tan \theta$.

Since a tangent meets a radius at a right angle, this gives us two triangles.

The first triangle on the left is our triangle with a radius 1 from the unit circle, and sides $\sin \theta$ and $\cos \theta$ from above. The triangle on the right has a base of the x-axis, a side of 1 from the unit circle and the other of $\tan \theta$ from our tangent. It has a right angle from where a tangent meets a radius.

Since both of these triangles all have the same angles ($90 - \theta - \left(90 - \theta\right)$, they are mathematically similar.

We can find the scale factors of these triangles. By looking at the length between the angle $\theta$ and the right angle, there is a scale factor of $\frac{1}{\cos} \theta$. The scale factor by looking at the sides between $90 - \theta$ and the right angle is $\tan \frac{\theta}{\sin} \theta$

Since this is the same scale factor in both cases:

$\frac{1}{\cos} \theta = \tan \frac{\theta}{\sin} \theta$

By multiplying both sides by $\sin \theta$

$\sin \frac{\theta}{\cos} \theta = \tan \theta$ as required.

3

## In my book there is a example of a Mountain with Windward and Leeward side so I searched if there's mountain like that but... Can you Help me please What is windward and Leeward side? Give me one of an Example of Windward and Leeward side?

Mark Kononov
Featured 7 hours ago · Earth Science

The windward side of a mountain is the side in which moisture in the air is released as it goes up in elevation, and the leeward side is the side where the air goes down into the land, now dry.

#### Explanation:

The best way to explain windward and leeward is an example. An example close to where I live is Mount Hood, Oregon. The Pacific Ocean is west of Mount Hood, and all air heading towards the east picks up moisture from the ocean. As the most air travels throughout the land, it rains so the area on the windward side of a mountain has more precipitation.

When the moist air heads towards the mountain (Mount Hood in this case), it gains elevation. Air pressure is the weight of the atmosphere, so as the air moves up there is less atmosphere above it and therefore less pressure. According to Gay Lussac's law pressure and temperature are proportional, so as the pressure drops the temperature also drops. The amount of water vapor air can hold is based on it's temperature, the cooler the air the less vapor. So as the air cools condensation occurs and precipitates out.

https://en.wikibooks.org/wiki/High_School_Earth_Science/Weather_and_Atmospheric_Water

The red line in the graph above shows the maximum amount of water vapor air can hold, and you can see it increases with temperature.

The side of the mountain where this happens is called the windward side, since this is the side the wind blows against. When the air travels over the mountain and descends that's called the leeward side.

On the leeward side, as the air loses elevation the pressure increases, and therefore the temperature increase. As the graph above shows as the air warms it's ability to hold moisture increases, but since the air lost moisture on the windward side, if there is no source for new moisture on the leeward side the air just gets drier and drier. The leeward area is often a desert or a dry place, since the air has much less moisture than it can actually hold.

Here's a map of Oregon to show an example:

The green area is the area west of the mountain, so the west side of the mountains is the windward area (Also Mount Hood is part of the Cascade Mountains). The orange area is then the desert area, and then that means the east side of the mountains is the leeward side.

1

## If angle A=20, side a=30, and side c=40, what are all possible values for angle C? .

Tom M.
Featured 8 hours ago · Trigonometry

$C = {27.1}^{\circ}$ or $C = {153}^{\circ}$

#### Explanation:

This question concerns itself with the ambiguous case of the sine rule.
Sometimes with the sine rule, you can get two values for the angle. One is acute, the other is obtuse.
This is partly because there are two triangles we can draw for the information given:

The way we calculate the angle is the same regardless. We use the sine rule:

$\sin \frac{A}{a} = \sin \frac{C}{c}$.

$\sin C = \frac{c \sin A}{a}$

$\sin C = \frac{40 \sin 20}{30}$

$\sin C = 0.45602 \ldots$

Leave this value in your calculator.

One of the answers for the angle will be the inverse sine of this.

