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4

The region of integration $x<y<\sqrt{\frac{\pi}{2}}$ and $0<x<\sqrt{\frac{\pi}{2}}$ is corresponding to $0<x<y$ and $0<x<\sqrt{\frac{\pi}{2}}$, therefore \begin{equation} \int_0^{\large\sqrt{\frac{\pi}{2}}}\int_x^{\large\sqrt{\frac{\pi}{2}}}\, \sin y^2\, dy\ dx=\int_0^{\large\sqrt{\frac{\pi}{2}}} \sin y^2\int_0^y\, dx\ ...


3

Define $g(x,y)=x+y$, and it is pretty clear that $\lim_{(x,y) \to (0,0)} g(x,y)=0$. Now $$ f(x,y)=\frac{\sin g(x,y)}{g(x,y)}=S(g(x,y)), $$ where $$ S(t)=\frac{\sin t}{t} $$ for $t \neq 0$. Since $\lim_{t \to 0} S(t)=1$, the theorem about limits of composition of functions gives that $\lim_{(x,y) \to (0,0)}f(x,y)=1$.


2

Using the linearity of the inner product, we get the following: $G(x+\epsilon h) = \langle A(x+\epsilon h),x+\epsilon h\rangle = \langle Ax+\epsilon Ah,x+\epsilon h\rangle$ $= \langle Ax,x \rangle + \langle Ax , \epsilon h \rangle + \langle \epsilon Ah , x \rangle + \langle \epsilon Ah,\epsilon h \rangle$ $= \langle Ax,x \rangle + \epsilon \langle Ax , ...


2

If $F(x) = f^T(x) g(x)$ the product rule gives $DF(x)(h) = (Df(x)(h))^T g(x) = f^T(x) Dg(x)(h)$. With $f(x) = Ax$, you have $Df(x)(h) = Ah$ and with $g(x) = x$, you have $Dg(x)(h) = h$. Substituting gives $DF(x)(h) = h^T A^T x + x^T A^Th = 2 x^T Ah$ (using $A=A^T$). This is sometimes written as $Df(x) = 2 x^T A$. Aside: This is easy to verify directly by ...


2

If you have two functions $f$ and $g$, then $\Delta(fg)=f\Delta g+g\Delta f+2\nabla f\cdot\nabla g$. Now we obviously want to choose $f(x)=|x|^{-3}$ and $g(x)=\mu\cdot x$. Recall that $\nabla |x|^\alpha=\alpha|x|^{\alpha-2}x$; to remember this rule, remember that the gradient is a vector and the formula should hold true in dimension one. These choices give ...


2

I'm going to answer by guesswork on what the German might be, generally agreeing with @Git Gud. 0. Your "that shows this is solvable" really should be "that shows that $(0,0,0)$ really is a point on the level-surface. "Local resolution" probably means "there's a function z = f(x, y)" such that $$ x^4 + e^y + sin( f(x, y) ) + f(x, y)^5 = 1 $$ for all ...


2

For a function on 3-space, it's a level-surface rather than a level curve, but you're right, it's at level 4. The other fact that you need to know is that hte gradient of your function is perpendicular to the level surface. Since $$ \nabla f (x, y, z) = ( (1+4y)/(5z^2 + 3), 2/(5z^2+3), -(x+2y+4xy)/(5z^2+3)^2) $$ we get $$ \nabla f (6, -1, 1) = ( (-3)/8, ...


2

Indeed, since you are trying to find the area, the integral can simply be weighted by 1. In polar coordinates, the area of the entire 3-petal graph is given by $$A=\int_{-\pi/2}^{\pi/2}\frac{1}{2}r(\theta)^2\,\mathrm{d}\theta=\int_{-\pi/2}^{\pi/2}\frac{1}{2}\cos^2(3\theta)\,\mathrm{d}\theta=\frac{\pi}{4}$$ and so the area of a single petal is $\pi/12.$ Here ...


2

One method you can apply is the lagrange-multiplier method. The lagrange function is: $\mathcal L=Q(L,K)+\lambda (B-C(L,K))$ $\mathcal L=L^{0.5}\cdot K^{0.5}+\lambda (800-25L-40K)$ The partial dervatives, w.r.t L and K, are: $\frac{\partial \mathcal L}{\partial L}=0.5\cdot L^{-0.5}\cdot K^{0.5}-\lambda\cdot 25=0 $ $\frac{\partial \mathcal L}{\partial ...


