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0

EDIT: It turns out the identity is incorrect. Consider the vector field $\vec{b}=\begin{pmatrix}xy\\0\\xy\end{pmatrix}$ (it doesn't work)


1

Firstly note that the integrand function is positive in your domain. Since $$\int_0^1 \frac{\sin y}{y} dy =L < +\infty$$ you can see that $$\int_1^{+ \infty} \frac{1}{x^2}\int_{e^{-x}}^1 \frac{\sin y}{y} dy \ dx \le \int_1^{+ \infty} \frac{L}{x^2} dx < + \infty$$ so the integral converges.


1

It is convergent. $\frac{\sin y}{y}$ is an integrable function on it's own. You could take the lower bound of the $y$ integral as 0 (overestimation), then the y integral will just be a number. And the x integral will be 1, so it will be finite.


1

The domain of $f$ consists of lines of the form $y_k = \frac{5x}{6} + k$, where $k \in [-\frac{1}{6},\frac{1}{6}]$, so since $f(x,\frac{5x}{6}+k) = \sin^{-1}(6(\frac{5x}{6}+k) - 5x)) = \sin^{-1}(5x + 6k - 5x) = \sin^{-1}(6k)$. As $k$ ranges from $\frac{-1}{6}$ to $\frac{1}{6}$, the value $6k$ ranges from $-1$ to $1$. You should verify this. So the range of ...


0

I have partial, but positive results. There is a fundamental book “Differential and Integral Calculus” by Grigorii Fichtenholz. This is a famous book for our students and it has many translations (but except English). I found in Appendix of the vol. I the following Theorem (I-II), with a long and complex proof. A simple curve without special points ...


0

The one sided (take two sided if more appropriate, it doesn't matter here) derivative is defined as $df(x,h) = \lim_{t \downarrow 0} {f(x+th)-f(x) \over t}$. It is not hard to show that if $f$ is differentiable at $x$, then $df(x,h) = \langle \nabla f(x), h \rangle$. Or you can compute directly from the formula above. Aside: If $u$ is a unit vector, then ...


-1

you can use Lagrangians ( Constrained Optimization). I give you an example here : http://mat.gsia.cmu.edu/classes/QUANT/NOTES/chap4.pdf


1

use the two equations as simultaneous equations at zero and solve the homogeneous equations. $4(x^3 - y) = 0$ and $4(y^3 - x) = 0$. Pretty sure that this gives a line in space rather than point that you might be use to. this is because there is no positive numbers without variables ie: $4(x^3 - y) = 3$ and $4(y^3 - x) = 5$ will give a single point extrema.


1

Hint: Look at the corresponding closed region $R$, so that the extrema will either be critical points in int$R$ or they will lie on the boundary of $R$. For the saddle points, use the second derivative test. Second hint: Step 1: Solve the following equations and use the points in the second derivative test to decide if they are local extrema or saddle ...


0

Hint: For each $y\in B(0,r)$, consider the function $g:[0,1] \to \mathbb R$, $$g(t) = f(ty).$$ Then apply mean value theorem to this $g$ (Chain rule should be used). Remark: After all these you can only show that $f$ is constant. Might not be zero though.


0

If $f$ is a given function of $\theta$ only, then we have the following: $\nabla f(\cos \theta,\sin \theta)=\frac{df}{d\theta}\nabla \theta=\frac{df}{d\theta} \hat \theta$ Where $\hat \theta$ is a unit vector that points in the direction of increasing $\theta$. Thus a unit vector parrallel to $\nabla f$ is indeed $\hat \theta$. In Cartesian coordinates ...


0

By the definition of the directional derivative $f_u=\nabla f \cdot \vec{u}$ i.e. the inner product of the vector $\vec{u}=u_1\hat{x}+u_2\hat{y}$ with $\nabla f$. Since you have calculated the vector $\nabla f$, the rest of the problem is geometric. You can solve it by using algebra, i.e. solve the algebraic system $$ \begin{cases} \vec{u}\cdot\nabla ...


0

let k = 1, then x^2+y^2 = c^2, where c is constant. then (x, y) which satisfy above equation will form circle with radius c. then level curve will be circular for f(x, y) because any (x,y) on unit circle evaluate to same f(x,y) k scale x, causing circle to become elliptic. so level curve become elliptic. analytically, you can solve value of k, which ...


