# Tag Info

0

Equation of tangent plan is given by $$z-g(x_0,y_0)=f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)$$ so $$z=g(x_0,y_0)+ f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)\\= g(-6,9)+ f_x(u(-6,9),v(-6,9))(x+6) + f_y(u(-6,9),v(-6,9))(y-9)\\= -6+ f_x(-5,-9)(x+6) + f_y(-5,-9)(y-9)\\= -6+ 5(x+6) + 8(y-9)$$

2

We don't need any calculus to find the minimum surface area - AM-GM works fine. Solution 1. AM-GM We have the surface area as $xy+2yz+2zx$ with the constraint of $xyz=216$.$$xy+2yz+2zx = xy+\frac{432}{x}+\frac{432}{y} \ge 3 \sqrt[3]{xy \cdot \frac{432}{x} \cdot \frac{432}{y}} = 3\sqrt[3]{432^2} = 108 \sqrt[3]{4}$$ The equality holds at $x=y=6\sqrt[3]{2}$ ...

1

Substituting $x=r\cos\theta$ and $y=r\sin\theta$, $$\int_0^{2\pi}\int_0^3\frac{5r}{\sqrt{5^2-r^2}}drd\theta=\int_0^{2\pi}\left[5\sqrt{5^2-r^2}\right]_3^0d\theta=10\pi$$ Using the already known formula for spherical cap (https://en.wikipedia.org/wiki/Spherical_cap) and $r=5,h=1$, $$A=2\pi rh=10\pi$$ Those two results match.

0

Let $X$ be the variable you want to calculate, and let $Y$ and $Z$ be two other variables. You have : $$X = \dfrac{(X+Y)+(X+Z)-(Y+Z)}{2}$$

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Assume that $A,B,C,D$ are real numbers. Suppose $A\leq B\leq C\leq D$. Then, $A+B\leq A+C\leq A+D \leq B+D \leq C+D$ and $A+C\leq B+C \leq B+D$. Hence, if we order the pair sums as $S_1\leq S_2\leq S_3 \leq S_4 \leq S_5 \leq S_6$, then we know $A+B=S_1$, $A+C=S_2$, $B+D=S_5$, and $C+D=S_6$. Note that $T:=S_1+S_6$ must be the same as $S_2+S_5$ and ...

0

Ampere's law states that $$\int_C \vec{B}\cdot d\vec{r} = \mu I.$$ If $C$ is a circle of radius $R$, it can be parametrized as follows: $$\vec{r}(t)=R\cos(t)\vec{i}+R\sin(t)\vec{j},\quad 0\le t \le 2\pi,$$ so Ampere's law can be rewritten as $$\int_0^{2\pi} \vec{B}(\vec{r}(t))\cdot \vec{r}'(t)dt = \mu I.$$ But if the magnetic field $\vec{B}$ is ...

1

The basic ideas are correct, but you have to be careful with the order of the integration(do not forget the roots and squares) First of all I think you meant this: $\int_{0}^{2}$$\int_{0}^{\sqrt{-y^2-z^2+4}}$$\int_{0}^{\sqrt{-x^2-z^2+4}}$ $xy, dxdydz$ But there is still the problem that you first integrate with respect to $x$ but at the same time you have ...

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First, $x^2=-y^2-z^2+4$ does not imply $x=-y-z+2$. The best we can do here is $x=\sqrt{-y^2-z^2+4}$, the positive square root because the region is in the first octant. If your innermost integral is with respect to $x$ then this is the suitable upper bound. Similarly, $y^2=-x^2-z^2+4$ and so $y=\sqrt{-x^2-z^2+4}$, but this cannot be the upper bound for the ...

2

For a problem like this, I think it helps to draw out as much as you can. Now clearly drawing this in three dimensions can get messy, but notice that the bounds for $x$ are given to be from $x=0$ to $x=1$. We can choose to integrate this last and then disregard it for a bit and deal with just $z$ and $y$. Now if we draw out the lines $z = y+1$, $y+z = 1$, ...

