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11

One example would be the gravitational potential energy of a point in relation to a pointwise mass in space. The closer you are to the point, the faster you go. http://en.wikipedia.org/wiki/Potential_energy#Potential_energy_for_gravitational_forces_between_two_bodies If you want simpler examples, take any basic equation that implies a linear connection ...


9

Physics has lots of examples, but it's already the closest field to math. (Plus, in order to observe asymptotic gravity, you'd need a black hole...) You could use Walmart. If shoppers arrive nondeterministically at rate $\lambda$ and are served at nondeterministically at rate $\mu$, the average wait time is $$\frac{1}{\mu − \lambda} − \frac{1}{\mu}$$ ...


6

From distance equals rate times time, you get $r=\frac{d}{t}$ For a fixed distance, the less time you take to cover that distance, the faster you go, with a vertical asymptote at $t=0$.


6

You can think of it also as, Where do $y=\sin{(3x)}$ and $y=-1$ intersect.


6

What you're looking for can be described as a parabola opening towards the positive $x$ axis. I'm going to refer to it as a "sideways parabola," since the "standard" parabola that people learn opens towards the positive $y$ axis. The bad news: You cannot express a sideways parabola as a function of $x$. Why? Let's go back and look at a restriction on ...


5

Throw a stone obliquely. Due to air friction, the trajectory follows a vertical asymptote. http://www.mathcurve.com/courbes2d/paraboleamortie/paraboleamortie.shtml


4

The graph you describe will have equation $x=y^2$. You cannot write it in the form "$y=\langle\hbox{function of $x$}\rangle$" because, just as you pointed out, every $x>0$ will correspond to two $y$ values, not just one.


3

You can use domain coloring: The two roots can be seen clearly: they are the points having a rainbow around it.


3

As pointed out in the comments, the graph is indeed an ellipsoid. In the event you don't have a 3D graphing program handy, here's one way to think about it: we have the equation $$x^2+2y^2+3z^2=12.$$ We'll set the coordinates $x,y,z=0$ (one at a time) and this will tell us what the graph looks like in the $yz$, $xz$, and $xy$-planes respectively. For ...


3

Newton's inverse square law of gravity! Can't much closer to real-life, everyday experiences than gravity! The law reads: $F = G\frac {m_1m_2} {d^2}$ Where $F$ is the gravitational force between two bodies, $G$ is the universal gravitational constant, $m_1$ and $m_2$ are the masses of the two bodies, and $d$ is the distance between the two bodies. ...


3

Yes, $F(x)=x$ for all $x\in[0,1]$. Sketch of proof: $\Phi(n)$ counts all primitive lattice points in the triangle with vertices $(0,0)$, $(0,n)$, and $(n,n)$. (Here a lattice point $(a,b)$ is primitive if $\gcd(a,b)=1$. The bijection sends $(a,b)$ to $a/b$.) You can count primitive lattice points by inclusion-exclusion: if $F(n)$ is the total number of ...


3

Your graph is correct. The vertical segments you see in the second graph may be a result of a graphing program that is set in continuous mode so that it connects the dots even when they should not be connected, as in a step function or at a vertical asymptote. A function $f$ associates a unique value $f(x)$ to each $x$ in its domain. When we graph a ...


3

Your logic is correct, and you have proved that the function is even. To draw a graph of the function, all you need to do is to draw the graph $y=|x|$ and shift it down by $3$ for the $-3$ factor in the equation. Here is the graph:


2

A straight parabola (i.e., one with its directrix parallel to the $\;y$- axis) has its maximal or minimal point at its vertex (depending on its leading coefficient's sign), and $$f(x)=y=(x-h)^2+k\implies f'(x)=2(x-h)=0\iff \color{red}{x=h}$$ and then $$f(h)=k$$ so the vertex is indeed at $\;(h,k)\;$ , and it is a minimum point iff ...


2

I suspect there is no natural sigmoid function that will "preserve" in some sense your original function $f$. In all generality, you want to "squeeze" the real line ($\mathbb{R}$) into an (open) interval of the form $(x_1,x_2)$. Let's try to squeeze both the $x$-axis and the $y$-axis so that we may fit one graph into a rectangle. Let this rectangle be, for ...


2

Have you tried to describe, how high is the aiming point at the wall in front of you when you rise a rifle at a given angle? (answer: $\mathrm{height} = \mathrm{distance} \times \tan(\text{angle})$ with a vertical ;) asymptote at $\tfrac \pi 2$) Another 'tan' disguise: how far away from the Earth you need to be to see half of it? Similar: $\sec x =\tfrac ...


2

Exponential functions have the formula $$f(x)=a^{x-h}+k$$ Since you base is $e$, this then translates to $$f(x)=e^{x-h}+k$$ You have two points then. So plug them both in to get $$20=e^{1-h}+k$$ $$200=e^{4-h}+k$$ Thus $$20-e^{1-h}=200-e^{4-h}$$ $$e^{-h}(e^4-e)=180$$ $$e^{-h}=\frac{180}{e(e^3-1)}$$ $$-h=\ln{\frac{180}{e(e^3-1)}}$$ ...