$C = 27.131 \ldots$
$C = {27.1}^{\circ}$

C can also be $180 - C$, (I will show why later).

so $C = 180 - 27.131 \ldots$
$C = 152.868 \ldots$
$C = {153}^{\circ}$

So why does this work? Consider the graph of $y = \sin x$ for ${0}^{\circ} \le x \le {180}^{\circ}$

The red curve is the graph of $y = \sin x$, and the green line is the line $y = \frac{4}{3} \sin 20$, which is what we obtained from the sine rule earlier.

Since we know that our angle is in a triangle, it cannot be greater than 180, and cannot be less than 0.

However, in this range, note how there are two solutions to these equations? One at ${27.131}^{\circ} ,$ and one at ${153.869}^{\circ}$. This means that our angle C can take either of these values, whether it is acute or obtuse.

For any angle, $\theta , \sin \theta = \sin \left(180 - \theta\right)$. It is necessary to consider both angles.

This does not arise for the cosine rule. The graph of $y = \cos x$ does not give the same value for the range $0 \le x \le 180$, so such a problem does not arise.

1

## What is x and y when 3x+y=6 and y=x-2?

Tony B
Featured 15 hours ago · Algebra

the point common to both plots is $\left(x , y\right) \to \left(2 , 0\right)$

#### Explanation:

Given:

$3 x + y = 6 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots E q u a t i o n \left(1\right)$
$y = x - 2 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(2\right)$

$\textcolor{b l u e}{\text{Determine the value of } x}$

Using $E q n \left(2\right)$ substitute for $\textcolor{red}{y}$ in $E q n \left(1\right)$ giving:

$\textcolor{g r e e n}{3 x + \textcolor{red}{y} \textcolor{w h i t e}{\text{d")=color(white)("d")6 color(white)("dddd") ->color(white)("dddd")3x+(color(red)(x-2))color(white)("d")=color(white)("d}} 6}$

$\textcolor{g r e e n}{\textcolor{w h i t e}{\text{ddddddddddd.d")->color(white)("dddddd") 4xcolor(white)("ddd")-2color(white)("ddd")=color(white)("d}} 6}$

Add $\textcolor{red}{2}$ to both sides

$\textcolor{g r e e n}{4 x - 2 \textcolor{w h i t e}{\text{d")=color(white)("d")6 color(white)("dddd") ->color(white)("dddd")4xcolor(white)("d")-2color(red)(+2)color(white)("dd")=color(white)("d}} 6 \textcolor{red}{+ 2}}$

color(green)(color(white)("ddddddddddddd.d")->color(white)("dddd") 4xcolor(white)("d")+color(white)("d")0color(white)("dddd")=color(white)("dd")8

Divide both sides by $\textcolor{red}{4}$

$\textcolor{g r e e n}{4 x \textcolor{w h i t e}{\text{d")=color(white)("d")8color(white)("dddddd.d")->color(white)("dddd") 4/color(red)(4)xcolor(white)("d")=color(white)("d}} \frac{8}{\textcolor{red}{4}}}$

$\textcolor{g r e e n}{\textcolor{w h i t e}{\text{ddddddddddddd.d")->color(white)("dddddd") xcolor(white)("d")=color(white)("d}} 2}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Determine the value of } y}$

Substitute for $\textcolor{red}{x = 2}$ in $E q n \left(1\right)$

$\textcolor{g r e e n}{y \textcolor{w h i t e}{\text{d")=color(white)("d")color(red)(x)-2 color(white)("dddd")->color(white)("dddd")ycolor(white)("d")=color(white)("d}} \textcolor{red}{2} - 2}$

$\textcolor{g r e e n}{\textcolor{w h i t e}{\text{dddddddddddddd")->color(white)("dddd")ycolor(white)("d")=color(white)("d}} 0}$

$\textcolor{m a \ge n t a}{\text{So the point common to both plots is } \left(x , y\right) \to \left(2 , 0\right)}$