New answers tagged

3

You start by writing $g_x$ as function of $f_u$ and $f_v$, thus $$g_x = \frac{ \partial g}{\partial x} = \frac{ \partial u}{\partial x} \frac{\partial f}{\partial u} + \frac{ \partial v}{\partial x} \frac{\partial f}{\partial v} = \frac{ \partial \frac{1}{y} -\frac{1}{x}}{\partial x} f_u + \frac{ \partial xy \exp\left(- \frac{z^2}{2} \right)}{\partial x} ...


1

The Lagrangian equations are $$\begin{cases}\begin{align}2x&=-\lambda y\\2y&=-\lambda x\\2z&=2\lambda z\\z^2&=xy+1\end{align}\end{cases}$$ A systematic approach is by noting that the first three form a parameteric linear system with determinant ...


3

If $z=0$ you have: $$ \begin{aligned} 2x&=-\lambda y\\ 2y&=-\lambda x\\ xy&=-1 \end{aligned} $$ and the first two of these imply that $\lambda^2=4$ and that $x=y$ or $x=-y$ and the first of these is incompatible with $x$ and $y$ being real and satisfying the third equation. So you are left with $x=-y$ and $xy=-1$... If $z \ne 0$ we have ...


1

Charles Pugh's Real Mathematical Analysis or Hubbard's Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach.


0

The projection into the $xy$-planes shows the outline formed by the planes $y=1$, $y=x$, and $y=-x$. You can see that if $x$ is the outer variable, then for $x\le0$, $-x\le y\le1$, whereas when $x\ge0$, $x\le y\le1$. You can determine that by drawing a vertical line (a curve of constant x) through the figure. That means we would have to break up the ...


1

We can write ${\bf A}\cdot {\bf e}_i = A_i$, so $\nabla({\bf A}\cdot {\bf e}_i)$ is the gradient of $A_i$, To pick the $i-$th component, we dot that with ${\bf e}_i$ again to obtain $\frac{\partial A_i}{\partial x_i} = \nabla({\bf A}\cdot {\bf e}_i)\cdot {\bf e}_i$, and the desired field is: $$\sum_{i=1}^n \frac{\partial A_i}{\partial x_i}\,{\bf e}_i = ...


2

The function $f(x,y,z) = e^y\cos x+z$ is a potential for the field $(-e^y\sin x,e^y \cos y, 1)$, so by the fundamental theorem of calculus we have $$\int_C -e^y\sin x\,{\rm d}x + e^y\cos x\,{\rm d}y + {\rm d}z = f(\pi,\pi,0)-f(0,0,1).$$


0

In the definition $$ \lim_{x \to a} \frac{\| f(x) - f(a) - L(x - a)\|}{\|x - a\|} = 0$$ It is clear that one must have $\lim_{x \to a} f(x) = f(a)$ since otherwise the numerator alone wouldn't even converge to 0 (as $x \to a$).


2

No, of course not: Think of the function $e^{x-y}$ which along the line $x=y$ is not only bounded, but even constant $1$ - yet it clearly is not bounded.


1

This not a complete answer, but I hope it still is helpful. As you said, you have a chart on $A$, so it only remains to construct charts around boundary points. By assumption, for each point $x\in FrA$, there is an open neighborhood $U$ of $x$ in $\mathbb R^n$ and a diffeomorphism $F$ from $U$ onto an open subset $V\subset\mathbb R^n$ such that $U\cap ...


1

$$ \frac{\partial}{\partial t}f(x,t)=\frac{\partial}{\partial t}\int_{0}^{g(x,t)} e^{-u^2} du=e^{-g^2(x,t)}\frac{\partial}{\partial t}g(x,t)\ , $$ where one uses the fundamental theorem of calculus $\frac{d}{dz}\int_0^z dt\ h(t)=h(z).$ Taking a second derivative (using the product rule) $$ \frac{\partial^2}{\partial t^2}f(x,t)=\frac{\partial}{\partial ...


2

Assuming $y=a x$ ($a\geq0$), the constraint leads to $$x=\frac{k}{\sqrt{3 a^2+2 a+2}+a+1}$$ and $$x^2y=\frac{a k^3}{\left(\sqrt{3 a^2+2 a+2}+a+1\right)^3}$$ Now, computing $$\frac{d}{da}x^2y=-\frac{(2 a-1) \left(3 a+\sqrt{a (3 a+2)+2}+2\right)}{\sqrt{a (3 a+2)+2} \left(a+\sqrt{a (3 a+2)+2}+1\right)^4}k^3$$ which cancels if $a=\frac 12$. All of this makes ...


