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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

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

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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