0

$$3a-6b=\frac65$$ $$a^2+b^2=1$$ $$b=\pm \sqrt{1-a^2}$$ Solve the equation $$(a,b)=(24/25,7/25)\ or\ (-4/5,-3/5)$$


-1

i do not think there is sufficient info to find gradient of f for f(x,y) where x and y are on unit circle. we only know f(x,y) = 17 on unit circle, but there is no info for f(x,y) infinitely close to the unit circle.


1

Hint: Change to polar coordinates and show $$\lim_{r\to 0}\frac{r^2\sin(r^{-2})-0}{r}=0$$ Hint 2: The sine function is bounded.


2

It is a closed region, so max and min must occur. They can only occur on the boundary or at critical points of the function. So you can use the following steps: Step 1: Find all the critical points of the function, and check whether they are in the constraint region. Step 2: Use regular Lagrange multiplier method on the boundary of the disk. Then ...


0

You want to get a Hamiltonian system \begin{alignat}{2} \dot q_k &=Q_k&&=\frac{∂H}{∂p_k}\\ \dot p_k &=P_k&&=-\frac{∂H}{∂q_k} \end{alignat} The Weierstraß theorem tells us that such a function $H$ defined by its partial derivatives exists if the problem is considered on a simply connected domain and those partial derivative functions ...


0

Usually when your integration domain is a sphere, you use spherical coordinate. If your integration domain is a circle or disk, you use polar coordinate. In this example, the domain of $z$ value is from $0$ to the plane $z=-4x+2y$, so apparently spherical coordinate wouldn't work well. Now if you look at the projection of the object onto the $xy$-plane, it ...


4

For any natural number $j\neq 0$ we have: $$\begin{eqnarray*}\sum_{\substack{k=0\\k\neq ...


1

Your series is not absolutely convergent, if only because it includes a divergent positive sub-series: take all terms for which $k \ge 0$ and $j = k+1$; then $\frac1{j^2-k^2} = \frac1{2k+1}$, whose sum is $+\infty$. Also, we know that in general, if $\sum_{ij} |a_{ij}| = +\infty$, we don't have $\sum_i\sum_j a_{ij} = \sum_j\sum_i a_{ij}$.


0

Partial derivative is the derivative of a function with several independent variables with respect to any one of them, keeping the others constant. The symbols $ \dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y} $ are used to denote such differentiations. And the expressions $ \dfrac{\partial u}{\partial x}, \dfrac{\partial u}{\partial y} $ ...


2

The $C$ is unfortunate notation that hides the distinction between a path and its range. One (of many suitable) parameterisations could be $C:[0,4] \to \mathbb{R}^2$, $C(t) = \begin{cases} (ta,0 ) , & t \in [0,1) \\ (a, (t-1)a) , & t \in [1,2) \\ (a-(t-2)a, a) , & t \in [2,3) \\ (0, a-(t-3)a) , & t \in [3,4] \end{cases}$. Then the integral ...


0

The other responses answer your question well, but I would like to add a partial derivative is also a 'total' derivative of some function, which is the one-variable function obtained from your original function by setting all the other variables (different from the one with respect to which you are differentiating) constant. So, if, for concreteness, $f : ...


0

Here's something that may help: If $x$ and $y$ are both independent functions of $t$ and $z=f(x,y)$ then $$\frac{dz}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$$ In a sense, to get the total derivative, you add up all of the partial derivatives. So to answer your question, the partial derivative isn't really ...


0

$\mathbf d \cdot \nabla$ is just an operator. Specifically it's the operator $$d_1\frac {\partial}{\partial x} + d_2\frac {\partial}{\partial y} + d_3\frac {\partial}{\partial z}$$ where $\mathbf d = d_1\mathbf e_1 + d_2\mathbf e_2 + d_3\mathbf e_3$. As you can probably tell from the three equations you wrote, this operator is related to the directional ...


0

You have the right idea, you just have to use the multivariate chain rule now. $$\frac{\partial g}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x} + \frac{\partial f}{\partial w}\frac{\partial w}{\partial x} = f_u + 0 - f_w.$$ Can you complete the others?


2

If you're going to apply multivariable calculus tools to the distance function, it's best to use the squared distance function: $$f(\omega)=\|x-\omega\|^2,\quad g(\omega)=\langle a,\omega\rangle$$ The minimum is attained in the same place, but this $f$ expands as inner product, allowing for simpler computations: $\nabla f(\omega) = 2(\omega-x)$. So, the ...