1

Discontinuous partial derivatives do not imply that the function is not differentiable. See this link for a counter example: http://mathinsight.org/differentiable_function_discontinuous_partial_derivatives


0

If a linear map preserves right angles then it preserves all angles and we can then deduce that all it has to be a multiple of an orthogonal matrix, ie. $AA^T=1$. See this: ALL Orthogonality preserving linear maps from $\mathbb R^n$ to $\mathbb R^n$?


2

No. Let $v_z = (0,0,x)$ and the region of integration be the unit circle. Clearly, by symmetry, $\int \int v_z \mathrm{d}A = 0$. However, $\frac{\partial v_z}{\partial x} = 1$ is positive everywhere, so its integral is not zero. (In fact, it's $\pi$.)


0

It's exactly the same as your example with the equation $F(x,y,z)=0$. The difference is that you have to change the number of dimensions. Let $F:{ \mathbb{R} }^{ 5 }\rightarrow { \mathbb{R} }^{ 3 }$ be: $$ F(x,y,z,u,v)=({ x }^{ 2 }+y\cos { (uv)+{ z }^{ 2 },\quad{ x }^{ 2 }+{ y }^{ 2 }-\sin { (uv) } +2{ z }^{ 2 }-2,\quad xy-\sin { (u)\cos { (v) } +z) } } ...


1

Let $f(\varphi, \theta) = ((b + a \cos\theta) \cos(\varphi),~ (b + a\cos\theta) \sin(\varphi),~ a\sin\theta)$. Then the torus $T$ with major radius $b$ and minor radius $a$ is given parametrically by $T = f([0,2\pi) \times [0, 2\pi))$. Now the surface integral of $T$ is, as always, defined by parametrizations as $$\oint_T 1ds = ...


0

HINTS $f(x,y,z)=1$ means $f$ maps any vector in $\mathbb{R}^3$ to $1$. When you say $g(x,y,z)=x$, this means that $g$ maps any vector in $\mathbb{R}^3$ to its first coordinate. Volume of the region $A$ is defined as $\iiint_A dV$.


0

I think you are mixing up two things. The value/point $(x,y)$ for which the $d(x,y)$ and $d²(x,y)$ are minimal and the actual value of $d$ and $d²$ for $(x_{min},y_{min})$. The value for which d and $d^2$ are minimal is the same, as distance is always positive $d>0$ hence multiplying it with $d$ will result in $d^2>0$ Both expressions are equivalent ...


0

There is a term called monotonic. A monotonic function doesn't change direction -- as its input grows, it either never-shrinks or never-grows. A "strictly monotonic" function is a monotonic function that never stays the same -- it always either grows or shrinks. On the positive real numbers, squaring is a strictly increasing monotonic function. There are ...


0

$d^2 = (x−3)^2+y^2+(16-x-y)^2$ \begin{align} \dfrac{\partial d^2}{\partial x} &= 0 \\ 2(x-3) -2(16 - x - y) &= 0 \\ x - 3 - 16 + x + y &= 0 \\ 2x + y = 19 \\ \end{align} \begin{align} \dfrac{\partial d^2}{\partial y} &= 0 \\ 2y - 2(16-x-y) &= 0 \\ y - 16 + x + y &= 0 \\ x+ 2y &= 16 \\ \end{align} ...


2

Set $y=x^2 $. Then $y^2 = x^4$. Your equation becomes : $108y^2 - 507y + 300 = 0$. Solve that and then replace $x = y^2$ for the solutions you found and you will find the solutions to your starting one.


2

Your idea is correct for total derivatives, but the equivalent of $$ \frac{\mathrm dx}{\mathrm dv}=\frac1{\frac{\mathrm dv}{\mathrm dx}} $$ is not $$ \def\p#1#2{\frac{\partial#1}{\partial#2}} \p xv=\frac1{\p vx} $$ but the matrix equation $$ \pmatrix{ \p xu&\p yu\\ \p xv&\p yv }= \pmatrix{ \p ux&\p vx\\ \p uy&\p vy }^{-1} $$ (which you ...


