# Tag Info

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Your picture shows that such a function is not uniquely determined, whatever the definition of "curving around". Even $f$ itself is "curving around" its local maxima. If you want a "simple" expression defining a function $g$ such that $g(x)\geq f(x)$ for all $x\in{\mathbb R}$ then note that $$-1\leq\cos{33\pi\over x}\leq1\qquad(x\in{\mathbb R})\ ,$$ and that ...

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Unfortunately the basic premise of your integral, namely that the integral of a whole circle is equal to $1$ does not work in general. One simple example would be the flat torus. Its scalar curvature is zero at all points, so no matter what curve you pick, the integral will be zero. Another example would be integrating a curve around a saddle point (https://...

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After some more thinking, I can now answer my own question. The key of the proposed solution is Descartes' rule of signs. Let $h(x)$ be the second derivative of the function considered, that is $h(x) = (d/dx)^2\; (1 - x^k) \mathbin{/} (1 - x^n)$. After some computations, $h(x)$ can be expressed as a generalised (the exponents are not necessarily integers) ...

1

Let me address the general problem where you have $n$ data points $(x_i,y_i)$ for $i=1,2,\cdots,n$ and you want to fit, in the least square sense, the model $$y=\frac {a x}{b+x}$$ which is nonlinear (because of $b$). So, at a time, nonlinear regression will be required which means that "reasonable" estimates will need to be provided. You can have these ...

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Let $y=\frac{ax}{b+x}$. Then for some input $x$ and output $y$ we have \begin{align} \frac{ax}{b+x} &= y\\ ax-by&=xy \end{align} If we want to fit the points $(1,65),(200,70),$ and $(800,75)$ we can set up the following system of $3$ equations and $2$ unknowns: \begin{align} a-65b &= 65\\ 200x-70b &= 14000\\ 800x-75b &= 60000\\ \end{...

2

It points inward because that is the direction that the tangent to $c$ is bending towards as we move along the curve. Since $\ddot{c}\approx\frac1{\Delta t}\left(\dot{c}(t+\Delta t)-\dot{c}(t)\right)$ it is a rescaled difference between two nearby tangents, and indicates the direction of their bending. The curvature measures deviation from the straight ...

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For a monic polynomial $P(x) \in \Bbb{R}[x]$ of degree $4$. Pick two points $(a,P(a)),(a+c,P(a)+d)$ on the curve $P(x)-y=0$, let $$h(t) =P(a+tc) - (P(a)+td) = c^4 t(t-1) (t^2+ft+g)$$ $g = (\frac{h'(0)}{c^4}=-\frac{c P'(a)-d}{c^4}$, $\frac{h(2)}{c^4} = 8+4f+2g, f= \frac{\frac{h(2)}{c^4}-8-2g}{4}$ Iff $f^2-4g>0$ then $t^2+ft+g = (t-u)(t-v)$ has two real ...

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One sufficient condition would be that there exists 2 local minima at $x_1$ and $x_3$ and a local maxima at $x_2$ such that $$x_1<x_2<x_3$$ $$f(x_1), f(x_2) < f(x_3)$$ Since $\lim_{|x|\rightarrow \infty} f(x) = \infty$, we can guarente a line will intersect $f$ at 4 points above $f(x_1)$ and $f(x_3)$ and below $f(x_2)$. I.e. there are 3 "humps". ...

2

Since $t^2 \leq \sqrt{t}$ for $t\in [0,1]$ and, $\sqrt{t} \leq t^2$ for $t \in [1,2]$, we can't just subtract the areas of the two functions, we need to incorporate the absolute value into our integrand. $$\displaystyle\int_0^2 |\sqrt{t}-t^2|dt$$ When we have an absolute value, we should use a piecewise function to solve the integral: \begin{cases} ...

3

Since $t^2\leq \sqrt{t}$ for $t\in [0,1]$, and $\sqrt{t}\leq t^2$ for $t\in [1,2]$; the area is given by $$\int_0^1 \big(\sqrt{t}-t^2\big)dt+\int_1^2 \big(t^2-\sqrt{t}\big)dt=\frac{1}{3}+\left(3-\frac{4\sqrt 2}{3}\right)=\frac{10-4\sqrt 2}{3}$$ For $t>1$ the area is enclosed by the two graphs and the line $x=2$.

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You need to handle parametrization all the way ( both differentiation and integration) $$y^{'}=\cfrac{dy}{dx} = \cfrac{dy/da}{dx/da} = \cfrac{\sin(a)}{1 - \cos(a)} =\tan(a/2)$$ $$A= 2 \pi \int _{a1} ^{a2} y\sqrt{1+y^{'2}} dx =2\pi \int _{a1} ^{a2} y\sqrt{1+y^{'2}} \frac{dx}{da} da$$ $$=2\pi \int _{a1} ^{a2} (1-\cos \alpha) .\sec(a/2). (1-\cos \alpha). ... 0 Parametric Equation of Cycloid is$$x=a-\sin(a)~, \quad y=1-\cos(a)$$Now from the above equation$$\cos(a)=1-y\implies\sin(a)=\sqrt{1-(1-y)^2}\implies\sin(a)=\sqrt{2y-y^2}\implies a=\sin^{-1}\left(\sqrt{2y-y^2}\right)$$So$$x=a-\sin(a)\implies x=\sin^{-1}\left(\sqrt{2y-y^2}\right)~-~\sqrt{2y-y^2}\implies x~+~\sqrt{2y-y^2}=\sin^{-1}\left(\...

