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0

Hint:Use $w = u^2$. Then $dw = 2u\ du$ so $$\frac{4}{3 \dot\ 2}\int_{0}^{\frac{1}{2}} w e^{-2w} dw $$ Now use integration by parts. I hope you can take it from here.


1

Let $t=-2u^2$. Then, $u\,du = -\frac{1}{4}dt$ so that \begin{align} \frac{4}{3}\int_0^{1/\surd2}u^3e^{-2u^2}\,du &= \frac{4}{3}\int_0^{1/\surd2}u^2e^{-2u^2}\,u\,du \\ &= \frac{4}{3} \int_0^{-1} \left(-\frac{1}{2}t\right)\,e^t \left(-\frac{1}{4}\right)dt \\ &= \frac{1}{6}\int_0^{-1} t\,e^t \, dt. \end{align} This last integral can be done ...


0

The region contained in the "shadow" of the intersection, being the interior of a cylinder, is a circle centered at $(a/2,0)$ and has radius $a/2$. A circle $(x-h)^2+(y-k)^2=r^2$ is parameterized by (there is more than one way, however) $$\begin{matrix}x=h+r\cos\theta\\ y=k+r\sin\theta\end{matrix}$$ where $0\le \theta < 2\pi$. This particular ...


0

The solution observes that the equation for the cylinder $x^2+y^2=ax$ is equivalent to $$(x-\tfrac{a}{2})^2+y^2=\tfrac{a^2}{4}$$ which should remind you of the equation of a circle (since we are working in three dimensions it is actually the equation of a cylinder; the $z$ variable is free). This is completing the square. The standard parametrization of a ...


0

Combine three standard results: If $x$ appears in an expression multiple times, we can differentiate the expression with respect to $x$ by differentiating with respect to each $x$ individually, holding the others constant, and adding up the results. (This is just the ordinary multivariate chain rule, applied to a function $G(x_1, x_2, \ldots x_n)$ by ...


1

No. You can only conclude that it is positive semidefinite. Take $f(x,y) = x^2$ for a simple example.


2

A nontrivial counterexample is the function $f(x,y) = x^4+y^4$. Clearly, $f$ has a local minimum at $(x,y) = (0,0)$. However, the Hessian at $(0,0)$ is the zero-matrix, which is not positive definite.


0

I know a lot of people who are confused by what a vector equation of a line mean. Even before you sit down to find the vector equation of a line, you must ask what does the equation really mean? It means that given the equation of the line you can tell just by manipulating a parameter the points through which the lines passes. Now how do we uniquely define ...


0

Ok. Now I understand what you're asking... You seem to be looking for a vector lying on the plane tangent to $z=f(x,y)$ at the point $(x,y)=(1,2)$ whose projection is ${\bf u}$. The equation of the tangent plane at $(x,y)=(1,2)$ is $z = 11+15(x-1)+6(y-2)$ since $f(1,2)=11$ and $\nabla f(1,2) = 15{\bf i}+6{\bf j}$. If we move $x$ and $y$ in the ${\bf ...


0

This is an easy application of the method of Lagrange multiplier. Suppose the box has width $a$, height $b$ and length $c$. You need to minimize the surface area $$f(a,b,c)=2ab+2ac+2bc$$ subject to the constraint $$g(a,b,c)=abc=V.$$ We have $$\nabla f=(2b+2c,2a+2c,2a+2b),\quad \nabla g=(bc,ac,ab)$$ and it is easy to see that a solution to $\nabla f=\lambda ...


1

The amount of material used to make the sides of the walls is proportial to the surface area of the box. If $l$ is the length, $w$ is the width, and $h$ is the height, then the volume is $$V = lwh$$ and the surface area (which we'd like to minimize) is $$S = 2(lw+lh+wh)$$ We can reduce the minimization problem to minimizing a function of two variables ...


0

A few years back I decided to have a look at Milnor's lovely proof, and blogged about it here. It's close to impossible to improve on anything that Milnor writes, but it's conceivable that a slightly different arrangement of his proof would clarify certain points. To answer the question: if you look carefully at how $f_t$ is defined, you'll see that it ...


1

If you note that $x\neq 0$ and $y\neq 0$ you can divide the second equation by the third and get: $$\frac{2ye^{2xy}}{2xe^{2xy}} = \frac{2\lambda x}{2\lambda y} \ ,$$ which after simplification says that $x^2=y^2$. Taking the first equation, you can then write $$x^2 + y^2 = 2x^2 = 16$$ This gives $x=\pm 2\sqrt{2}$ and therefore four possible solutions ...