3

Be careful: you should consider $x$ and $y$ both to be functions of $u$ and $v$. Take the partial derivative with respect to $v$ of both equations in the system to get (don't forget the chain rule!): $$\left\{ \begin{array}{rcl} \displaystyle 2x\frac{\partial x}{\partial v}+2y\frac{\partial y}{\partial v}&=&0 \\[7pt] \displaystyle \frac{\partial ...


2

First you need to find the point(s) where you have to give the tangent plane. For the gradient you have: $$\nabla u = \left( \frac{\partial u}{\partial x},\frac{\partial u}{\partial y} \right) = \left( \frac{1}{x+1/y},\frac{-1/y^2}{x+1/y} \right)$$ And you're looking for the points $(x,y)$ where this is equal to $\left( 1,-\tfrac{16}{9}\right)$, so: ...


1

This can be done in a pretty standard way. $$\vec r = (x, y)$$ $$ x= u$$ $$ y = \frac vu$$ $$ \vec r_u=(1, -\frac v{u^2})$$ $$ \vec r_v=(0, \frac 1{u})$$ $$||\vec r_u ||=\sqrt{1+{(\frac v{u^2})}^2}$$ $$||\vec r_v ||=\sqrt{{(\frac 1{u})}^2}$$ $$\vec e_u = \frac {(1, -\frac v{u^2})}{\sqrt{1+{(\frac v{u^2})}^2}} $$ $$\vec e_v = \frac {(0, \frac ...


1

If you only care about the locations of minima and maxima, the derivative chain rule can make this easier. Say you have a function $f(x)$ that you want to maximize/minimize. In your case, this would be the distance formula $d = \sqrt{\Delta x^2 + \Delta y^2}$. Very often, there is another function that considerably simplifies the derivative. Call that ...


0

I think it is best to use some elementary geometry. The volume of a cone is one third of the product between the base area and the height, by Cavalieri's principle. The distance of the vertex of the cone (i.e. the origin) from the plane $x+4z=a$ is $\frac{a}{\sqrt{17}}$ by a well-known formula. The boundary of the base is the set of points $(x,y,z)$ such ...


0

You should have $$f(x,y)= f(1,1) +f_x(1,1).(x-1)+f_y(1,1).(y-1)+f_{xx}(1,1).\frac{(x-1)^2}{2!}+2f_{xy}(1,1).\frac{(x-1)(y-1)}{2!}+f_{yy}(1,1).\frac{(y-1)^2}{2!}$$ It's $2f_{xy}(1,1).\frac{(x-1)(y-1)}{2!}$ instead of $f_{xy}(1,1).\frac{(x-1)(y-1)}{2!}+f_{yx}(1,1).\frac{(x-1)(y-1)}{2!}$ because $f_{xy}=f_{yx}$ In your case the fact that $f_{xy}=0$ makes your ...


1

The explicit value of the integral may be difficult to find, but its convergence depends only on the convergence of: $$\iiint_{x^2+y^2+z^2\geq 1}\frac{d\mu}{(x^2+y^2+z^2)^p}=\frac{4\pi}{2p-3}\qquad \left(p>\frac{3}{2}\right)$$ since $\exp\sin(x+y+z)$ is always bounded between the positive constants $\frac{1}{e}$ and $e$. By assuming $p>\frac{3}{2}$ ...


1

Simply put, $$ \frac{d(y^2)}{dx} = 2 y \frac{dy}{dx} = 0 $$ and $$ \frac{d y}{dx} = 0 $$ have the same latter condition to be fulfilled to find an extremum. And that goes for any square function of x !!


0

In spherical coordinates, you are looking for the volume of the following region: $$ E=\{(\rho,\theta,\phi)\;|\;0\le \rho\le 4\cos\phi,0\le \theta \le2\pi, 0\le \phi\le\frac{\pi}{4}\} $$ So your volume equals $$ V(E)=\iiint_E \rho^2\sin\phi\; d\rho d\theta d\phi $$


0

Hint: If there were a vector field $G$ defined in the complement of the origin such that $F = \nabla \times G$, then Stokes' theorem would imply $$ \iint_{S} F \cdot n\, dS = 0 $$ because the sphere has empty boundary. On the other hand, you can calculate $$ \iint_{S} F \cdot n\, dS $$ explicitly (without calculus, even!), since on the unit sphere you have ...


