Tag Info

Hot answers tagged

10

We differentiate the entire expression with respect to $x$ $$3x^5-5y^3=5x^2+3y^5$$ becomes $$15x^4-15y^2\frac{\mathrm{d}y}{\mathrm{d}x}=10x+15y^4\frac{\mathrm{d}y}{\mathrm{d}x}$$ Rearranging to solve for $\frac{\mathrm{d}y}{\mathrm{d}x}$ and factoring out we get ...


6

Say that $0 < \sqrt{x^2+y^2} < \delta$. We can conclude that $|x|<\delta$ and $|y|<\delta$. Now $x^4 + 3y^4 \geq x^4 + y^4 \geq 2x^2y^2\geq x^2y^2 $ Thus, $$ \left| \frac{x^3y^2}{x^4 + 3y^4} \right| \leq \frac{|x|^3y^2}{x^2y^2} = |x| < \delta$$ Now given any $\varepsilon > 0$ if we choose $\delta = \varepsilon$ this would verify the ...


4

Hint: We can write the integral as - $$\displaystyle \int_{-\infty}^{\infty}{{e}^{-{y}^{2}}} dy \int_{-\infty}^{\infty}{{e}^{-{z}^{2}}} dz \int_{-\infty}^{\infty}{{e}^{-{x}^{2}}} dx$$ We know that Gaussian integral $$\displaystyle \int_{-\infty}^{\infty}{{e}^{-{a}^{2}}} da = \sqrt{\pi}$$ So, we can write the integral as - $$\displaystyle ...


4

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


3

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


3

Use the chain rule. You have on the left-hand-side: $$\begin{bmatrix} \frac{\partial x}{\partial \alpha} & \frac{\partial x}{\partial \beta} \\ \frac{\partial y}{\partial \alpha} & \frac{\partial y}{\partial \beta} \end{bmatrix} \begin{bmatrix} \frac{\partial \alpha}{\partial z} & \frac{\partial \alpha}{\partial w} \\ \frac{\partial ...


3

Another approach is to use the fact that for $x\gt0$, we have $\left(\sqrt{x}-\frac1{\sqrt{x}}\right)^2\ge0\implies x+\frac1x\ge2$: $$ \begin{align} \left|\frac{x^3y^2}{x^4+3y^4}\right| &=\frac{|x|}{\sqrt3}\frac{x^2(\sqrt[4]3\,y)^2}{x^4+3y^4}\\[9pt] ...


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}, ...


2

Hint: Look at $F(x,y,z) = x\sin z+y\cos z - e^z$ and compute $\frac{\partial F}{\partial z}(2,1,0)$.


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 ...


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 ...


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 ...


2

$\phi(x,y) = x+iy$, so $f(\phi(x,y)) = (x+iy)^2 = x^2-y^2+2 xy i$. All $\phi^{-1}$ does is to map $x+iy$ to $(x,y)$, so we have $F(x,y) = \phi^{-1}(f(\phi(x,y))) = (x^2-y^2, 2 xy)^T$. Now differentiate $F$. The resulting matrix is the linear transformation. I am guessing that the problem is trying to show the connection to the Cauchy Riemann equations.


2

HINT : $$\int_{-\infty}^{\infty}e^{-t^2}dt=\sqrt{\pi}$$ And separate your integrals. For the sake of completeness: $$\int e^{-t^2}dt=\dfrac{\sqrt{\pi}}{2}\text{erf}\ t+C$$ where $\text{erf}\ t$ is the error function, for which $$\lim_{t\to\infty}\text{erf}\ t=1$$ $$\lim_{t\to -\infty}\text{erf}\ t=-1$$


2

$$ T(t)\cdot N(t)={1\over\|T'(t)\|}T(t)\cdot T'(t)={1\over2\|T'(t)\|}{d\over dt}\left[ T(t)\cdot T(t)\right]=0$$ Because $T(t)\cdot T(t)={r'(t)\cdot r'(t)\over \|r'(t)\|^2}\equiv 1$


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 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 ...


1

Note that along $y=1/x,x>0, f(x,y) = e^{-1}.$ So as $x^2 + y^2 \to \infty $ along this curve, $f(x,y) \not \to 0.$ Since we know that $f \to 0$ on lines through the origin, this shows that $\lim_{x^2+y^2\to \infty}f(x,y)$ does not exist.


1

There's some misunderstanding. It is correct that you fixed a line $y = kx$ (that is, $k$ is the slope of the line), and then you let $x \to \infty$ (So by that equation, $y \to \infty$ also). Now you have $$f(x, kx) = \frac{xkx}{e^{x^2 k^2x^2}} = \frac{kx^2}{e^{k^2 x^4}}$$ So there are two case: $k=0$: Then $f(x, 0) =0$ for all $x$ and so ...


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 ...


1

Your parameterization doesn't look right. In particlar, note that if $x a\cos u \sin v$, $y = a\sin u \cos u$, and $z = a\cos v$, then $x^2+y^2+z^2 = a\cos^2u\sin^2v+a^2\sin^2u\cos^2u+a^2\cos^2v$, which does not simplify to $a^2$. One correct parameterization of the sphere would be $\vec{r}(u,v) = (a\cos u \sin v, a\sin u \color{red}{\sin v}, a\cos v)$, for ...


1

$$\lim\limits_{(x, y)\to(0, 0)}\frac{\sin(x+y)}{x+y}$$ The Taylor series of $\sin(x+y)$ at around $(0, 0)$ is $$ (x+y)-\frac16 (x+y)^3+O\left((x+y)^5\right) $$ Therefore $$\lim\limits_{(x, y)\to(0, 0)}\frac{\sin(x+y)}{x+y}$$ $$=\lim\limits_{(x, y)\to(0, 0)}\frac{(x+y)-\frac16 (x+y)^3+O\left((x+y)^5\right)}{x+y}$$ $$=\lim\limits_{(x, y)\to(0, ...


1

The surface $S$ in question is a level surface of the function $F(x,y,z):=x^2yz-4xyz^2$. The tangent plane to $S$ at $P$ is orthogonal to the gradient $\nabla F(P)$. Therefore it has an equation of the form $$\nabla F(P)\cdot {\bf r}=c\ ,\tag{1}$$ where ${\bf r}=(x,y,z)$. Now fix the constant $c$ in such a way that the point $P=(1,2,1)$ satisfies the ...


1

Yes, find the gradient $\nabla{f}=\left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},\frac{\partial f}{\partial z},\right) $ at point P Then find two vectors that are orthogonal to the gradient at P. You have a base for the tangent plane. Now you need make the appropriate displacement (which is P) in order to get the equation of the ...


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 $\sum_i\sum_j a_{ij} = \sum_j\sum_i a_{ij}$.


1

Given what you know, $|\sin (x-y)| \leq |x-y| \leq |x| + |-y| = |x|+|y|$, so that $$ g(x,y) = \frac{\sin^2 (x-y)}{|x|+|y|} \leq \frac{(|x|+|y|)^2}{|x|+|y|} = |x|+|y|$$ What happens as $(x,y) \rightarrow (0,0)$ to this upper bound? What is a lower bound to this expression (hint: that's deceptively easy). CONTINUATION: Through the work above we have found ...


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.



Only top voted, non community-wiki answers of a minimum length are eligible