2

It is an ellipsoid. Source: http://www.wolframalpha.com/input/?i=x^2%2B2y^2%2B3z^2%3D12


2

Here is a sketch of a common algorithm. While basic, there are some obvious ways to improve it. For concreteness sake, suppose we wish to plot the contour $f(x,y)=0$, where $$f(x,y) = x^3 - x y + y^4.$$ Let us start with a rectangular grid in the plane. Something like so: This particular grid is $10\times10$ and lives in the square ...


2

Let $M\in\mathbb{R}$ be an absolute maximum on $(-\infty,\infty)$ for a function $f$ defined in $\mathbb{R}$. Then $\forall x\in\mathbb{R},\,f(x)\le M$. Let $x_0\in\mathbb{R}$ and let $\varepsilon>0$. $\forall x\in ]x_0-\varepsilon,x_0+\varepsilon[,\,x\in\mathbb{R}$ and so $\forall x\in ]x_0-\varepsilon,x_0+\varepsilon[,\,f(x)\le M$ thus $M$ is a local ...


1

Define the graph of a function $f:D\rightarrow\mathbb R$ to be the set of points $$ G(f)=\{(x,f(x))\ \mid\ x\in D\} $$ Then the function $g(x)=f(x-a)$ is defined whenever $x-a=y\in D$ so that $x\in E=\{y+a\ \mid y\in D\}$ and the graph is $$ \begin{align} G(g)&=\{(x,g(x))\ \mid\ x\in E\}\\ &=\{(x,f(x-a))\ \mid\ x\in E\}\\ &=\{(y+a,f((y+a)-a))\ ...


1

Pick a point $t$ on horizontal axis, and look at it's value $f(t)$ . As you add a non zero number to $t$ to the horizontal axis, say $1$, if you look at the value in that point $f(t+1)$, you are only adding a multiple of that number to vertical axis - $f(t+1)=c*(t+1)=c*t + c*1$. If the constant is $0$, you are not adding anything as you go along the ...


1

Recall the definition of the slope of a line. The slope of a line that passes through the points $(x_1, y_1)$ and $(x_2, y_2)$ is $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ Since we cannot divide by zero, the slope is only defined if $x_1 \neq x_2$, that is, the slope is only defined if a line is not vertical. Suppose that a line non-vertical passes through ...


1

your $u$ has a horizontal asymptote. so you you know the derivative should have a horizontal asymptote too. you derivative will have a positive local max and then decrease and approach zero. your derivative should look a little bit like the reflection on the $x$-axis of $u.$


1

You should graph $y = \sin(3x) + 1$ and look at the points where $y = 0$, i.e. the points where the graph crosses the y-axis. Edit: As Edward Jiang pointed out in the comments, the above is technically incorrect. You are asked to graph $\sin(3x) = -1$. Hence, to find the values of $x$ that solve this equation, just graph $y = \sin(3x)$ and find the points ...


1

There is an example from physics. The tension in a rope hung between two trees. http://web.mit.edu/8.01t/www/materials/InClass/IC_Sol_W03D1-1.pdf In order for the rope to be absolutely flat, the tension at the ends must be infinite.


1

It is surprising that no one has mentioned Ford circles. The heights of the centers of Ford circles are exactly proportional to the reciprocals of the denominators. Let us suppose that the content of the section on total area of Ford circles is correct. Then if a point is randomly chosen in the interior of one of the Ford circles corresponding to a ...


1

Because the vertex of a parabola is the point $(\bar{x},\bar{y})$ such that $\bar{y}$ is maximum (if $a < 0$) or minimum (if $a > 0$), and these situations both happen for $\bar{x} = h$, which ultimately leads to $\bar{y} = k$. You can also think of the symmetry of the parabola in the $y$-axis. Meaning, find the roots of $a(x-h)^2+k$ and take their ...


1

Use what you have computed: On $(-\infty,2)\cup(3,\infty)$ the polynomial $x^2-5x+6$ is positive. So, the function is equal to $$\log_{1/4}(x^2-5x+6)$$ On $(2,3)$ the polynomial $x^2-5x+6$ is negative. So, the function is equal to $$\log_{1/4}((x-2)(3-x))$$ Now, apply to each of these the procedure to determine the main elements of the graph. For example ...


1

The slope between the given points is $$ m = \frac{3 - 17}{11 - 4} = -2 $$ The slope of the perpendicular line is $$ m' = -\frac{1}{m} = \frac 12 $$ Define $$ \Delta x = \cos(\arctan(m')) = \frac{1}{\sqrt{(m')^2 + 1}} = \frac{2}{\sqrt{5}}\\ \Delta y = \sin(\arctan(m')) = \frac{m'}{\sqrt{(m')^2 + 1}} = \frac{1}{\sqrt{5}} $$ The coordinates of the points you ...



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