0

Suggestion: Using ideas that Helmholtz proved, the gravitational vector field can be decomposed into the sum of an irrotational vector field (call it $\phi$) and divergence-less vector field (call it $A$). This means a vector field $F$ can be written as $F = -\nabla \phi + \nabla \times A.$ Obviously the gravitational field is curl free (conservative), so ...


0

Let $r = x^2 + y^2$. Note that if the values are $a = \frac{x}{r}$ and $b = \frac{y}{r}$, then we will have: $w = \frac{x}{r}$ $dx$ + $\frac{y}{r}$ $dy$. Then, the following follows: $ w(X) = \frac{1}{r} (-xy +xy)$ = $0$,$ w(Y) = \frac{1}{r} (x^2+y^2)$ = $\frac{1}{r} r = 1.$ This contradicts the conditions given ($w(X) =1$, $w(Y) =0$). Thus, the solution in ...


0

If it helps, you may write : $\forall t \in \mathbb{R}, \, X(t) = \begin{bmatrix} x_{1}(t) & x_{2}(t) \end{bmatrix}^{\top} \quad \text{and} \quad Y(t) = \begin{bmatrix} y_{1}(t) & y_{2}(t) \end{bmatrix}$. So that : $$ \forall t \in \mathbb{R}, \, g(t) = f\big( x_{1}(t),x_{2}(t),y_{1}(t),y_{2}(t) \big). $$ The chain rule writes : $$ \small{ g'(t) = ...


1

From the two equations concerning $f_x$, $f_y$ you obtained it immediately follows that $$\sin y=\sin(x-y)=\cos x\ .$$ Now $\sin y=\cos x$ is equivalent with $${\rm (i)} \quad y={\pi\over2}+x\qquad\vee\qquad{\rm (ii)} \quad y={\pi\over2}-x$$ (all angles are modulo $2\pi$). In case (i) we then obtain $\cos x=\sin(x-y)=-1$, with consequences. In case (ii) ...


0

Use this trigonometric formula (and the ones with $\cos$) until you only have a product of sines and cosines. Then, say it is null iff one of them is null


1

Sometimes a picture's worth a thousand words. But as it has been pointed out, some words may also be useful. The blue plot is the distance function. It may seem redundant, but when two points are distant from $\;x\;$, their distance is $\;x\;$. So here, $\;d\;$ is the distance function, and $\;d(x) = x\;$. The green plot is the distance squared (when ...


2

The first thing you should notice is that a distance is a positive number. Suppose the minimum distance is $d_{\text{min}}$. The statement means that any other distance $d$ is larger that $d_{\text{min}}$: $$d\geq d_{\text{min}}.$$ As distances are positive, the squares are ordered in the same way $$d^2\geq d_{\text{min}}^2.$$ You can convince yourself by ...


1

The fine point is in the phrase "regardless of $\theta$". Let $f$ be the characteristic function of the set $$A:=\left\{{1\over k}\biggl(\cos{1\over k},\sin{1\over k}\biggr)\>\biggm|\>k\in{\mathbb N}_{\geq1}\right\}\ .$$ Then $\lim_{r\to0}\tilde f(r,\theta)=0$, regardless of $\theta$, but the $\lim_{(x,y)\to(0,0)} f(x,y)$ does not exist. If, however, ...


1

I would like to point you to the answer I gave here for an elaborate explaination on the use of $\lim\limits_{r\to0}$. To sum up what is said there; The problem is not that $\lim\limits_{r\to 0}f(r,\theta)$ only considers straight lines, the problem is that $\lim\limits_{r\to 0}f(r,\theta)$ is not what you probably think it is. $f(r,\theta)$ is a function ...


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.


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 $$ u_y(x,x^2)=-\frac{1}{2}.$$


0

The OP asks: What am I doing wrong with this method? When should I not use polar coordinates to find limits of multivariable functions? The answer to the second question is somewhat unsatisfactory: If you find a limit, then you can. If you don't find a limit, then you can't. So now, let's just leave that behind us and focus on the first question: ...


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.


15

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 ...


44

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.


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 $x$. ...


26

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 ...



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