# Tag Info

3

The winding number will be $1$ inside a simple closed curve and zero outside. If you have a well-behaved simple closed curve imagine a line (ray i.e. one direction) from your point to infinity. If the line crosses the curve an odd number of times the point is inside the curve. For polygons see, for example, this link which deals with both ideas.

3

Hint: For the first function use the identity $$\cos(x)=\sin\left(x+\dfrac\pi2\right),$$ for the second it is useful to note that $\arcsin(\sin(x))=x$, since $\arcsin x$ and $\sin x$ are inverse functions. Also be careful about the domains and ranges of those functions, and so $\arcsin(\sin(x))=x$ isn't true for all $x$.

3

The implicit equation is \begin{align} r^2 &=36\cos(2\phi)\\ &=36[\cos^2(\phi)-\sin^2(\phi)]\\ r^4 &=36[r^2\cos^2(\phi)-r^2\sin^2(\phi)]\\ x^4+2x^2y^2+y^4 &=36x^2-36y^2 \end{align} But parametrically, to plot the curve, \begin{align} x&=6\cos(\phi)\sqrt{\cos(2\phi)}\\ y&=6\sin(\phi)\sqrt{\cos(2\phi)} \end{align} Since we ...

2

For the first : The graph is symmetry both about $x$-axis and $y$-axis. So, we only need to consider the case when $x\ge 0$ and $y\ge 0$. Then, note that $$|||a|-2|-1|=\begin{cases}-a+1&\text{if 0\le a\lt 1}\\a-1&\text{if 1\le a\lt 2}\\-a+3&\text{if 2\le a\lt 3}\\a-3&\text{if a\ge 3}\end{cases}$$ For the second : Since ...

2

When it comes to combining transformations, you should think of it as a replacement procedure. Shifting two units to the right means replacing $x$ with $x - 2$. A reflection over the $y$-axis means replacing $x$ with $-x$. So if we apply these transformations in the given order, then we would get $y = f(-x - 2) = \sqrt{-x - 2}$. On the other hand, the ...

2

As you observed that $f(2-x) = f\big(-(x-2)\big)$. Since $$f(x) \quad \stackrel{(1)}{\longrightarrow} \quad f(-x) \quad \stackrel{(2)}{\longrightarrow} \quad f\big(-(x-2)\big),$$ we can obtain $f(2-x)$ from $f(x)$ by transformations that correspond to (1) and (2), where (1) is reflection across the $y$-axis; (2) is translation to the right by $2$ units.

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The upper (or lower) half of any circle with center in the first quadrant and radius < distance center-axes, like $$y = 2 + \sqrt{1-(x-2)^2}.$$

2

How about $y=\frac{1}{\sqrt{x}}$?

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You could call them 'level sets'. Imagine you have a function $f(x,y) = |y| - |x| - \tan(y)\frac{-\sin(1/x)}{\tan(1/y)}$ then you're looking for the set of points where $f(x,y) = 0$. In python: import matplotlib.pyplot import numpy as np delta = 0.05 xrange = np.arange(-10.0, 10.0, delta) yrange = np.arange(-10.0, 10.0, delta) x, y = ...

2

In order you have a solution, the function $$f(x)=k\sqrt{x^2-1}-\log(x)$$ must go though a minimum and the value of the function at that point must be negative. So, $$f'(x)=\frac{k x}{\sqrt{x^2-1}}-\frac{1}{x}$$ The minimum occurs when $$\frac{k x}{\sqrt{x^2-1}}=\frac{1}{x}$$ (this reduces to a quadratic in $x^2$ and we need to only consider the positive ...

1

It depends what you mean by similarity. In mathematics similarity usually means the norm of the difference between them. This then begs the questions - what is a norm and what does difference mean. One norm of a function f(t) between a and b is ${{\left\| f \right\|}^{2}} = \int\limits_{a}^{b}{f{{(t)}^{2}}dt}$ The size of the difference between two ...

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Mathematica, follwing @Alan's advice. ($x$ is going left to right, $y$ is going bottom to top, like usual).

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It looks like the problem doesn't have a closed form answer, but I think I have a slightly more elegant way of formulating the problem, anyway. What you're trying to do is to maximize $k$ subject to the equations $y = k\sqrt{x^2-1}$ and $y = \ln x$ for $x > 0$. Solving for k gets you $k = \frac{y}{\sqrt{x^2 - 1}} = \frac{\ln x}{\sqrt{x^2 - 1}}$, so just ...