1

I think the first limits are correct, but not the second, the angle would be only between 0 and pi/2 based on the area if you draw on a graph.

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Ampere's Law: $$\oint \vec B \cdot \, d \vec l = \mu_0 I$$ We assume that the magnetic field $\vec B$ is constant, therefore we take it out of the integral: $$B \oint d l = \mu_0 I$$ $$B \cdot (\text{circumference of circle}) = \mu_0 I$$ $$B \cdot (2 \pi r) = \mu_0 I$$ $$B = \frac{\mu_0 I}{2 \pi r}$$

1

Note that $z$ is the composite map $$z: (x,y) \mapsto x^{2}-y^{2} =: \xi \mapsto \int_{1}^{\xi}\int_{0}^{u}\sin t^{2} dt du;$$ so by chain rule and fundamental theorem of calculus we first have $$D_{2}z(x,y) = \int_{0}^{\xi}\sin t^{2}dt\cdot (-2y) = \int_{0}^{x^{2}-y^{2}}\sin t^{2}dt \cdot (-2y),$$ and then we have $$D_{1,2}z(x,y) = (-2y)\cdot ... 1 As I noted in the comment section, I am a bit puzzled at your confusion, since you seem to arrive at \partial z / \partial y - however, maybe this will help.... Write$$ F( y ) = \int_1^{x^2 -y^2} f ( u ) \,du,$$where one views x as a constant. Then, by the FTC and the chain rule, one has$$F'(y) = f( x^2 - y^2)\cdot (-2y).$$In this question,$$ ...

1

From elementar trigonometry we see that the area is $A=\dfrac {\pi}{6}-\dfrac{\sqrt{3}}{8}$ where $\dfrac {\pi}{6}$ is the area of the circular segment of central angle $\theta =\dfrac{\pi}{3}$ and $\dfrac{\sqrt{3}}{8}$ is the area of the triangle of basis $\dfrac{1}{2}$ and height $\dfrac{\sqrt{3}}{2}$. We can find this with a double integral in polar ...

3

$g'$ has $n$ components, and $f$ has $n$ partial derivatives. You need to multiply each component of $g'$ with the corresponding partial derivative of $f$ (evaluated at $g(0)$) and then take the sum. In general, the derivative of a function $\mathbb R^m\to\mathbb R^n$ is an $m\times n$ matrix, and the chain rule is a matrix multiplication.

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HINT: $$\int_{\frac{1}{2}}^{1}\left[\int_{0}^{\sqrt{1-x^2}}1\space\text{d}y\right]\space\text{d}x=\int_{\frac{1}{2}}^{1}\left[\left[y\right]_{0}^{\sqrt{1-x^2}}\right]\space\text{d}x=\int_{\frac{1}{2}}^{1}\sqrt{1-x^2}\space\text{d}x=$$ For the integrand $\sqrt{1-x^2}$, substitute $x=\sin(u)$ and $\text{d}x=\cos\space\text{d}u$. Then ...

2

$$\int_{\frac 12}^{1} \int_{0}^{\sqrt{1-x^2}} 1 \quad dydx=\int_{0}^{\frac{\pi}{3}} \int_{1-0.5\sec\theta}^{1} \quad rdrd\theta$$ some details: $$x=0.5$$ $$r\cos\theta=0.5$$ $$r=0.5\sec\theta$$

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$\frac{1}{2}≤x≤1$ $0≤y≤\sqrt{1-x^2}$ Sketch the region which satisfies the above. Then change the region from $dy dx$ to $dx dy$. The limits become $\frac{1}{2}≤x≤\sqrt{1-y^2}$ and $0≤y≤\frac{\sqrt3}{2}$ As suggested above integrating it without changing limits is probably easier.

