Wikipedia:Reference desk/Archives/Mathematics/2012 November 6

= November 6 =

algebra inequality
If

$$\sqrt{N-1}+(N-1)\sqrt{K-1}<(N-1)\sqrt{K}$$

then does this mean that $$K<\frac{N^2}{4(N-1)}$$ ?

I am reading that this is the case but I am having trouble getting from a to b. Is there some inequality trick involved? Thorstein90 (talk) 03:43, 6 November 2012 (UTC)


 * Do we know anything about the possible values for N and K ? Specifically, can they be 1 or less ? StuRat (talk) 02:51, 6 November 2012 (UTC)


 * N and K are integers greater than or equal to 2 Thorstein90 (talk) 03:43, 6 November 2012 (UTC)


 * Try squaring both sides, and work from there. I got a different answer, but I might have made a mistake in the algebra. Looie496 (talk) 03:35, 6 November 2012 (UTC)


 * That's what I've tried doing a few times but it never comes out right. That's why I'm thinking it might involve something like Jensen's inequality... or perhaps the answer is wrong? Thorstein90 (talk) 04:17, 6 November 2012 (UTC)
 * There is no trick. Square both sides to get an inequality which simplifies to $$2\sqrt{(K-1)(N-1)}<N-2$$, then square it again.—Emil J. 12:42, 6 November 2012 (UTC)

Derivative word problems
Hello! I have three problems that I need help on:

1) A rectangle has its base on the x-axis and its upper two vertices on the parabola y = 12-x^2. What is the largest area the rectangle can have, and what are its dimensions?

2) A 500ft^3 tank is to be made by welding thin stainless steel plates together using as little weight as possible. How do I use derivatives to solve this?

3) A 1125-ft^3 open top rectangular tank with a square base x ft on a side and y feet deep is to be built with its top flush with the ground to catch runoff water. The costs associated with the tank involve not only the material from which the tank is made but also an excavation charge proportional to the product xy:


 * If the total cost is C = 5(x^2 + 4xy), what values of x and y will minimize it?

If you could offer any assistance, I would greatly appreciate it. --Colonel House (talk) 01:20, 6 November 2012 (UTC)


 * I fixed your numbering (auto-numbering doesn't work with blank lines). StuRat (talk) 01:55, 6 November 2012 (UTC)


 * 1) Start by coming up with a formula for the area. Let's define one vertex on the X-axis as (j,0).  Since it's going to be symmetric about the Y -axis, that gives us (-j,0) for the other vertex on the X-axis.  The width, therefore, is 2j.  Plugging either coord in to the parabola equation, we get (±j,12-j2) for the upper coords, with a height for the rectangle of 12-j2.  Multiplying by the width, we get an area of 24j-2j3.  Now that you have an area formula, you can maximize for that by varying j.  Specifically, find the derivative of the area formula and set it equal to zero, then solve for j.  Next plug that value of j back into the area formula to find the answer.  As a check I suggest trying values for j which are .01 more or less than the optimal value.  When you plug those into the area formula, you should get a slightly smaller area.  Show your work here and we will help with any problems. StuRat (talk) 02:07, 6 November 2012 (UTC)


 * (2) doesn't tell us enough about the constraints. —Tamfang (talk) 02:29, 6 November 2012 (UTC)


 * Yea, I'm thinking it will be a cube, and they want us to prove it, but it is confusing, as stated. StuRat (talk) 02:45, 6 November 2012 (UTC)


 * Without any relevant information to the contrary, one can probably assume that the steel is of uniform density and thickness and can be cut in any shape/size. Disregarding those factors would make this a surface area-minimizing problem (yielding a cube, undoubtedly). -- Kinu  t/c 02:58, 6 November 2012 (UTC)


 * Addendum: assuming it is a closed-top tank and the problem as worded constrains it to a rectangular shape. -- Kinu  t/c 18:04, 6 November 2012 (UTC)


 * A cube? Undoubtedly...? Assuming we can cut the steel sheets in &apos;any shape and size &apos; pieces, we can make a tank spherical (well, at least as close to spherical as we want). See Eindhoven Water Towers or San Juanico disaster. --CiaPan (talk) 06:53, 8 November 2012 (UTC)


