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Jumping over a log

I vaguely remember seeing the following in a mathematics contest:

There is a cylindrical log of radius R lying on its side on the ground.
An ant is trying go from one side of the log to the other in one jump.
How should the ant jump (from what position relative to the log, and at
what speed and angle) in order to do this with the minimum speed?
Use  $V\,$  for initial velocity (and  $V_{x}\,$  and  $V_{y}\,$  for its component vectors)
Use  $v\,$  for the velocity variable
Use  $g\,$  for the gravitational constant.

My approach: As I understand the problem, it can be reduced to two dimensions, as the ant is merely jumping over a circular cross-section of the log. So basically superimpose a negative parabolic trajectory over a circular log, with two symmetric points of intersection, come up with a cartesian coordinate system, and try to minimize $V={\sqrt {V_{x}^{2}+V_{y}^{2}}}$ .

For a parabolic trajectory whose midpoint is on the y-axis and whose start and end points are on the x-axis, the formula (derivation left out) is $y={\frac {-g}{2V_{x}^{2}}}x^{2}+{\frac {V_{y}^{2}}{2g}}$ .

The formula for the circle, with center $(0,R)\,$ and radius $R\,$ is $x^{2}+(y-R)^{2}=R^{2}\,$ , simplified as $x^{2}+y^{2}=2ry\,$ .

However, it gets extremely messy when I attempt combine these two equations, to the point that I cannot continue, so I suspect this approach is not the best way to solve the problem. Does anybody know the right way to solve this problem? JianLi 01:19, 29 July 2006 (UTC)[reply]

I'm not well-versed on the mathematics involved, but you might be able to construct the parabola by treating the log as an osculating circle. Then you'd get a formula for the parabola, and you could find the velocity and angle from that. -Maelin 03:38, 29 July 2006 (UTC)[reply]

If the osculating circle has radius 1, the parabola is given by 2−x²; but this does not give a satisfactory solution. The parabola does graze the circle at the top, but from the inside. For example, at x = 1⁄2 the parabola height is 1.75, but the sphere height is about 1.866. Instead, the circle must nestle inside the parabola, most likely touching the sides tangentially. --KSmrq^T 06:03, 29 July 2006 (UTC)[reply]

Not 2 - (1/2)x^2? Melchoir 07:03, 29 July 2006 (UTC)[reply]

Yeeps, how did I lose the 1/2? And it changes the conclusion as well, because it appears the circle fits inside nicely, with the parabola crossing zero at 2 and −2. So that someone can triple check (!) my computations, the standard formula for curvature of a parametric curve (x(t),y(t)) is

\kappa =\left|{\frac {x'y''-y'x''}{((x')^{2}+(y')^{2})^{3/2}}}\right|.

Letting x = t and y = 2−a·t², with a > 0, we obtain

\kappa (t)={\frac {2a}{(1+(2at)^{2})^{3/2}}}.

At t = 0 this reduces to 2a, which we set equal to 1. Thus a = 1⁄2, and the desired parabola would appear to be 2−1⁄2x². (The additive constant 2 is the height of the circle, and parabola, at zero.) At −2, the left jump-off point, the tangent vector is (1,2).

Another check is the Maclaurin series of the upper semicircle,

1+{\sqrt {1-x^{2}}}=2-{\frac {x^{2}}{2}}-{\frac {x^{4}}{8}}-{\frac {x^{6}}{16}}-{\frac {5x^{8}}{128}}+\cdots .

Since the higher-order terms are all negative, we have additional corroboration that the circle remains under the correct parabola.

Sorry for the slip, and thanks to Melchoir for covering my back. --KSmrq^T 16:18, 29 July 2006 (UTC)[reply]

Sure, my own reasoning is that the parabola's curvature is greatest at its apex, so away from that it "opens out" and stays outside its osculating circle, which of course has constant curvature. The German osculating circle article has a picture, while our own article could use some attention. (I'm not necessarily advocating this as a solution to the original question.) Melchoir 19:15, 29 July 2006 (UTC)[reply]

Now that we have osculation, perhaps it's worth reminding ourselves that we have not converted this to jump parameters, nor shown that this is the minimum speed solution. If we jump from farther away we will need more horizontal speed, but no less vertical speed, so we're fine there. But if we jump from closer, we'll need more vertical speed, but also less horizontal speed, so perhaps we can do better. We might consider a calculus of variations approach, but surely algebra will suffice.

Incidentally, we're ignoring all fluid dynamics of the air, though that would likely affect a real ant. Not only is the ant small enough that we'll have a high Reynolds number, but also the log will have a boundary layer on the scale of the ant.

