Hot answers tagged

24

If you want to figure out which of $\sqrt{2}$ and $\sqrt{3}$ is the smallest, you can do this simply by comparing $2$ and $3$ instead. More generally, if you want to make a square root as small as possible, you can do this by making what's under the square root as small as possible.


24

This has nothing to do with derivatives, nor epsilons; it's pure logic. If you have a function $f:\>P\to{\mathbb R}_{\geq0}$ defined on some set $P$ (like a parabola in the plane) and a strictly increasing function ${\rm sqr}:\>{\mathbb R}_{\geq0}\to{\mathbb R}_{\geq0}$ then the function $$g:={\rm sqr}\circ f:\quad P\to{\mathbb R}_{\geq0},\qquad ...


12

If $x \ge 0$ and $y \ge 0$ then $x \le y \iff x^2 \le y^2$. So as distances are greater or equal to 0, a distance is smaller or equal to another if and only its square is smaller or equal to the other's square. So as $\min distance$ (in terms of $x$) is the smallest possible value, it will occur at the same $x$ where the $\min distance^2$, the smallest ...


6

Suppose I have two boxes, one with $M>0$ dollars, one with $N>0$ dollars. I don't tell you what $M$ and $N$ are, but only that $M^2 > N^2$. Which box has more money? The box with $M$ dollars, because when comparing two positive numbers, it suffices to compare their squares. The one with the larger square is larger.


5

This isn't a rigorous explanation, but one that will hopefully intuitively make sense. Suppose that you are minimizing the distance between some curve $f(x)$ and a point $(x_0,y_0)$. We might not know what that minimum distance is, but we know that it is some value $d$, where $d\in\mathbb{R}$. Since we are assuming that $d$ is the minimum value, we know ...


4

Distances are positive real numbers. $f(x) = x^2$ is strictly increasing on the set of all positive real numbers. So $d_1 \le d_2 $ if and only if $d_1^2 \le d_2^2$. Using derivatives. Suppose $d(x) = \sqrt{f(x)}$. If $d(x_0) = 0$ for some $x_0$, then we know that a minimum is at $x_0$. So, from here on, we can assume that $d(x) > 0$ for all $x$. ...


3

$f(0)>0$ by hypothesis, and since $f(x)\to 0$ as $|x|\to\infty$, there exists $R$ such that $f(x)\leq f(0)$ for all $x$ with $|x|\geq R$. $f(x)$ has a maximum $M$ on the set $B=\{x:|x|\leq R\}$ since $f$ is continuous and $B$ is closed and bounded, and $M\geq f(0)$ since $0\in B$. Therefore the choice of $R$ guarantees that $M$ is a global maximum.


3

The volume of the cone is $\pi r^2 h/3$ where $h$ is the height and $r$ is the radius. We also know that $r + h = P$, which leads to the constraint $r + h - P = 0$. So the expression for which to find extrema is: $$F(r, h, \lambda) = \frac{\pi r^2 h}{3} - \lambda(r+h-P).$$ Taking the partials with respect to $r, h, \lambda$ and setting each equal to zero ...


3

$\displaystyle\int_0^\infty \left( \int_y^\infty \cdots \,dx\right) \,dy$ means $y$ is a positive number, and for any particular value of $y$, $x$ must be bigger than $y$. $\displaystyle\int_0^\infty \left( \int_0^x \cdots\,dy\right) \,dx$ means $x$ is a positive number, and for any particular value of $x$, $y$ must be smaller than $x$ but still positive. ...


2

The interchanging of two $\int$ signs is justified using the Fubini theorem. Moreover, notice that $$(x,y)\in [y,\infty)\times[0,\infty)\iff 0\le y\le x\iff (x,y)\in[0,\infty)\times[0,x]$$ and this explains how the domain has changed.


2

Do you have to use Lagrange? The volume of a cone is given by $$V=\frac 1 3 \pi r^2 h.$$ You are given that $r+h=P$ and you need to maximize $V$ given this restriction. Because you are not given any real values for the restriction, your answer will be expressed in terms of $P$. Put $h=P-r$ into the equation for the volume above: $$V=\frac 1 3 \pi r^2(P-r)$$ ...


2

The thing is like in the following picture. of wikipedia. For $f=d$ you increment $d$ until you touch $g=c$. In the moment of contact you take a minimum. If you go on, just before $f=d$ leaves the contact, you take the maximum. Thinking it well it is like parking! Really, the idea is so productive that is the base of the Morse's theory.


2

Yes $$\int_M d\omega =\int_{\partial M} \omega=0$$ since $\partial M=\emptyset$ beause $M$ is closed.


2

You can parametrize the surface by the function $G(x,y) = (x,y,y^2 - x^2)$. Then you can cross $G_x = \frac{\partial G}{\partial x}$ with $G_y = \frac{\partial G}{\partial y}$ which are both parallel to the tangent plane to give you a normal.


2

Imagine a simple example in 1 dimension $$f(x)=\left\{\begin{array}{c}1/|x|,\quad x>1\\ 0,\quad \mbox{otherwise}\end{array}\right.$$ This function is bounded, and its limit at infinity is zero. However, $\int_{-\infty}^{\infty}f(x)dx=\infty$. the condition $\lim_{x\to\pm\infty}=0$ is a necessary condition for the converging of the integral, but it is not ...


2

No, you cannot conclude that the integral is finite. As an example take a function like $$f(x,y)=\min\{1, 1/\sqrt{xy}\}$$ This will be bounded between $0$ and $1$, but for large $x$ and $y$ the function will be $1/\sqrt {xy}$ and its integral behaves like $\sqrt{xy}$ and thus is not finite.