2

The change of variables $u = xy$, $v = x^2 - y^2$ works. Check that it's one-to-one in the first quadrant. (I assume you want only the first quadrant region bounded by those curves.) EDIT: First we prove that the the transformation is one-to-one on the first quadrant. Let $x, y > 0$, and define $u$, $v$ as above. $x^2$ and $-y^2$ are the roots of the ...


1

Notice that you have a miscalculation in your first equation (the $-xy$ term should be $-2xy$) and that each of your two equations factor: $$\begin{cases} x(3 - x - 2y) = 0 \\ y(3 - 2x - y) = 0 \end{cases}.$$ Thus you have to consider four possibilities: (1) $ x = y = 0$ (2) $ x = 3 - 2x - y = 0$ (3) $ 3 - x - 2y = y = 0$ (4) $ 3 - x - 2y = 3 - 2x - y ...


1

Consider the variables \begin{align*} u = xy, && v = x^2-y^2. \end{align*} We obtain that $x=\frac{u}{y}$. Therefore: $$y^2=x^2-v,$$ $$y^2 -\frac{u^2}{y^2}=-v,$$ $$\frac{y^4-u^2}{y^2} =-v,$$ $$y^4+vy^2-u^2=0,$$ $$y^2 = \frac{-v\pm \sqrt{v^2-4(-u^2)}}{2} = \frac{-v \pm \sqrt{4u^2+v^2}}{2}.$$ From which we obtain: $$x^2=\frac{-v \pm ...


1

First, the subtraction order does not matter for this problem. Second, hint: can you write down a vector which gives the direction of the first line? Note that it is not $$(-2-2t, 6t, -3+4t)\ ,$$ this is a vector from the origin to the line.


1

Your answer is correct, but the logic is wrong. You should consider the above surface as a level surface of the function: $$ f(x, y, z) = x^2 + 2y^2 + z^2 - 2x - 2z - 2$$ Then the gradient should be a vector with 3 components. $$ \nabla f(x, y, z) = (2x - 2, 4y, 2z - 2) $$ If the tangent plane is horizontal, the gradient must point in the $z$-direction, ...


1

Turns out that one answer is also on Wikipedia, in the article on the arg function $$\mbox{atan2}(x,y) = 2\tan^{-1} \left({ y \over {\sqrt{x^2+y^2}+x}}\right)$$ where the values of $\tan^{-1}$ are taken in $(-\pi/2,\pi/2)$. The right-hand side is well-defined and has continuous partial derivatives in the domain in the question. Verifying the formula ...


1

I'm referring to the following figure: If $(x,y)=(r\cos\theta,r\sin\theta)$ then $$\tan{\theta\over2}={y\over r+x}\ .$$ It follows that the principal value of the argument ("atan2" in some circles) is given by $${\rm Arg}(x,y)=\theta=2{\theta\over2}=2\arctan{y\over\sqrt{x^2+y^2}+x}\ .$$ Note that $\sqrt{x^2+y^2}+x\ne0$ in the domain in question. ...


1

If $\log (*)$ means $\log_e (*)$ or $\ln (*)$ then we have \begin{align*} \int_{\mathbf{c}}{f(x,y,z)ds}&=\int_{1}^{e}f(x(t),y(t),z(t))(||c′(t)||)dt\\ &=\int_{1}^{e}{\frac{1}{t^3}\sqrt{\frac{1}{t^2}+1}\;dt}\\ &=\int_{1}^e{t^{-3}\sqrt{t^{-2}+1}\,dt}\\ &=\left.-\frac{1}{3}\left(t^{-2}+1\right)^{3/2}\right|_1^e\\ ...


1

if you are integrating over an annulus, then by symmetry $$ \iint_{D} (\frac{x^2}{x^2+y^2})dA = \iint_{D} (\frac{y^2}{x^2+y^2})dA $$ so by adding you have $$ \iint_{D} (\frac{x^2+y^2}{x^2+y^2})dA = \iint_{D} dA = \pi(b^2-a^2) $$


1

$$\int\limits_0^R \int\limits_0^{\sqrt{R^2-x^2}} e^{-x^2-y^2} \; dy \; dx = \int\limits_0^{\pi/2} \int\limits_0^{R} e^{-\rho^2} \rho \; d\rho \; d\theta = \frac{\pi}{4} \left(1 - e^{-R^2}\right).$$ The domain of integration is a quarter circle of radius $R$, so when one converts to polar coordinates one sees that $\rho$ goes from zero to $R$ and ...