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You can project the solid onto the xy - plane. You'll find that the projection onto the xy - plane is a right - angled triangle which is given by $$D = \{ (x, y) \ \epsilon \ R^2 \ | \ 0 \le x \le 1, 0 \le y \le x \}$$ and therefore the solid can be represented as $$\{ (x, y, z) \ \epsilon \ R^3 \ | \ 0 \le z \le 1 - x^2, (x, y) \ \epsilon \ D \}$$ Then you ...

4

I think you're falling into a trap many new students to of notation first, meaning second. You are writing things down and then asking what they mean. This is akin to writing a bunch of words and then asking "what does this sentence mean?". I'm not going to go through each of your 7 integrals to say which does or does not make sense, nor will I give you a ...

1

The region of integration represented by D can be expressed as $D= D_1 \ U \ D_2$ where $$D_1 = \{ (r, \theta)| 0 \le r \le 1, {\pi\over 3} \le \theta \le {\pi\over 2} \}$$ and $$D_2 = \{ (r, \theta) | 0 \le \theta \le {\pi\over 3}, 0 \le r \le {1\over {2cos \theta}} \}$$ Then you can split the double integral over the two non - intersecting domains and ...

0

$R \in [0,1]$ because the radious isn't affected by the change of coordinates but you should take some care about $\theta$. When $x=\dfrac{1}{2}$, then $y=\dfrac{\sqrt{3}}{2}$ then $\theta=tan^-1(\sqrt{3})=\dfrac{\pi}{3}$ so $\theta \in [0, \dfrac{\pi}{3}]$

0

With this kind of question its good to sketch the region. We have $0 \leq y \leq \sqrt{1-x^2}$, and $0 \leq x \leq 1/2$. The first chain of inequalities can be re-written as $y\geq 0, y^2 \leq 1 - x^2$ or $1 \geq y^2 + x^2$, $y\geq 0$. This is the upper-half unit disk. Then we have the vertical strip between $0$ and $1/2$ for $x$. Try to draw a picture ...

2

I don't see anything wrong in your calculation and I think it correct.

5

By Tonelli's theorem, you can interchange the order of this iterated integration: \begin{align} & \int_0^1\int_{3y}^3 e^{x^2} dx dy \\ = & \int_0^3 \int_0^{x/3} e^{x^2} dy dx \\ = & \frac{1}{3}\int_0^3 xe^{x^2} dx \\ = & \frac{1}{6}\left.e^{x^2}\right|_0^3 \\ = & \frac{1}{6}(e^9 - 1). \end{align}

1

Switch the order of integration.