1

I think I will post this picture, I think students here undervalue the skill of drawing; in this case, I am putting a computer picture, but it can be substantiated. The calculation part, for you to finish, is to correctly identify the two points where the graph is perfectly vertical, one in the first quadrant, one in the second. Also verify that the apparent ...


0

I agree, what I mean is that if you want to use the formula $\frac{dF}{dx}(x,y)= f(x,x) $+ 2nd term, then the $f(x,x)$ is the full expression after the integral symbol, but divided by the $dy$ Therefore, applying this method formally, you should write $\frac{dM}{dt} = e^{-(t-u) } \frac{dS}{S dt}$ + 2nd term.


1

An elementary way: assume $(x,y_1)$ and $(x,y_2)$ lie on the curve. Then $$x^2y_1^3-3xy_1^2-9y_1+9=x^2y_2^3-3xy_2^2-9y_2+9$$ $$0=x(y_1^3-y_2^3)-3x(y_1^2-y_2^2)-9(y_1-y_2)$$ $$0=x(y_1^2+y_1y_2+y_2^2)-3x(y_1+y_2)+9$$ Now find when a real solution $y_2$ exists by taking a discriminant. As I'm working it out, it appears that we need to use the quadratic ...


2

No, this is not true, the basic idea for a counterexample already surfaced in your own comments on the question, it is a function which jumps at irrational points. Let $f:\mathbb{Q} \to \mathbb{Q}$ be defined by $$ f(x) = \begin{cases} x & \text{ for } x \notin (0,\sqrt{2})\\ x+\frac{1}{2^n} & \text{ for } x \in \left(\frac{\sqrt{2}}{n+1}, ...


0

even if would apply the classical calculus rules, then I think you did not use correctly the last formula your write. Namely, in the first term of the last equation, there is no $dy$. Hence, following your logic, you need to divide the first term in your equation by $dt$. This then gives the correct equation. Hope this helps.


0

Consider the sequences $$ (x_n, y_n) := \left( \frac{1}{n}, \frac{1}{n^5} \right) \quad \text{and} \quad (x_n', y_n') := \left( \frac 1 n , - \frac{1}{n^5} \right) \; .$$ They both converge to $(0,0)$, i.e. $$ \lim_{n \to \infty} (x_n, y_n) = \lim_{n \to \infty} (x_n', y_n') = (0,0) \; . $$ Now calculate $h(x_n, y_n)$ and $h(x_n', y_n')$, and from that, ...


3

hint:Two paths: $y = x^5$, and $y = 2x^5$ will do.


0

Let $f:t\in \mathbb{C}^n\rightarrow I\otimes t^*t^T$. Then the derivative is $Df_t:h\in\mathbb{C}^n\rightarrow I\otimes (h^*t^T+t^*h^T)$. EDIT. Answer to hughhouse. No, the derivative is the linear function s.t. $f(t+h)=f(t)+Df_t(h)+o(||h||)=I\otimes (t^*t^T+h^*t^T+t^*h^T+o(||h||))$.


0

Well if you assume that the object is acted on by a central force (gravity) then $\mathbf r(t)$ is parallel to $\mathbf a(t)$ for all times $t$. And we know that $$\dfrac{d}{dt}(\mathbf r \times \mathbf v) = \require{cancel}\color{red}{\cancelto{0}{\color{black}{\mathbf v \times \mathbf v}}} + \color{red}{\cancelto{0}{\color{black}{\mathbf r \times \mathbf ...


1

Using Green's identity $$\iiint_V u \Delta v dV = - \iiint_V \nabla u \cdot \nabla v \; dV + \iint_S u \frac{\partial v}{ \partial n} dS $$ We define the new function $U = V- V'$ and replace both u, and v in the above $$\iiint_V U \Delta U dV = - \iiint_V \nabla U \cdot \nabla U \; dV + \iint_S U \frac{\partial U}{ \partial n} dS $$ Which simplifies ...


0

Your expression $$\kappa(t) = \frac{\|{\bf r}'(t)\times{\bf r}''(t)\|}{\|{\bf r}'(t)\|^3}$$ is actually good for curves with any parametrization, not only for unit speed curves (i.e., curves that are parametrized by arc-length). When I read the question earlier, in a bit of a hurry, I thought that you were asking about a deduction of said formula - as I ...