1

Solve the equation $x^3=\sqrt[3]{x}$: $$x^3=x^{\frac13}\implies\\ \left(x^3\right)^3=\left(x^{\frac13}\right)^3\implies\\ x^9=x.$$ Divide both sides by $x$ and observe that $x=0$ is a solution: $$\frac{x^9}{x}=1\implies\\ x^{9-1}=1\implies\\ x^8=1\implies\\ x=\pm1.$$ So, the solution set consists of three elements: $\{-1,0,1\}$. And those are the $x$ ...

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To find the intersection between the graph, just find the solutions of $x^3=x^{1/3}$. $$x^3=x^{1/3} \Rightarrow x^9-x=x(x-1)(x+1)(x^2+1)(x^4+1)=0 \Rightarrow x=-1,0,1$$ Since $x^3\geq x^{1/3}$ for $x\in[-1,0]$, and $x^{1/3}\geq x^3$ for $x\in[0,1]$, the area between the two graphs is given by $$\int_{-1}^0 (x^3-x^{1/3})dx+\int_0^1 (x^{1/3}-x^3)dx$$

3

$$x^3 = x^{\frac{1}{3}} \rightarrow x^9 = x \rightarrow x\left(x^8 - 1\right) = 0$$ And so we have $x = 0$ or $x^8 - 1 = 0$. For the later we employ the identity $$a^2 - b^2 = (a + b)(a - b)$$. Thus, $$x^8 - 1 = 0 \rightarrow (x^4 + 1)(x^4 - 1) = 0$$ Assuming you are seeking Real Solutions only we see ...

3

$$x^4+y^4=c$$ $$4x^3+4y^3 \frac {dy}{dx}=0$$ $$\frac {dy}{dx}= \frac {-x^3}{y^3}$$ For the orthogonal curves we get $$\frac {dy}{dx}=\frac {y^3}{x^3}$$ $$\frac {dy}{y^3}= \frac{dx}{x^3 }$$ $$\frac {-1}{y^2}=\frac {-1}{x^2}+c$$ $$y^2=\frac {x^2}{1+cx^2}$$

3

The gradient of a function $f$ is already perpendicular to all contour curves $f(x, y) = c$. So what you really want is for $r'$ and $\nabla f$ to be parallel, not perpendicular. And yes, a straight line $r(t) = (t, t)$ will do that, as $r'(t) = (1, 1)$ is parallel to $\nabla f(x, y) = (4x^3, 4y^3)$ at any point along $r$ (except perhaps at the origin, that'...

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This is true for most "nice" curves but isn't true in general, at least under the usual definition of a linear continuum, particularly for space-filling curves as @PeterShor mentioned in their comment. If a space-filling curve could be approximated by straight lines, as I understand your condition, it would be differentiable everywhere, and there exists no ...

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You can check if all control points are co-planar. If yes, the Bezier curve is a planar curve (deduced from Bezier curve's convex hull property) and therefore all points on the curve belong to the same plane.

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Thanks to everyone, I'll provide the solution from my teacher which seems, at least for me, a bit more easier to understand. We have to show that $Int(\Gamma) = Int(\gamma_0 - \gamma_1) \subset U$, or in other words $ind_{\Gamma}(z) = ind_{\gamma_0 - \gamma_1}(z) = 0 \ \forall z \notin U$. Let us define the mapping $$\eta_s(z) : [0, 1] \to \mathbb{R}, s \... 2 The discussion in the comments shows that some concepts need to be clarified. Here is a pragmatic approach. You start with curves \gamma_j : [0, 1] \to U, j \in \{0, 1\}, which are continuously differentiable. Let us assume that the homotopy H : I \times I \to U from \gamma_0 to  \gamma_1 is a continuous map which is partially differentiable with ... 1 Quantity which is to be find \cfrac{dy}{dx},$$ \cfrac{dy}{dx} = \cfrac{dy/da}{dx/da} = \cfrac{\sin(a)}{1 - \cos(a)} 

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Hint: $\dfrac{\operatorname dy}{\operatorname dx}=\dfrac {\frac{\operatorname dy}{\operatorname da}}{\frac{\operatorname dx}{\operatorname da}}$. (This is Leibniz's suggestive notation.)

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Hint: consider $x\longmapsto x^\alpha$, $\alpha\in(0,\infty)$. Define $x\longmapsto f(x,a)$ as a piecewise function using the symmetry condition. Write $a$ as function of $\alpha$.

2

Hint: Since $||\beta(s)||^2=\beta(s)\cdot\beta(s)$ you can use the product rule to show $\frac{\partial}{\partial s}||\beta(s)||^2=0$.

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Note that a curve lies on a circle if $\beta(s)\cdot \beta(s) = c = r^2$ for some (necessarily positive) constant $c$. Since $I$ is connected, it suffices to check that the derivative of $\beta(s)\cdot \beta(s)$ is $0$. However the derivative of $\beta(s)\cdot \beta(s)$ is $2\beta(s)\cdot \beta'(s)$, and we already know this is $0$ by the given information. ...

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