# Tag Info

5

In polar coordinates: $$r^4\cos^4t+r^2\cos^2t+r^4\sin^4t+r^2\sin^2t=2$$ $$r^4\left(\cos^4t+\sin^4t\right)+r^2-2=0$$ $$r^2=\frac{-1+\sqrt{1+8\left(\cos^4t+\sin^4t\right)}}{2\left(\cos^4t+\sin^4t\right)}$$ Now the arc length for polar coordinates is $$\int_0^{2\pi}\sqrt{r^2+\left(\frac{dr}{dt}\right)^2}dt$$ So it appears you have a challenging integral to ...

3

Using the polar form, Maple evaluates the integral to 30 digits as $6.68187645290337621429065046080$. The Inverse Symbolic Calculator and Maple's identify function both come up with nothing. I'd guess that the probability of this having a "closed form" solution is rather small.

2

This link in wikipedia explains anything you want about finding envelopes. God bless wiki! :) The solution to this problem is a piece of parabola. The procedure is thoroughly explained in the link I mentioned.

2

Simply ensure that all three points are on the plane. A point and two direction vectors are required to define a plane. By stating "head of the three vectors", this implies that the vectors are coming from the origin (with the tail at 0,0,0). The "head" essentially represents a point on the plane.

1

It should just mean through the three points, although it does seem to be a rather bizarre way to phrase it.

1

If you know how to find the normal line to the graph of a function at a point, convert this problem to such a problem. Convert those parametric equations to a function equation $y=f(x)$ by eliminating $t$. Also find the relevant point on the curve $(x_1,y_1)$ by substituting $t=2$ into the parametric equations. Then continue from there. There are other ...

1

If you write your vectors in the basis of eigenvectors $f_1,\ldots,f_k$ of $C$, then if $u=\sum t_jf_j$ we have $$u^TCu=\sum t_j^2\lambda_j,$$ where $\lambda_1,\ldots,\lambda_k$ are the eigenvalues of $C$ (counting multiplicities). So, in each direction, you are stretching the unit circle by $\sum t_j^2\lambda_j$ a convex combination of the eigenvalues. In ...

1

It should be $0$ instead of $4$; to see a why notice: Let $C := \{ (x,y,z) \in \Bbb{R}^{3} \mid 2x^{2}+3y^{2}-z^{2} = 4 \}$; let $f: (x,y,z) \mapsto 2x^{2}+3y^{2}-z^{2}$ on $\Bbb{R}^{3}$. Then $f^{(-1)}\{ 4 \} = C$; but $\nabla f(x,y,z) = (4x, 6y, -2z)$ for all $(x,y,z) \in \Bbb{R}^{3}$, which is a vector normal to $C$ at $(x,y,z)$ for all $(x,y,z) \in C$; ...

1

The equations you get are $$b=ca^k$$ $$m=cka^{k^-1}$$ dividing the first equation by the second yields $$\frac{b}{m}=\frac{a}{k}$$ which can be solved for $k$, and you can then solve for c.

1

You can remember that $\vec a\times\vec b$ is perpendicular to both $\vec a$ and $\vec b$, so that $\vec n_0\times\vec n_1$ gives the direction of the intersection line. In addition, you know that both planes pass through $(0,0,0)$ and so does the intersection line.

1

I don't think calculating the angle between two lines will help you to find the equation of the line of intersection of two lines. It should be clear that the line of intersection is the line which is perpendicular to the normal of both the given planes. Then the line will be along the cross product of the normal vector of both the planes. $\therefore$ ...

1

$\newcommand{\Cpx}{\mathbf{C}}$Let $\alpha:[a,b] \to \Cpx$ be a continuous curve (such as a curve of class $C^{1}$), and let $\{V_{c}\}_{c \in C}$ be an arbitrary (finite or not) covering of the image $\alpha([a,b])$ by open subsets of $\Cpx$. By continuity of $\alpha$, each preimage $U_{c} = \alpha^{-1}(V_{c}) \subset [a, b]$ is (relatively) open, and the ...

1

Suppose $K$ is a compact subset of a metric space $X$ and $\{V_\alpha\}$ is an open cover of $K$ in $X.$ Then there exists $\epsilon>0$ such that if $E\subset K$ and $\text {diam } E < \epsilon,$ then $E$ is contained in some $V_\alpha.$ Proof: Suppose not. Then for each $n \in \mathbb {N},$ there exists $E_n \subset K$ with \$\text {diam } E_n < ...

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