0

I would suggest finding the axes of the ellipse using purely geometry (plus coordinate algebra) ways. This may require some knowledge of how to manipulate bilinear forms but I think most precalc textbooks cover it for conics, so I assume fair game.


0

Hint: You can minimize/maximize $r^2=x^2+y^2$ instead of $r$. You have $r^2=y^2+xy-1$ Use your equation to get an expression for $x$ in terms of $y$ and substitute it in. Take the derivative with respect to $y$, set to zero, etc.


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Answer: Yes it does, and there is a much simpler method using translational symmetry. We begin as before, but then substitute $\zeta=\xi-x$ to remove the $x$-dependence of the boundary: \begin{align} \frac{\partial \hat T_D}{\partial t} &= \frac{1}{m(D)}\int_{x+D} \frac{\partial T}{\partial t}\!(\xi,t)\,d\xi \\ &= \frac{1}{m(D)}\int_{x+D} ...


2

Okay it is ABUNDANTLY clear you are not very confident with notation (as it is awful) so I shall teach you all you need to know about doing stuff on surfaces. Lets tackle this intuitively, I will assume you are happy with the area of a parallelogram being related to the vector cross product, if not it is literally by definition (remember the sin rule for ...


2

Assuming that you have a point source light, Find the equation of the line that passes through the light source and one of the vertexes of the box. Then project this line to the appropriate plane. Repeat for all vertexes, and you have the shadow outline. This is known in computer graphics as stencil shadows. If you want some example math, I can do that. I ...


1

Your euclidean norm is wrong, you most likely forgot about the square root. It should be: $$ \left\| \begin{pmatrix} -\cos\varphi\sin^2\theta\\ -\sin\varphi\sin^2\theta\\ -\sin\theta\cos\theta \end{pmatrix} \right\| = \sqrt{(\cos^2\varphi+\sin^2\varphi)\sin^4\theta + \sin^2\theta\cos^2\theta} = \sqrt{(\sin^2\theta + \cos^2\theta)\sin^2\theta} = ...


2

You are not expected to actually compute these line integrals using the parameterizations; this would be a rather painful procedure. The question appears to be testing your knowledge of the winding number. The curve $\gamma$ has winding number $1$ about the origin, since it travels once counterclockwise. The curve $\alpha$ has winding number $-1$, being ...


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0

You can use $\displaystyle V=\int_0^{2\pi}\int_{\frac{\pi}{2}}^{\frac{3\pi}{4}}\int_0^{\sqrt{6}}\rho^2\sin\phi\; d\rho\; d\phi\; d\theta$


0

You can find it the same way as two lines in 3D. Parameterize your planes by base points $p_0, p_1$ and pairs of vectors $v_0^0, v_0^1, v_1^0, v_1^1.$ You want to find the four parameters $\alpha_i^j$ so that the vector $(p_0 + \alpha_0^0 v_0^0 + \alpha_0^1 v_0^1)-(p_1 + \alpha_1^0v_1^0+\alpha_1^1v_1^1)$ is perpendicular to both planes. This leads to a ...


0

You answer question 2 correctly, the Cauchy estimates show that $\frac{\partial f}{\partial z}$ is uniformly bounded on $\mathbb{R}\times D_r(0)$ for all $r > 0$ such that $\overline{D_r(0)} \subset \mathcal{U}$. To see the continuity of $\frac{\partial f}{\partial z}$, fix a $z_0\in\mathcal{U}$. Then for $r > 0$ small enough that $\overline{D_r(z_0)} ...


1

A functional $F$ is a map that takes functions from an appropriate functional space and returns numbers. If the functional $F$ is represented through an integral like in the OP, then the Lagrangian $f=f(x,y,y')$ is generally seen as a function of the variable $x$ and the functions $y$, $y′$. The Lagrangian can be dependent of $x$ and $y$, i.e. $f=f(x,y)$, or ...


1

You have a function $f(x,y,z)$ presumably, and then you take a composition $h(x,y) = f(x,y,g(x,y))$. The chain rule here is $$ \frac{\partial h}{\partial x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial z}\frac{\partial g}{\partial x} $$ and similarly for $y$. You should verify that on your own, and check a couple examples to convince ...