 * 3) Find the equation for the volume: x2y = 1125. Solve for y: y = 1125/x2.  Now plug that into the cost equation: C = 5(x2 + 4xy) = 5(x2 + 4x(1125/x2)) = 5(x2 + 4500/x) = 5x2 + 22500/x = 5x2 + 22500x-1.  As before, find the derivative, set it equal to zero, and solve for x.  From this you can use y = 1125/x2 to find y, then plug x and y into the original cost equation.  Check, as before, by trying values of x which are .01 higher or lower.  Find y for each case, then plug that x and y into the original cost equation.  The cost should be slightly higher, in each case.  StuRat (talk) 02:42, 6 November 2012 (UTC)


 * Thank you so much everyone! I am working on another assignment now, but will post any work if I need more help. Thanks again!--Colonel House (talk) 02:51, 6 November 2012 (UTC)


 * You're quite welcome. StuRat (talk) 03:12, 6 November 2012 (UTC)

The following discussion is regarding problem 2:
Why is a cube better than something approximating a sphere? (e.g. a dodecahedron). (not op) --91.120.48.242 (talk) 07:07, 6 November 2012 (UTC)


 * It isn't, but the differences between radically different shapes don't invite solutions involving derivatives. —Tamfang (talk) 08:18, 6 November 2012 (UTC)


 * If we assume unit volumes the surface areas of the Platonic solid are tetrahedron: 7.21, cube: 6, octahedron: 5.72, dodecahedron: 5.31, icosahedron: 5.14. By comparison the sphere is 4.83. Beyond that geodesic domes do a good job at minimising surface area yet still being constructible from flat faces.--Salix (talk): 08:33, 6 November 2012 (UTC)

This tank too is constructible from sheets of steel plate.


 * {(x,y,z) : x2+y2≤1, y2+z2≤1, z2+x2≤1}

What is the volume and surface area? Bo Jacoby (talk) 12:11, 6 November 2012 (UTC).


 * A sphere makes sense, but given that this problem as worded is similar to what is seen in an introductory calculus course, it most likely indicates that such an object should be a rectangular prism in shape based on information that might be omitted in the OP's question. Indeed, these three problems appear to be paraphrased from #6, 11, and 12 from section 4.4 of Calculus by Finney et al.; the "steel plates" one states that it is square-based. Having the problems exactly as worded would be helpful. -- Kinu  t/c 18:02, 6 November 2012 (UTC)


 * Thanks for finding it. Can you write to exact question here, please ? StuRat (talk) 19:08, 6 November 2012 (UTC)
 * woot* I solved that one differently. Not sure if it's rigorous, though.
 * I tried the simpler problem to construct a box-shaped 500 cu.ft. tank of width x, length y, and given height z. Because z is given, this amounts to minimise xy+xz+yz while xy=500/z. Unsurprisingly, x and y are equal.
 * Now I could have investigated what the optimal hieght is if z is not given, but I reasoned in the same way that x=z and y=z, thus, x^3=y^3=z^3=500.
 * I hope my contribution is helpful rather than confusing. Headache. - ¡Ouch! (hurt me / more pain) 14:54, 12 November 2012 (UTC)

integration problem
if I = integral from 0 to pi of " log(1 + cosα .Cosx ) / cosx " with respect to x ,then how we can find I. I have started as by differentiating both sides with α ,but further i find no way to proceed. — Preceding unsigned comment added by 182.187.101.249 (talk) 11:15, 6 November 2012 (UTC)


 * Try . Bo Jacoby (talk) 12:21, 6 November 2012 (UTC).


