# Tag Info

13

Call $t=2x^2+y^2$. Clearly $t \ge 0$, and your equation can be rewritten as $$2t+ \cos t + y^2=1$$ Now, it is easily checked that the function $2t+ \cos t$ is strictly increasing (simply compute its derivative, and check that it's $>0$), so that we have the following inequalities $$1= 2t+ \cos t + y^2 \ge 2 \cdot 0 + \cos 0 + y^2 = 1 + y^2 \ge 1$$ which ...

6

Your mistake is thinking in terms of "dependent" / "independent" symbols without introducing precise mathematical meaning for that. You can't say that $\frac{\partial p}{\partial x}=1$ implies $\frac{\partial x}{\partial p}=1$. $p$ is a function of two variables $(x,y)$, and to calculate $\frac{\partial x}{\partial p}$, you should introduce another variable,...

5

In polar coordinates you have $$\nabla g = \frac{\partial g}{\partial r} \hat r + \frac{1}{r}\frac{\partial g}{\partial \theta } \hat \theta$$ where $\hat r$ and $\hat \theta$ are the unit orthogonal vectors at any point. So you can calculate $$\nabla g \cdot \hat r = (\frac{\partial g}{\partial r} \hat r + \frac{1}{r}\frac{\partial g}{\partial \theta } \... 4 By using \frac{\partial x}{\partial p}\neq 0 and \frac{\partial y}{\partial p}\neq 0, you make x and y depending on p ! Therefore, it has no sense to consider p(x,y). Indeed, if p(x,y) would have sense, then p would be dependent and independent of x and y, which is impossible. 4 The function f is of three variables, so you should be using$$\frac{\partial f}{\partial x},$$or f_x or f_1 to denote the partial derivative of f with respect to x. Here df/dx has no meaning because f is a function of three variables. You would only use this notation (or f') if f was a function of x alone. (You might also use f' to ... 3 For a ternary function f(x,y,t), the expression \dfrac{df}{dz} would usually be read as the total derivative of f with respect to the exogenous argument z. In general, you have for this total derivative that$$ \frac{df}{dz} = \frac{\partial f}{\partial x} \cdot \frac{dx}{dz} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dz} + \frac{\...

3

Let us rewrite the product rule as follows: $$(fg)'=f'g+g'f=\frac{f'}{f}fg+\frac{g'}{g}fg=\left(\frac{f'}{f}+\frac{g'}{g}\right)fg$$ Yours is just the generalization to $n$ factors, but is handled in the exact same way.

3

I don't see why you think the coordinate transformation is troublesome... Take radial and spherical coordinates, but forget about the spherical bit... i.e. $$||x|| = r \quad \& \quad dx = \omega(\theta) r^{d-1} dr$$ where $\omega(\theta)$ and doesn't depend on $r$ (and doesn't depend on $\delta$). Thus $$\int_{ ||x|| \geq \delta} \frac{ dx}{||x||^{... 3 To show the statement is false, notice that for all x\neq 0: f(x,x)=3/2 but f(x,2x)=12/17, so the limit does not exist. 3 Instead of polar coordinates, split into \iint 1\,dxdy + \iint xy\,dxdy. The latter integral is 0 by symmetry; the former is just the area of D. 2 First we parameterize our surface by two families of curves \{u=c\} and \{v=d\}: The we can see that$$\mathbf r_u = \frac{\partial \mathbf r(u_0,v_0)}{\partial u} = \lim_{h\to 0} \frac{\mathbf r(u_0+h,v_0)-\mathbf r(u_0,v_0)}{h}$$will be tangent to the curve u=u_0 at the point (u_0,v_0) and thus also tangent to the surface. Likewise for \... 2 Hint: You are trying to prove something that is false. 2 Decompose the domain D as a triangle T=\{(x,y): 0\leq x\leq 3, 0\leq y \leq x\} and a subgraphic$$ S=\left\{(x,y): 3\leq x\leq 6, 0\leq y\leq\frac{9}{x}\right\}.$$Then we have:$$ \iint_D 3x^2\,d\mu = \iint_{T}3x^2\,d\mu+ \iint_{S}3x^2\,d\mu = \int_{0}^{3}3x^3\,dx+\int_{3}^{6}27x\,dx=\color{red}{\frac{1701}{4}}. $$2 The first question is: Under what circumstances is$$ \frac \partial {\partial\theta} \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \bullet\bullet\bullet $$the same as$$ \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \frac \partial {\partial\theta} \bullet\bullet\bullet \text{ ?} $$The next question is: Why is$$ \frac \partial {\partial\theta} \...

2

But if I take this curve (that I found simply equaling the limit to 1) $$x = \sqrt{\frac{y^2}{y-1}}$$ But $\varphi(y)=\left(\sqrt{\frac{y^2}{y-1}},y\right)$ is not a valid path to $(0,0)$, Namely, it is only defined for $y>1$, so you cannot follow $\varphi(y)$ while having $y\to 0$.