1

The conversion between cylindrical and Cartesian coordinates is $x = r\cos\theta$, $y = r\sin\theta$, $z = z$. Thus, the density $\rho(x,y,z) = 3-z$ (Cartesian) becomes $\rho(r,\theta,z) = 3-z$ (cylindrical).


0

Normally, just like $$\iiint_\Omega dV = \text{Volume}(\Omega),$$ so too $$\iiint_\Omega \rho(x,y,z) dxdydz = \text{Mass}(Omega),$$ where $\rho$ denotes density and $\Omega$ denotes your region.


2

Convexity implies $f(y)\ge f(x)+\langle \nabla f(x),y-x\rangle$ for all $x,y$. Specializing this to the points of your set, you will find they are points where $f$ attains its global minimum, say $m$. Argue that $\{x:f(x)=m\}$ is convex.


0

The general form of a lagrange multipliers problem is that you want to maximize/minimize an objective function $f(x,y,z)$ with some constraint $g(x,y,z) = 0$. To do this, you have to solve the equations $\nabla f(x,y,z) = \lambda \nabla g(x,y,z)$ and $g(x,y,z) = 0$ to get the possible points $(x,y,z)$ at which a maximum/minimum is obtained. In this ...


4

Main Idea In short, reverse the order of integration. Then it works very simply.


11

Some good $($introductory$)$ sources, in general, for all things smooth manifolds: Topology from the Differentiable Viewpoint, by Milnor Differential Topology, by Guillemin-Pollack Differential Forms and Applications, by Do Carmo A Comprehensive Introduction to Differential Geometry, Vol. 1, by Spivak Introduction to Smooth Manifolds, by Lee Brian Conrad's ...


6

I'll explain the relationship between trivialization of the tangent bundle and having $n$ linearly independent vector fields: A vector field is a smooth map $X: M \rightarrow TM$ such that $X(p) \in TM_p$. That is, it's a smooth assignment of a tangent vector at $p$ to each point $p \in M$. Suppose we have $n = \dim(M)$ vector fields $\{X_i\}_{i=1}^n$ ...


4

Every parallelizable manifold is obviously orientable, hence you get an easy to check obstruction : non-orientable manifolds are not parallelizable. This immediately shows that, for example, all even-dimensional projective spaces $\mathbb P^{2n}(\mathbb R)$ are not parallelizable. Moreover there is a perfect (albeit a bit more advanced) criterion for ...


1

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1

From $f=u+iv$ we obtain $|f|^2=u^2+v^2$, which implies $${\partial\over\partial x}\bigl(|f|^2\bigr)=2uu_x+2vv_x$$ and then $${\partial^2\over\partial x^2}\bigl(|f|^2\bigr)=2(u_x^2+v_x^2)+2uu_{xx}+2vv_{xx}\ .$$ Since $u$ and $v$ both are harmonic (a consequence of the CR equations) we therefore obtain $$\Delta\bigl(|f|^2\bigr)=2(u_x^2+v_x^2+u_y^2+v_y^2)\ ...


0

Let $\nabla f(x,y)=(\frac{\partial f}{\partial x}(x,y),\frac{\partial f}{\partial x}(x,y))$ be gradient of $f$, then: Gradient at point $(1,1)$: $$\nabla f(1,1)=(\frac{\partial f}{\partial x}(1,1),\frac{\partial f}{\partial x}(1,1))$$ Length of gradient at point $(x,y)$: $$\|\nabla f(x,y)\|=((\frac{\partial f}{\partial x}(x,y))^2+(\frac{\partial ...


0

The gradient is a vector-valued function. You evaluate at the point and you obtain a vector. For example, if $f = x^2 + y^2$, then $\nabla f : \mathbb{R}^2 \to \mathbb{R}^2$ is given by $$(\nabla f)(x, y) = \left(\frac{\partial f}{\partial x}(x, y), \frac{\partial f}{\partial y}(x, y)\right) = (2x, 2y).$$ So at the point $(1, 1)$ you obtain the ...


0

When your Hessian determinant is equal to zero, the second partial derivative test is indeterminant. So then you could simply look at the equation or you can develop contours around possible mins and maxs and use Gauss's Theorem to see if there are mins and maxs within them. So essentially, if the Hessian is equal to zero, you are screwed or the question is ...