1

We want to integrate $$I = \iint_D e^{-(x^2+y^2)} \, dA $$ with $D$ being the region bounded by $0 \le y \le \sqrt{R^2-x^2}$ and $ 0 \le x \le R$. This is a quarter circle of radius $R$ in the first quadrant. In polar coordinates this is equivalent to $0 \le r \le R $ and $0 \le \theta \le \pi/2$. Using a change of variables: $$ I = \int_0^{\pi/2} \int_0^R ...


1

Since the line is perpendicular to the plane, the normal vector $\langle 1,3,1\rangle$ for the plane is parallel to the line, so the line has vector equation $\langle x,y,z \rangle=\langle 1,0,6\rangle + t\langle 1,3,1\rangle$ and parametric equations $x=1+t, \;\;y=3t, \;\;z=6+t$.


1

All the functions given by the OP are $C^\infty$ smooth. The derivative of $g$ is the identity matrix and it is invertible. It is a good exercise to prove that the derivative of a linear map is the map itself; that is, if $f(x)=Ax$ for a matrix $A$, then $f'(x)=A$ for all $x$. Now $g(s,t)=(g_1(s,t),g_2(s,t))=(0,2+2s-t)$. This is again a linear mapping, but ...


1

We have $$\begin{eqnarray}x =& a \cosh \mu \cos \nu,\\ y =& a \sinh \mu \sin \nu.\end{eqnarray}$$ For part (a), if we choose a value of $\mu$ and hold $\mu$ constant at that value, then $a\cosh \mu$ and $a\sinh \mu$ are simply two constants. If we label those constants $m$ and $n$, the curve that we trace as we vary $\nu$ is just the curve described ...


1

I'm going to suggest that you restrict your attention to the partial derivatives, i.e., how $\ln(f(x_1, \ldots, x_n))$ varies when you vary $x_i$. For this, you can think of this as a one-variable calculus problem, and the result is $$ \frac{d\ln(f(\mathbf x))}{dx_i} = \frac{1}{f(\mathbf x)} \frac{df(\mathbf x)}{dx_i}. $$ By long tradition, this derivative ...


1

Hint: What is the area of integration? You're integrating over the triangle bounded by the lines $y=0$, $x=\sqrt{\pi/2}$ and $y=x$. Try to reformulate the same integral with the integration order changed for the same triangle.


1

Your limit-of quotient definition of derivative doesn't work well in the multi-dimensional case. Note that something goes wrong already at your first displayed equation where you want a limit of quotients of reals to be a matrix. That is not going to go well. Rather you're probably looking for a definition that says $f:\mathbb R^k\to\mathbb R^n$ is ...


1

Hint If you assume that the length of basis is $x$ ($0<x<1$) then its weight is $y=1-x^2.$ So, its area is $x\cdot (1-x^2).$ You need to find the maximum of this function.


1

You want to maximize $xy$ with the costraints $x\ge 0$, $y\ge 0$, $x^2+y=1$. Solve $x^2+y=1$ for $y$ and replace in the function to be maximized.


1

One possible way is using Lagrange Multiplier. We want to maximize $f(x,y)$ Subject to $g(x,y)=c$ In this case, $f(x,y)=xy$ , $g(x,y)=x^2+y$ and $c=1$ We'll define $L(x,y,\lambda)=f(x,y)+\lambda(g(x,y)-c)=xy+\lambda(x^2+y-1)$ We want to find $x$ and $y$ such that: $\partial L / \partial x = 0$ $\partial L / \partial y = 0$ $\partial L / \partial \lambda = ...


1

As you showed, the vector field is conservative, so it doesn't matter which path you take, the only thing you need are the starting and end point. First, as $\mathrm{F}$ is conservative, you have to calculate a function $f$ such that $\nabla f=\mathrm{F}$. An easy way to do this is using this formula: $$\displaystyle f(x,y) = ...



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