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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$$2 Take limit along curve y=x^5. 2$$ 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}}$$So there are two case: k=0: Then f(x, 0) =0 for all x and so ... 1 use the two equations as simultaneous equations at zero and solve the homogeneous equations. 4(x^3 - y) = 0 and 4(y^3 - x) = 0. Pretty sure that this gives a line in space rather than point that you might be use to. this is because there is no positive numbers without variables ie: 4(x^3 - y) = 3 and 4(y^3 - x) = 5 will give a single point extrema. 1 Hint: Look at the corresponding closed region R, so that the extrema will either be critical points in intR or they will lie on the boundary of R. For the saddle points, use the second derivative test. Second hint: Step 1: Solve the following equations and use the points in the second derivative test to decide if they are local extrema or saddle ... 1 Your parameterization doesn't look right. In particlar, note that if x a\cos u \sin v, y = a\sin u \cos u, and z = a\cos v, then x^2+y^2+z^2 = a\cos^2u\sin^2v+a^2\sin^2u\cos^2u+a^2\cos^2v, which does not simplify to a^2. One correct parameterization of the sphere would be \vec{r}(u,v) = (a\cos u \sin v, a\sin u \color{red}{\sin v}, a\cos v), for ... 1$$\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.

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