1

"Health is determined by" is the key phrase in your question. If we can say that health is a SUM, like $$H = w_1 + w_2 + \ldots + w_{20}$$ where $w_i$ is the fraction of the RDA, but limited to a maximum of $1$, then this is a constrained optimization problem, and pretty well adapted to standard techniques like the simplex method. If "Health" is some ...

1

A uniform limit of real-analytic functions certainly need not be real-analytic. Any continuous function on $[a,b]$ is a uniform limit of polynomials (and is hence the sum of a uniformly and absolutely convergent series of polynomials). I can't think of any "simple" criteria. Even uniform convergence of a sequence of functions together with uniform ...

1

Yes, that doesn't make sense. I guess, I was being very sloppy, there...

1

Hint: $f(x,y)$ is uniformly continuous on any $[a,b]\times [0,1].$

1

It is wrong at very beginning. $p$ is a function change with $x$ and $y$. If only $x$ changes and $y$ is invariant, $\frac{\partial x}{\partial p}=1$ and $\frac{\partial y}{\partial p}=0$ because $\frac{\partial x}{\partial p} = \lim_{\triangle p->0} \frac{\triangle x}{\triangle p}$ If only $y$ changes and $x$ is invariant, $\frac{\partial y}{\... 1 It's not hard, you just need to work with piecewise definitions. Derivatives in general are a local notion, so really all that matters is you're not explicitly in the knife-edge case where$t_0 = t_1$(aka the diagonal of your domain). As long as$t_0 \neq t_1$you can always find a sufficiently small open nbhd around your point where you can just treat the ... 1 a)$Im(f)$is unbounded, so no absolute min/max. b)$Im(f)=[0, \infty)$the absolute minimum is$0$but there is no max. c) The image of a continuous real function defined on a compact set is compact so there is an absolute max and min. d) the absolute max/min doesn't exist because$Im(f)=(2, \infty)$1 You have three equations from the first order conditions: $$2xy^2z^2=2\lambda x \qquad 2x^2yz^2=2\lambda y\qquad 2x^2y^2z=2\lambda z$$ Suppose$x=0$. Then the first equation is satisfied, and the other two equations imply that$\lambda=0$(since we cannot have$x=y=z=0$as that does not satisfy the constraint). This gives us infinitely many solutions with ... 1 How is the interior derivative a derivative? I wouldn't say it is. My background is in Clifford algebra, and that discipline's equivalent of this operation is universally referred to as a product operation, not a derivative operation. What is the geometric content of Hodge duality? Short version: you're finding the orthogonal complement of whatever ... 1 No.$G$is not a function of$H$, so it doesn't make sense to talk about the derivative of$G$with respect to$H$in any sense.$G$is a function of$x$and$y$. 1 Hint: consider$g:R^2\times R\rightarrow R^3$defined by$g(x,y,z)=(f(x,y),z)$shows that the rank of the differential is 3 and deduce that it is a local diffeomorphism by using the local inversion theorem, then consider the composition of$g$with the projection (which is an open map) on$R^2$which is$f$. 1 Hint: Step: Draw the region Step: Hyperbola$y=9/x$and$y=x$have an intersection. What is the$x$-value of this intersection? Call this value$x_0$. Step: Divide the Integral into two pieces $$\int_{x=0}^{x_0}\int_{y=0}^{x}3x^2 dy dx$$ $$\int_{x=x_0}^{6}\int_{y=0}^{9/x}3x^2 dy dx$$ 1 Your notation is terribly confusing, so let me do things properly for you. You have a point$(a,b)$such that$F(a,b) = 0$. You also have that$\frac {\partial F} {\partial y} (a,b) \ne 0$. The implicit function theorem guarantees the existence of a differentiable function$g$such that$y = g(x)$in some neighbourhood of$a$and$F(x,g(x)) = 0$on it. You ... 1 The interchanging of$\partial x, \partial y$and$\partial F$cannot be done without making additional presumptions. You can achieve the result by using the total differentiation.$dF=\frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial y}dy=0$The equation has to be set equal to zero, because this is the condition for a stationary point$(x_0,...

1

Another approach that is sometimes helpful: use polar coordinates $$\begin{cases}x=r\cos t\\y=r\sin t\end{cases}\implies\frac{3x^2y^2}{x^4+y^4}=\frac{3\cos^2t\sin^2t}{\underbrace{\cos^4t+\sin^4t}_{=(\cos^2t+\sin^2t)^2-2\cos^2t\sin^2t}}=$$ $$=\frac{\frac34\sin^22t}{1-\frac12\sin^22t}$$ Either way, it is clear that $\;r\to0\implies\;$ the limit depends on ...

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