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## Hot answers tagged multivariable-calculus

7

Note that for $x \neq 0$ and $y \neq 0$ we have $$\left | \frac{x^2y}{x^2 + y^3} \right| = \left| \frac{y}{1 + \frac{y^3}{x^2}} \right| = \left| \frac{1}{\frac{1}{y} + \frac{y^2}{x^2}} \right| \leq \left| \frac{1}{\frac{1}{y}} \right| = \left| y \right|.$$ If $x = 0$ or $y = 0$ (but not $x = y = 0$), then we also have $$\left | \frac{x^2y}{x^2 + y^3} ... 5 If 64= (3y^2)^2 + 4x^4, then 3y^2 =\sqrt{ 64 - 4x^4}. Plugging that to x^2+y^2 we get:$$x^2+\frac{\sqrt{64 - 4x^4}}{3}.$$Now you have to maximalize a function of one variable. Remember that if 64 = 4x^4+9y^4, then (setting y = 0) we obtain 16 \ge x^4, therefore |x| \le 2. 5$$0\le\left|\frac{x^2y}{x^2+y^2}\right|\le \left|\frac{x^2y}{x^2}\right|=|y|\xrightarrow[(x,y)\to (0,0)]{}0$$4 I think you made a mistake for the last part. And if you apply polar coordinate properly ( in general it may not simplify the problem though), it will give you the correct answer. \lim_{(x,y) \to (0,0)}\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}=\lim_{r \to 0} \frac{r^2}{\sqrt{r^2+1}-1}=\lim_{r \to 0}\frac{r^2(\sqrt{r^2+1}+1)}{r^2+1-1}=\lim_{r \to ... 4 You were almost there. Use conjugates:$$\frac{r^2}{\sqrt{r^2+1}-1}=\frac{r^2(\sqrt{r^2+1}+1)}{r^2}=\sqrt{r^2+1}+1\xrightarrow[r\to 0]{}2$$3 Note that \frac{x^2}{x^2+y^2} \leq 1 so your function is bounded by \pm y. 3 Notice that$$ |f(x,y)|=\left|\frac{\sin(xy)}{\sqrt{x^2+y^2}}\right|\le \frac{|xy|}{\sqrt{x^2+y^2}}\le \sqrt{x^2+y^2} \quad \forall (x,y)\ne (0,0). $$Therefore \lim_{(x,y)\to(0,0)}f(x,y)=0 and so you can define a continuous function by setting$$ \tilde{f}(x,y)=\begin{cases} f(x,y) &\mbox{ if } (x,y)\ne (0,0)\\ 0 & \mbox{ if } (x,y)=(0,0) ...

3

The statement is not true. If a function has continuous partial derivatives on an open set U, then it is differentiable on U. But a differentiable function need not have continuous partial derivatives. A standard example is the function $f(x)=x^2\sin(\frac1x)$ which is differentiable but its partial derivative with respect to x ...

3

Hint: Choose a function such that the overall limit at $(0,0)$ does not exist (but the limit along some path does). Example: take $$f(x,y) = \begin{cases} \frac{x^2}{x^2 + y^2} & (x,y) \neq (0,0)\\ 0 & x = y = 0 \end{cases}$$ Verify that $\Gamma_f$ is path-connected, but $f$ is not continuous at $(0,0)$.

3

Additional assumptions are necessary. Consider in ${\mathbb R}^2$ the example $$f(x,y)=g(x,y):=x^2\ .$$ Then $D^2f(0,0)=D^2g(0,0)\ne0$ (as quadratic forms), but the limit $$\lim_{y\to0}{f(0,y)\over g(0,y)}$$ does not exist. Therefore let's assume that $D^2f(0)=c\>D^2 g(0)$, where $H:=D^2 g(0)$ is positive definite. Then for $x=r\>u$, with $r>0$ and ...

2

Let $u = \dfrac{y}{x}, v = \dfrac{y}{x^2} \to x = \dfrac{u}{v} , y = \dfrac{u^2}{v} \to \dfrac{x^2}{y^3} = \dfrac{v}{u^4} \to I = \displaystyle \int_{1}^2 \int_{1}^2 \dfrac{v}{u^4}\cdot \left|\begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} ... 2 A (nonempty) open set contains an interval or ball, and given that these have positive content (a slightly tricky proof, but presumably covered in your book or class), you know that the open set itself cannot be zero content. The empty set, however, is a valid open set with zero content. If a set is covered by finitely many bounded sets, then it is bounded. ... 2 What is$U_y$? The jacobian matrix is$J = \begin{pmatrix} 1 & 0 \\ 3v & 3u\end{pmatrix}$It's determinant is$3u$2 $$4 + \frac{1}{t^2} + 4t^2 = \frac{4t^2 + 1 + 4t^4}{t^2} = \frac{(2t^2 + 1)^2}{t^2} = \left(\frac{2t^2 + 1}{t}\right)^2$$ So the square root is as given. 2 By AM-GM inequality we have: $$2|xy|\leq x^2+y^2$$ So: $$0<\frac{|xy|}{x^2+y^2}\leq\frac{1}{2}$$ Multiply both sides by$|x|$: $$0\leq \frac{|x^2y|}{x^2+y^2} \leq \frac{|x|}{2}$$ Now$\frac{|x|}{2}$goes to zero when$(x,y) \to 0$, so$\frac{|x^2y|}{x^2+y^2}$also goes to zero. 2 You’ll probably want to to apply the following proposition. Note that it does not depend on$f$and$g$being continuous. Prop. Suppose$\lim_{x\to p} f(x) = c$and$\lim_{y\to c} g(y) = L$. Then$\lim_{x \to p} g(f(x)) = L$. Proof. Our goal is to make$|g(f(x))-L|$arbitrarily small. So fix$\epsilon>0$. Suppose that$|g(y) - L| < \epsilon$for all ... 2 $$t:= \alpha'(s) \in \mathbb{R}^2,\ t'=\kappa n\in \mathbb{R}^2$$ And$|t|=1$implies that $$t\cdot n=0$$ Then $$\cos\ \theta = t\cdot e_1,\ -\sin\ \theta =n\cdot e_1$$ so that $$-\sin\ \theta \theta ' = t'\cdot e_1=\kappa n\cdot e_1$$ 2 If$\theta(s)$is the angle ´twixt the$x$-axis and the tangent line at the point$\alpha(s)$, the unit tangent vector$\mathbf T(s) = \alpha'(s)$is given by$\mathbf T(s) = (\cos \theta(s), \sin \theta (s)); \tag{1}$then we have$\mathbf T'(s) = (-\sin \theta(s), \cos \theta (s)) \theta'(s); \tag{2}$setting$\mathbf n(s) = (-\sin \theta(s), \cos ...