1

Taking a look at your data, I guessed that maybe a function of the form $f(x)=a/x^p+b$ might work. If we then parameters $a$, $b$, and $p$ to minimize $$\sum_{(x,y)\in \text{data}} (f(x)-y)^2,$$ we find $f(x)=\frac{2106.91}{x^{2.11436}} + 0.463718$. The sum of the squared errors is $0.000480789$, which seems pretty good. The graph of $f$ with the data ...

1

I had to do my own version, using @Tharindu's knowledge for my assignment. Here it is if anyone wants a more detailed version. The coordinates are different, since I had a different question: We need to find an equation that will create a line that is perpendicular to the equation $x=3.58$, given that one point on the line B. As shown in steps beforehand, ...

1

Assume your simple closed curve is the zero set of a function $$C: f(x,y) = 0$$ and $f$ also takes positive and negative values. (thanks @Federico Poloni for drawing attention to the need of this last condition). Now the complement of the curve has two connected components by Jordan's theorem so none of these components can intersects both of the non-void ...

1

Well... at $x=5$, it is clear that $$f(5)=(5-5)^2(5+6)=0^2*11=0$$ So from here, you need to observe that if you pick some number greater than $5$ for $x$ (we can write this as $x=5+x_0$, where $x_0 >0$) then we get $$f(5+x_0) = ((5+x_0)-5)^2((5+x_0)+6)=(x_0)^2(11+x_0)>0$$ The $>$ above is true because $(x_0)^2$ is always positive and $11+ x_0$ ...

1

you want to use Y = Asin(BX+C) such that the upper part (positive) will be your dayTime, OK: sin(x) varies in the positive Y-axis range when x varies between [0, $\pi$] in radians Assume the day is regulated to fit [0, 24] hours $$sunrize = sr$$ $$sunset = ss$$ length of the day (bright side) will be [ss-sr] (whatever the unit of time you use) this ...

1

It's a problem of definition of the function. Every function is not given by the relation $f$ but by three things: 1) a set $A$, the domain; 2) a set $B$, the codomain; 3) a relation $R$ which is a subset of $A\times B$ and has the property that it is a function, i.e. for each $x\in A$ there is exactly one couple $(x,y)\in R$. So a function is really a ...

1

A quotient is defined when both its numerator and denominator are defined and the denominator is nonzero. The quotient $\frac1{\frac1x}$ is defined when $1$ is defined (always true) and $\frac1x$ is defined, and $\frac1x$ is nonzero (always true). The quotient $\frac1x$ is defined when $1$ is defined (always true) and $x$ is defined (always true), and $x$ ...

1

A simple solution would be to plot (for a suitable range of $y$) $$x(y) = \pm\cos^{-1}\left(2 - \tfrac{1}{2}({e^y} + y{e^y} + {e^{ - y}})\right) +2\pi n$$ for $n = 0, \pm 1, \pm 2,...$

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a) Critical values are places where the derivative is zero or where the derivative fails to exist. Your graph shows three such points. b) A relative extreme is a point where the graph changes from increasing to decreasing, or vice versa. In terms of the derivative, it is a place where the derivative switches from positive to negative or vice versa (i.e., ...

1

Using $r^2=x^2+y^2,x=r\cos\phi,y=r\sin\phi$, and the trig identity $\cos(2\phi)=\cos^2\phi-\sin^2\phi$, we have $$r^2=36\cos(2\phi)$$ $$\implies r^4=36r^2(\cos^2\phi-\sin^2\phi)=(6r\cos\phi)^2-(6r\sin\phi)^2$$ $$\implies(x^2+y^2)^2=36x^2-36y^2$$ $$\implies {x}^{4}+2\,{x}^{2}{y}^{2}+{y}^{4}-36\,{x}^{2}+36\,{y}^{2}=0$$ Now let $x^2=a,y^2=b$. We now have ...

1

Keep in mind that the domain of the arcsin function is $[-1,1]$ and its range is $[-\pi/2,\pi/2]$. This is important because even though $\sin x$ and $\arcsin x$ are inverse functions, it's not correct to say that $\arcsin(\sin x)=x$ for all $x$. This might explain why your graphing calculator is giving you "pointed" curves.

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