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Choose the coordinate axes and introduce polar coordinates such that the $1^{st}$ electron is located at $\vec{p}_1 = (0,0,r_1)$. the $2^{nd}$ electron is located at $\vec{p}_2 = (x,y,z) = (r \sin\theta\cos\phi, r \sin\theta\sin\phi, r\cos\theta)$ It is known that $\frac{1}{r_{12}} = \frac{1}{|\vec{p}_1 - \vec{p}_2|} = ... 0 hint: consider the Lagrange-function $$f(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^2-z^2-1)$$ 1 As mentioned in the comment, the Leibniz integral rule is not for this situation. Instead, just integrate it outright: $$\int_0^{y^2} e^{-xy}\,dx={e^{-xy}\over -y}\Bigg|_{x=0}^{x=y^2}={e^{-y^3}\over -y}-{1\over -y}={1-e^{-y^3}\over y}.$$ Perhaps you were asked to find${d\over dy}\int_0^{y^2} e^{-xy}\,dx$? Now that would summon the Leibniz rule... 0 You cannot use Green's theorem here, because the surface is not planar. You must start by parametrizing the surface. Since it belongs to the sphere, and that$\phi_2 < \frac{\pi}{2}$, you can write: $$x=x, \quad y=y, \quad z=\sqrt{R^2-x^2-y^2},$$ with$(x,y)$belonging to the projection of the surface in the$xy$plane, namely the ring (in polar ... 1 This limit exists if there exists$L$such that for every$a_n$and$b_n$, it can be proved if$\lim_{n\rightarrow \infty}a_n=0, \forall n (a_n\not =0)$and$\lim_{n\rightarrow\infty}b_n=0,\forall n(b_n\not = 0)$, $$\lim_{n\rightarrow \infty}\frac{|a_n|^{e^{\frac{1}{a_n}}}b_n}{a_n-b_n}=L$$ If we choose$a_n=b_n=\frac{1}{n}$, because$a_n-b_n=0, there for ... 1 Personally, I would suggest "Vector Calculus" from Jerrold E. Marsden & Anthony J. Tromba. A very good book which contains all of what you are requiring (surface integrals, the theorems of vector calculus and vector analysis), the vector identities (for example for the usage of divergence and curl); changing order of integration or setting the bounds ... 1 Hint: the linear system \eqalign{ \lambda(- 12 y + 2 x + 20) = F_x' &= 0 \cr \lambda(126 y - 12 x + 60) + 9 =F_y'&= 0\cr } has a unique solution for12\lambda^2 + 16\lambda + 1\ne 0$(why?). What happens when$12\lambda^2 + 16\lambda + 1 = 0$? 1 One thing we need to realize is$z^2$is maximized when$h(x,y,z)=z$is either maximized or minimized. By taking the partial derivatives with respect to$x,y,z$, you have $$2x\lambda+u=0\implies x=-{u\over2\lambda}$$ $$18y\lambda+3u=0\implies y=-{u\over6\lambda}$$ $$-4z\lambda+3u+v=0\implies z={3u+v\over4\lambda}$$ Now sub in to$x+3y+3z=5$and ... 1 By Cauchy-Schwarz, we have $$4z^2=(1+1)(x^2+9y^2) \ge (x+3y)^2 = (5-3z)^2$$ giving $$(z-1)(z-5) \le 0$$ which implies $$1 \le z \le 5$$ $$1 \le z^2 \le 25$$ The equality holds at$(x,y,z)=(1,\frac{1}{3},1)$and$(x,y,z)=(-5,-\frac{5}{3},5)$1 you may be able to do this without the use of lagrange method. let$n_1, n_2$are the unit normal vectors of the two planes. first you can deal with the easier case of$n_1 = \pm n_2.$so wlog we take$n_1 \neq \pm n_2.$suppose that the planes $$\frac{a_0}{\sqrt{a_1^2 + a_2^2 + a_3^2}}+ x\cdot n_1\, = 0, \frac{b_0}{\sqrt{b_1^2 + b_2^2 + b_3^2}} + x \cdot ... 1 I would like leave a comment. Being a beginner in StackExchange, my reputation is too low to do so. You may solve the equations so that one variable is eliminated and parametrize the solution. As the distance between (0,0,0) and the solution set is given by Pythagoras theorem. you may find the required point by completing square or differentiation. 1 I would like to give you a hint because it is pretty easy. First of all, you need to find the intersection of the two planes (a line). Because you have 2 equations with 3 variables, you can reduce it to 1 equation with 2 variables. Do not forget to impose conditions to make sure two planes do not parallel each other. Secondly, when you have a line in which ... 