2

Let $f(x_1,x_2,x_3)= x_1^2-2(x_2^3+x_3^3)$, which is a polynomial, hence continuous. Then the set in question is $f^{-1}([0,\infty))$ which is closed, hence by the definition of continuity, the inverse image of this closed set is closed.


2

I think the confusion comes from using $x$ in two different ways. I will use $x = (x^1,.., x^n)$ as coordinates on $U$. Let $y = (y^1,...,y^n)$ be any particular point in $U$. Consider the curve $\gamma:[0,1]\rightarrow U$ with $\gamma(t) = p + t(y-p)$. In terms of the $x^i$ coordinates, we have $x^i(t) = p^i + t(y^i - p^i)$. Now, consider the ...


2

Two ways: First: Compactifying $\mathbb{R}^m$ with one point and defining $f(\infty)=0$ makes this function continuous on a compact set, hence achieves a maximum somewhere. Since it is positive on $\mathbb{R}^m$ and $0$ on $\infty$, it must be on $\mathbb{R}^m$ and not on $\infty$. Therefore, the restriction back to $\mathbb{R}^m$ has a maximum. Second: ...


2

Here is the idea, which I leave to you to make rigorous. If $f$ is identically $0$, then the conclusion is clear, so suppose there is some $\alpha \in \mathbb{R}^{n}$ such that $L := f(\alpha) > 0$. Since $f(x) \to 0$ as $|x| \to \infty$, there exists an $r > 0$ such that $|x| > r$ implies $f(x) < L/2$; clearly, $r > \alpha$ since $f(\alpha) = ...


2

$$\int_0^\pi \int_0^\pi |\cos(x+y)| dx dy$$ $$=\int_0^{\pi\over2} \int_0^\pi |\cos(x+y)| dx dy+\int_{\pi\over2}^\pi \int_0^\pi |\cos(x+y)| dx dy$$ $$ =\int_0^{\pi\over2} \int_0^{\frac\pi 2-y} |\cos(x+y)| dx +\int_{\frac\pi 2-y}^\pi |\cos(x+y)| dx dy$$$$ +\int_{\pi\over2}^\pi \int_0^{\frac{3\pi}2-y} |\cos(x+y)| dx +\int_{\frac{3\pi}2-y}^\pi ...


2

We compute using Leibniz rule $$ 0 = \frac{d}{dx} u(x,x^2) = u_x(x,x^2) + u_y(x,x^2) \cdot 2x = x + 2x \cdot u_y(x,x^2) = x (1+ 2u_y(x,x^2)).$$ Hence $$ u_y(x,x^2)=-\frac{1}{2}.$$


2

Your expression for $y$ in terms of $v$ are not correct. The expressions should be $$x=\frac u2-\frac v3, \qquad y=\frac{2u}3+\frac v3$$ The original, source integral has the outer bounds $$0\le x\le 1$$ and the inner bounds $$0\le y\le 1-x$$ Substituting the correct expressions for $x$ and $y$ give $$0\le\frac u3-\frac v3\le 1, \qquad ...


2

In this case: you want distance between $\;(x,2x)\;,\;\;(1,4)\;$ $$\begin{cases}d(x)=\sqrt{(x-1)^2+(2x-4)^2}\implies d'(x)=\frac{(x-1)+2(2x-4)}{\sqrt{(x-1)^2+(2x-4)^2}}\\{}\\d^2(x)=(x-1)^2+(2x-4)^2\implies (d^2)'(x)=2\left[(x-1)+2(2x-4)\right]\end{cases}$$ Can you see now that searching for the point where the derivative vanishes is the same in both ...


2

If $g(x)$ is bounded in a neighbourhood of $0$, then this is true. If $g(x)$ is not bounded in a neighbourhood of $0$, then this in not true in general. Take for instance $f(x)=x$ and $g(x)=\ln(x).$ Then $\lim_{x\to 0} f(x)g(x)=0$. On the other hand we can take $f(x)=\dfrac{1}{\ln(x)}$ and $g(x)=\big(\ln(x)\big)^2$ and we have $\lim_{x\to ...


1

$y=\sqrt{1-x^2}$ is the upper semicircle of radius $1$ and $y= x$ is the line through the origin. The intersection between the curves is $\sqrt{1-x^2}=x$ that is $x=\frac{\sqrt 2}{2}$. So the domain of integration is the sector in the picture for $r\in[0,1]$ and $\theta\in\left[0,\frac{\pi}{2}\right]$: So your integral in polar coordinates will de $$ ...


1

The integral becomes, with your parametrization: $$\int_0^\pi(2\sin t,-2\cos t)\cdot(-2\sin t,2\cos t)dt=\int_0^\pi-4\;dt=-4\pi$$ so yes: I'd say it seems to be you got it right!


1

Since the problem is a special case, the answer is given in Figure without any integration. If the curve (semicircle) lies to the positive $x$ as in above Figure, then $$ \int_{C+} \mathbf{v}\circ d\mathbf{r}=\int_{C+}\| \mathbf{v}\|ds=\int_{C+}2\cdot ds=2\int_{C+}ds=2\cdot\left(\pi \rm R \right)=+4\pi \tag{01} $$ while if lies to the negative $x$, then ...


1

The integrand is the normal vector of the curved surface of the cylinder, and is orthogonal to the normal vector of the flat surface; thus the integral is simply the surface area of the curved surface.


1

The original region of integration is a triangle with vertices $(0,0),\,(0,1)$ and $(1,0)$. The linear transformation $L(x,y)=(x+y,y-2x)=(u,v)$ will move straight lines to straight lines, so the transformed region will also be a triangle, but its vertices will be $L(0,0)=(0,0),\,L(0,1)=(1,1)$ and $L(1,0)=(1,-2)$. Draw the triangle of the new region of ...



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