2

I see you have skipped a lot of details in your question. But I will still try my best to answer your question. You need to tell us what kind of a non-linear PDE you are trying to solve and your ICs/BCs. From your shock speed calculation, it looks like you are solving the Burgers Equation (Inviscid). As you say, the shock speed as soon as it forms can be ...

2

Take $f(\frac{1}{2},y)$ to be the Dirichlet function in $y$ and $f(x,y) = 1$ otherwise. $f$ is Riemann integrable on $[0,1] \times [0,1]$ but $g(y) = f(\frac{1}{2},y)$ is not Riemann integrable on $[0,1]$. $f$ is Riemann integrable on the unit square because the partition of $x$ containing $x = \frac{1}{2}$ can be made arbitrarily small. Since the infimum ...

2

$x^2=s,\ y^2=t \geq 0$ then $$(2s)^2+ (3t)^2=8^2 \Rightarrow 2s=4\cos\ \theta,\ 3t=4\sin\ \theta,\ 0\leq\theta \leq \frac{\pi}{2}$$ Then $$s+t=2\cos\ \theta + \frac{4}{3}\sin\ \theta=\sqrt{2^2+ \bigg(\frac{4}{3}\bigg)^2} \sin\ (\theta+\alpha),\ 0< \alpha < \frac{\pi}{2}$$ Let $\theta =\frac{\pi}{2}-\alpha$

2

First thing, $\vec c\cdot\vec r$ isn't a vector but a scalar function, so with $\vec r$ being the variable, $\vec c\cdot\vec r$ can be seen as a function from $\mathbb R^n$ ($n$ is the number of components of $\vec r$: if you're looking at it as a position vector, it might be $n=1$, $2$ or $3$) to $\mathbb R$. Since $\vec c\cdot\vec r=\sum_{i=1}^nc_ir_i$, ...

2

firstly, you must sketch the region $$\int_{\pi /4}^{\pi /2}\int_{0}^{2}\sin(r^2)rdrd\theta$$

1

Below is the region over which you are integrating. Can you now fix the limits of $r$ and $\theta$?

1

For Green's Theorem $$\oint_C (Mdx+Ndy)=\int \int_R \left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right) dxdy$$ Here, $M=(x-y)$ and $N=x$ such that \begin{align}\oint_C (Mdx+Ndy)&=2\int \int_R dxdy\\\\ &=2\int_{-1}^{1} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}dxdy\\\\ &=4\int_{-1}^{1} \sqrt{1-y^2}dy\\\\ ... 1 I would suggest as the follows, but you better double check !!! The plane equation is3x+3y+2z=12$$Then, integration over the tetrahedron may be written as$$ \int_{T}f(x,y,z)dxdydz = \int_0^6\int_0^{\frac{12-2z}{3}}\int_0^{\frac{12-3y-2z}{3}}e^zdxdydz=\frac{4}{9} \left(e^6-25\right). $$1 If \lim_{y\to 0}f(x,y) exists for x\ne 0, then the iterated limit will exist and equal f(0,0). But it's possible for f to be continuous at (0,0) and not have that first limit existing: Let g be the characteristic function of the rationals on \mathbb {R}. Set f(x,y)=x^2g(y). Then f is continuous at (0,0) but for x\ne 0,\lim_{y\to ... 1 Hint: Observe that$$\frac{\partial}{\partial y}\left(\frac x{x^2+y^2}+3x^2y+y^2+2x\right)=\frac{\partial}{\partial x}\left(\frac y{x^2+y^2}+x^3+2xy\right)$$and also$$\frac{\partial}{\partial y}\left(\frac{-y}{x^2+y^2}\right)=\frac{\partial}{\partial x}\left(\frac x{x^2+y^2}\right) so that $\;g\;$ is a conservative field in the given integration ...

1

Consider a simple example like: $C$ is a circle of radius 2 and $f(x,y)=1$. In this case $\int_C f(x,y)\,ds=4\pi$ since it computes the area of under the portion of $f(x,y)=1$ which lies above the circle (this is a cylinder of radius 2 and height 1). Parameterize $C$ by $(2\cos(t),2\sin(t))$ with $0 \leq t \leq 2\pi$. The if you neglect $s'(t)$ (the $ds$ ...

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