1 You have to find your bounds for d\theta and then find the bounds for dr as a function of \theta. The angle of your triangular region changes from 0 to \arctan(\sqrt{3}) = \pi/3, and your r changes from 0 to the line x = \sqrt{3}, or r = { \sqrt{3} \over \cos \theta}, so the iterated integral should be \\ \int _0 ^{\pi /3} \int _0 ... 1 In a general manner, consider \big( f_n \big)_{n \in \mathbb{N}} a series of continuous functions such that f_n \to f uniformly in a compact interval [a,b] . In your example, $$f_n = \sum\limits_{i = 1}^n F_n(x)$$ represent the partial sums with f_n(x) \to \sum\limits_{i = 1}^{\infty} F_n (x) = F(x). We want to prove ... 0 Rewrite your function as$$2x^2-3y^2-2x = -3(x^2+y^2) + 5x^2-2x = -3(x^2+y^2) + 5\left(x^2-\frac 25x+\frac{1}{25}\right)-\frac{1}{5} = -3(x^2+y^2) + 5\left(x -\frac{1}{ 5}\right)^2-\frac{1}{5}.$$Let the variable r=\sqrt{x^2+y^2}, then you obtain$$F=-3r^2 + 5\left(x -\frac{1}{ 5}\right)^2-\frac{1}{5}$$with r\in[0,1] and x\in [-r,r]. If you want ... 1 Your function F is defined on a disc of radius one, which is a compact set. Therefore, the maximums and minimums are either on the boundary of the disc, or strictly inside the disc. Case 1: suppose they are on the boundary. In this case, rewrite your problem in polar coordinates as follows:$$ F(r,t)=2r^2\cos^2(t)-3r^2\sin^2(t)-2r\cos(t), \quad ... 0 i think this can be done without using lagrange multipliers. here is a way to do this using basic geometry. suppose the point$P(a,b)$on the ellipse$x^2 +4y^2 = 4$is either the least or greatest distance from the line$x+y=4.$then the slope of the tangent to the ellipse is$-1.$that is $$\frac{dy}{dx}\Big|_{(a,b)} = -\frac a b \text{ and } a^2 + 4b^2 ... 1 Any point on the ellipse can be represented as (2\cos t,\sin t) The perpendicular distance$$=\dfrac{|2\cos t+\sin t-4|}{\sqrt2}$$Now 2\cos t+\sin t=\sqrt5\cos\left(t-\arccos\dfrac2{\sqrt5}\right) \implies-\sqrt5\le2\cos t+\sin t\le\sqrt5 \implies-\sqrt5-4\le2\cos t+\sin t-4\le\sqrt5-4 \implies maximum distance =\dfrac{\sqrt5+4}{\sqrt2} ... 2 First of all, all norms on \Bbb R^{n} are equivalent (do you know what that means?). As a consequence, if a function is continuous with respect to one norm, it is continuous with respect to every norm, so you can just pick the norm that makes the proof of continuity easiest based on the given function. Also, just as we prove continuity in the ... 0 The concept that you are thinking of is sequence continuity (which is equivalent to continuity in metric spaces such as R^n but that requires proof). If you check the condition "for every given open set containing the image there exists an open set (containing the original point) that is mapped into the given open set" then you have continuity by ... 1$$x+y=kx^2+4y^2=(k-y)^2+4y^2=5y^2-2ky+k^2=45y^2-2ky+k^2-4=0D=k^2-5(k^2-4)=20-4k^2\geq0k^2\leq5-\sqrt5\leq k\leq\sqrt5x+y=4 \quad vs.\quad x+y=\sqrt5\text{Distance between above two lines }=\frac{4-\sqrt5}{\sqrt2}=2\sqrt2-\frac{\sqrt10}{2}$$0 HINT: Any point on the straight line can be represented as (a,4-a) As the distance will vary with its square, check with$$f(x,y)=(x-a)^2+\{y-(4-a)\}^2-\lambda(x^2+4y^2-4)$$1 You need to parametrize the surface S, not the curve. You can proceed as follows: Since the surface is part of the cone z=\sqrt{x^2+y^2}, might as well use$$ x=x, \quad y=y, \quad z=\sqrt{x^2+y^2},$$with$x,y \in D:=\{(x,y)\; |\; x^2 +y^2 \le ax \}$. Next, compute,$\|r_x \times r_y \|=\sqrt{2}\$. You now have everything you need. The mass is ...

1

Differential forms are an appropriate generalisation of derivation and integration (i.e of calculus) for arbitary (within the scope of integration) manifolds and spaces. In order to arrive at such a generalisation (e.g like E. Cartan did) one starts with the basic definitions and operations of derivation and integration, how they affect the space they are ...

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