 * So you have
 * $$\frac{dI}{d\alpha} = -\sin(\alpha) \ \int_0^{\pi} \frac {dx}{1 + \cos(\alpha) \cos(x)}$$
 * Solve the integral on the right hand side ...
 * $$\int \frac {dx}{1 + \cos(\alpha) \cos(x)} = \frac{\arctan\left(\frac{2 \sin(\alpha) (1 - \cos(\alpha)) \tan \frac{x}{2}}{\sin^2(\alpha) - (1 - \cos(\alpha))^2 \tan^2 \frac{x}{2}}\right)}{\sin(\alpha)}$$
 * Now this expression is only piecewise true, because it has a discontinuity when the denominator inside the arctan is zero. That is when
 * $$\tan \frac{x}{2} = \frac{\sin(\alpha)}{1 - \cos(\alpha)}$$
 * So the right value for the definite integral is (value at x=&pi;) - (value at discontinuity, limit from above) + (value at discontinuity, limit from below) - (value at x=0)
 * The values at x=0 and x=&pi; are 0 (in the limit), so all that remains is
 * $$\int_0^{\pi} \frac {dx}{1 + \cos(\alpha) \cos(x)} = \frac{1}{\sin(\alpha)} \left(\lim_{x \to \frac{\sin(\alpha)}{1 - \cos(\alpha)}^+}\arctan\left(\frac{2 \sin(\alpha) (1 - \cos(\alpha)) \tan \frac{x}{2}}{\sin^2(\alpha) - (1 - \cos(\alpha))^2 \tan^2 \frac{x}{2}}\right) - \lim_{x \to \frac{\sin(\alpha)}{1 - \cos(\alpha)}^-}\arctan\left(\frac{2 \sin(\alpha) (1 - \cos(\alpha)) \tan \frac{x}{2}}{\sin^2(\alpha) - (1 - \cos(\alpha))^2 \tan^2 \frac{x}{2}}\right)\right)$$
 * That's just an arctan at +&infin; and one at -&infin;, so substituting the result back into the derivative of the integral leaves us with
 * $$\frac{dI}{d\alpha} = -\pi$$
 * Therefore, the answer to the original problem is
 * $$I = - \alpha \pi + c$$
 * for some constant constant c. We can determine c by calculating the integral at a value of &alpha; that lets us do so easily, e.g. $$\alpha = \frac{\pi}{2}$$:
 * $$\int_0^{\pi} dx \frac {\log(1 + 0 \cdot \cos(x))}{\cos(x)} = 0$$
 * thus $$c = \frac{\pi^2}{2}$$
 * $$I = \frac{\pi^2}{2} - \alpha \pi$$
 * (for a limited range of &alpha;)
 * Icek (talk) 14:41, 6 November 2012 (UTC)

Semi-protection statistics
I'm not sure if this is the correct reference desk for this question, or if this belongs in the Computing reference desk, but here it goes. Of all semi-protected articles:

1. How many are BLPs? Also, what percentage of semi-protected articles are BLPs?

2. Of those BLPs, how many are semi-protected indefinitely? Again, what percentage of semi-protected BLPs are semi-protected indefinitely?

3. Of BLPs semi-protected for long periods (>1 month), how many articles are only available on English? - that is, those that do not have articles on the German, French, Italian, Spanish etc. Wikipedias?

Thanks. Narutolovehinata5 tccsdnew 13:44, 6 November 2012 (UTC)
 * You'll want WP:DBR, probably. --Izno (talk) 22:28, 11 November 2012 (UTC)

File:Sinusoidal_spirals.svg
I've just realised that the lower graph (rectangular coordinates) of several sinusoidal spirals should have the n = -2 and n = 2 curves also reflected about the x-axis (i.e. have both positive and negative r values). Is that correct?

If so, am I correct to conclude that all graphs with even (including negative) values of n have this property? Thanks, cm&#610;&#671;ee&#9742;&#9993; 19:25, 6 November 2012 (UTC)


 * One may interpret it this way, but it could be quite inconvenient in some cases. For some curves, like a hyperbola with its two branches, it would make r(θ) a multi-function, having one positive and one negative value of r for the same theta. Even a circle r(θ) = R would become r(θ) = (R or &minus;R). --CiaPan (talk) 08:03, 8 November 2012 (UTC)


 * Thanks, CiaPan. My question was probably unclear, as I wasn't referring to a function r = f(θ) where r has only one value for any θ. What I meant was basically this question:


 * If yn = f(x), is the graph of y vs x symmetrical about the x-axis if and only if n is even, for any function f except f(x) = 0?


 * cm&#610;&#671;ee&#9742;&#9993; 19:24, 8 November 2012 (UTC)