0

If you're familiar with the $\epsilon-\delta$ definition of a single variable limit, it's very easy to adapt to the higher dimensional case. If we write $|x-c|$ as $d_\mathbb{R}(x,c)$, the definition of a one dimensional limit becomes: $\lim_{x\to c}f(x) = L$ iff $\forall \epsilon > 0.\exists\delta>0:d_\mathbb{R}(x,c)<\delta\implies ...


1

The natural extension of the classical $\epsilon-\delta$ definition of limit to a function of two variable is: The function $f(x,y)$ has limit $l$ as $(x,y)\rightarrow (x_0,y_0)$ if for every $\epsilon>0$ there exists $\delta>0$ such that $|f(x,y)-l|<\epsilon$ whenever $0<\sqrt{(x-x_0)^2+(y-y_0)^2}<\delta$. Note that this means that ...


2

The definition of the limit is as follows: $\lim_{(x,y)\to (x_0,y_0)}f(x,y)=L\,$ if and only if for all $\epsilon>0$ there exists a $\delta >0$ such that $$|f(x,y)-L|<\epsilon$$ whenever $0<\sqrt{(x-x_0)^2+(y-y_0)^2}<\delta$. NOTE: Alternatively, $\lim_{(x,y)\to (x_0,y_0)}f(x,y)=L\,$ if and only if for all $\epsilon>0$ there ...


1

The projection of the solid on the $xy$ plane is the region enclosed by the parabolas $y = x^2$ and $x = y^2$. These intersect at $(0,0)$ and $(1,1)$ and form the shape of a leaf. Then: $0 \le x \le 1$ and $x^2 \le y \le \sqrt{x}$. $z$ moves from $0$ towards the plane $x + y + 21 = z$. Therefore: $$V = \int_0^1 \int_{x^2}^{\sqrt{x}} \int_0^{x + y + 21} ...


1

Write $$ f \equiv x^3y + 2y^3z + 3xz^3 $$ Recall $$ \nabla = \left(\begin{matrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{matrix}\right)$$ We apply the gradient operator to $f$ which will give us the normal vector at some $(x,y,z)$ $$\nabla f = \left(\begin{matrix} ...


0

Hint: Given a surface $f(x,y,x)=0$ the gradient at a point $P$ is the vector orthogonal to the surface, so the equation of the tangent plane in $P$ is: $$ \langle \mbox{grad} f(x_P,y_P,z_P),(x-x_P,y-y_P,z-z_P) \rangle=0 $$ where $\langle \cdot,\cdot\rangle$ is the inner product.


0

The flux is given by $$\text{Flux}=\int_S \vec F\cdot \hat ndS$$ Here, $\hat n=(\hat xx+\hat yy+\hat zz)/\sqrt{x^2+y^2+z^2}$ in spherical coordinates and $$\vec F\cdot \hat n=(\hat x 4xz+\hat z2y)\cdot (\hat xx+\hat yy+\hat zz)/\sqrt{x^2+y^2+z^2}=2z(2x^2+y)/3$$ Now, we will convert to spherical coordinates with $x=3\sin \theta \cos \phi$, $y=3\sin ...


0

Focusing point is the definition of limits existance. Take a look at that: Let a function $f(x,y):R^2\rightarrow R$ and $x,y \in R$ $f$ is a scalar field and if all the paths that going to $(x_0,y_0)$ approaches the same value, the limit exists. Keyword is all paths. If you want to approach that point using polar coordiates, you have two ...


1

Try the two paths $t \mapsto (t,t)$ and $t \mapsto (t,t^5)$. The limits will be different. Explicitly: Consider the sequence $x_n = {1 \over n}, y_n = {1 \over n}$. Then show that $h(x_n,y_n) \to 0$. Now consider the sequence $x_n = {1 \over n}, y_n = {1 \over n^5}$. Then show that $h(x_n,y_n) \to {1 \over 3}$ (in fact, $h(x_n,y_n) = {1 \over 3}$).


0

An idea: $$\frac{\sin^2(x-y)}{|x|+|y|}=\sin(x-y)\frac{\sin(x-y)}{x-y}\frac{x-y}{|x|+|y|}$$ Now observe that $$\begin{align}&\sin(x-y)\xrightarrow[(x,y)\to(0,0)]{}0\\{}\\&\frac{\sin(x-y)}{x-y}\xrightarrow[(x,y)\to(0,0)]{}1\\{}\\&\left|\frac{x-y}{|x|+|y|}\right|\le\frac{|x|+|y|}{|x|+|y|}=1\end{align}$$ Thus we're done...



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