4

It's because \begin{align}df(x)\xi &= \frac{d}{dt}|_{t = 0} f(x + t\xi)\\ &= \frac{d}{dt}|_{t = 0} (x + t\xi)^T A(x + t\xi)\\ &= \frac{d}{dt}|_{t = 0} (x^T + t\xi^T) A(x + t\xi)\\ &= \frac{d}{dt}|_{t = 0} (x^T + t\xi^T)(Ax + tA\xi)\\ &= \frac{d}{dt}|_{t = 0} (x^TAx + t(\xi^T Ax + x^TA\xi) + t^2\xi^TA\xi)\\ &= \xi^T Ax + x^TA\xi\\ ...


2

There's a nice answer linked by alexjo using coordinates. Here's an answer without coordinates, using the fact that we know the derivative of a linear map: Consider $h(x,y) = x^T A y$, with $h: \mathbb{R}^k \times \mathbb{R}^k \rightarrow \mathbb{R}$. We can restrict $h$ to each factor with $h(x,-) : \{x\} \times \mathbb{R}^k \rightarrow \mathbb{R}$ and ...


0

Assuming that your mapping $$L : \mathbb{R^2}\times\mathbb{R^2} \to \mathbb{R^2}$$ maps $$((x_{11}, x_{12}),(x_{21}, x_{22})) \mapsto (y_1,y_2)$$ the first order derivative should be something like $$ \left( \begin{array}{ccc} \frac{\partial y_1}{\partial x_{11} } & \frac{\partial y_2}{\partial x_{11} } \\ \frac{\partial y_1}{\partial x_{12} } ...


0

If you have two functions of one variable, say $f(x)$ and $g(x)$, then you just write $f(x)=g(x)$ or, equivalently, $f(x)-g(x)=0$ and solve this equation. If $f$ and $g$ are linear functions then it's really easy. If they are quadratic polynomials, you know the formulas for its solutions. In fact there are also some formulas for the case when $f$ and $g$ are ...


0

This is correct, but the difference is that typically you will have level curves. That is, if you define a function $$f:\mathbb R\times\mathbb R \rightarrow \mathbb R$$ by $$f(x,y) = P(x,y)-Q(x,y)$$ then the sets $$S_c\equiv\left\{(x,y) : f(x,y) = c\right\}$$ are the level curves of $f$ (of course, adjust the domain for your particular functions as ...


0

Yes, finding the intersection between $P(x,y)$ and $Q(x,y)$ is done by simply setting $P(x,y) = Q(x,y)$. Note that it won't always be just a point (or a finite number of points) though!


1

Separating your variables appropriately, we have $$\int_0^T \delta(t-t_j)e^{-(s+\mu\lambda^2)t}\,dt\int_0^l \delta(x-R)\varphi(x)\,dx.$$ Since $0 < R < l$, the sifting property of the Dirac delta gives us $\varphi(R)$ for the second integral. The first is very similar since $0 < t_j < T$.


1

To your question about choosing the bounds of integration, it is customary to use $\theta \in [0,\pi]$. It is done because of the issue with integrating the $\sin (\theta)$ as you mentioned. What you are dealing with is the 3D equivalent of integrating a sine-function. If you evaluate: $$\int_0^{2\pi} \sin(\theta)d\theta$$ You obtain: $$\left[ \cos(\theta) ...


1

Well, I guess you compute the first and second fundamental forms: Here are a couple of terms of the first fundamental form: $$ g_{ss} = \| \dot{\gamma(s)} \|^2 \\ g_{st} = \dot{\gamma(s)} \cdot \gamma(s) ... $$ Clearly $X$ is a generally nice map, but if $X_s$ and $X_t$ are parallel, then the derivative is singular...so it can't be an immersion. In ...


4

Going off what you have, $X_t$ and $X_s$ are linearly independent if and only if $\gamma$ and $\dot{\gamma}$ are. This means that $X$ is immersed if and only if the curve is never moving directly toward or away from the distinguished point. Now, to compute curvature, we will find the eigenvectors of the second fundamental form. Notice that $X$ is invariant ...


1

Here is an animation of the surface of interest in three dimensions: Here is a contour plot, looking down from the $z$-axis, clearly showing that there is a relative extremum around $(2,2)$: It is not difficult to show mathematically that this extremum exists and is a relative minimum; and furthermore, that there are no other critical points.



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