Finally, most ants are not great jumpers, though Harpegnathos saltator seems to be a notable exception. Unfortunately, I have been unable to determine anything quantitative about its abilities. --KSmrq^T 00:06, 30 July 2006 (UTC)[reply]

Thanks for your thorough and interesting answer :) JianLi 02:34, 30 July 2006 (UTC)[reply]

Can anybody prove conclusively that this is the least-velocity answer? JianLi 17:11, 30 July 2006 (UTC)[reply]

Consider any point on the parabola, take the line perpendicular to the tangent, its intersection point with the symmetry axis must have equal distances to the parabola and the ground. This gives an equation having the solution r = d · (1 − 1/m) where d is half the total jumping distance and m > 2 the starting slope. For starting velocity v, we have d = v²/g · m/(1 + m²). The numerator of the derivative in question is 2 − (m − 1)², hence the maximum should be at m = 1 + √2. We excluded m < 2, but that's obviously not optimal according to the osculating circle argument. Hope I didn't make any stupid mistakes.--gwaihir 18:42, 30 July 2006 (UTC)[reply]

You say consider any point on the parabola; but your first assertion requires a point where the parabola is tangent to the circle. Assuming a radius of 1, the contact coordinates are (x,1+√(1−x²)). I'm not sure how this relates to your unstated equation solved for r, whatever r is.--KSmrq^T 23:32, 30 July 2006 (UTC)[reply]

Take an equation for the parabola, I used

y=h-x^{2}/(2b)

. For any point

(x_{0},y_{0})

on the parabola, we can compute the intersection point of the line perpendicular to the tangent line in

(x_{0},y_{0})

and the

y

-axis, and I get

(0,y_{0}-b)

. Now

(x_{0},y_{0})

is a point where the parabola is tangent to a circle with center on the

y

-axis, tangent to the

x

-axis, iff our intersection point has equal distances to

(x_{0},y_{0})

and to the

x

-axis, i.e., iff

r=y_{0}-b

and

x_{0}^{2}+(y_{0}-r)^{2}=r^{2}

with some number

r

which happens to be the radius of the circle in question. Putting all this together, you can eliminate

x_{0}

and

y_{0}

from the system of equations to get a single equation for the radius

r

, depending only on

h

and

b

, or, equivalently, on

d

and

m

.--gwaihir 00:05, 31 July 2006 (UTC)[reply]

Use the symmetry: the ant movement on the ascending part of the trajectory is symmetrical to that on the falling part. So it's enough to calculate the latter part, and apply the landing conditions symmetrically to the starting point.
The symmetry 'axis' is the circle's vertical diameter. Assume the ant is already on top of the cylinder and ask what is the minimum horizontal velocity $V_{x}$ it must start with, to separate immediately from the cylinder's surface?
(Some calculations removed by author --CiaPan 06:02, 3 August 2006 (UTC) )[reply]

V_{x}\geq {\sqrt {Rg}}

(More calculations removed by author --CiaPan 06:02, 3 August 2006 (UTC) )[reply]
The start/landing sites are where $y=0$ , which gives $x=\pm 2R$ .
The starting/landing angle is given by (removed) $\tan \alpha =\mp 2$ which means the angle about $\alpha \approx \mp 63^{\circ }26^{\prime }$ .
The starting/landing velocity is $V=\quad (removed)\quad ={\sqrt {5Rg}}.$
--CiaPan 07:17, 31 July 2006 (UTC)[reply]

You assume that in the optimal case, the parabola and the circle have only one point of tangency. Under this assumption, KSmrq gave the answer already (16:18, 29 July). But as I argued above, this assumption is apparently wrong.--gwaihir 07:53, 31 July 2006 (UTC)[reply]

You're right. However it was not just an assumption, but rather the result of an error in calculations (which I did not put here). After your note I made the same analysis other ways, taking the parabola apex height, the tangency point height and once more the horizontal velocity as an independent variable. They gave different results, so I checked them twice, and found errors. Thank you. --CiaPan 06:02, 3 August 2006 (UTC)[reply]

OK, I'll take a stab at this. As has been stated, the case of one point of tangency has been covered, and we know the optimal subsolution (the minimum enclosing parabola tangent at its vertex), so let's consider the other case: a parabola with height h and half-width d (so $y=h(1-(x/d)^{2})$ ). We want this to be tangent to the circle; we can accomplish this by finding its intersections with the circle and demanding that we have a double root (the limit of glancing incidence is two coincident intersections). So, with $e:=d^{2}$ and $z:=x^{2}/e$ , we have $ez+(h(1-z)-R)^{2}=R^{2}$ , or $h^{2}z^{2}+(e-2h^{2}+2Rh)z+(h^{2}-2Rh)=0$ . There are two conditions to satisfy: the double root, and that the resulting $z^{2}$ be nonnegative. The double root implies $b^{2}-4ac=-4h^{3}(h-2R)+(e-2h^{2}+2Rh)^{2}=0$ ; with that, $z^{2}\geq 0$ implies ${\frac {-b}{2a}}>0$ or (since $a\equiv h^{2}>0$ ) $e-2h(h-R)<0$ , so $e<2h(h-R)$ . Proceeding with the double root condition, we observe a quadratic in e: $e\in \left\{2\left(h(h-R)\pm {\sqrt {h^{4}-2h^{3}R}}\right)\right\}$ . Using the solution with the square root positive would violate the positivity condition, so we have $e(h)=2\left(h(h-R)-{\sqrt {h^{4}-2h^{3}R}}\right)$ . Now from conservation of energy $V_{y}(h)={\sqrt {2gh}}$ , and the time of flight to the apex is just $V_{y}/g$ , so $V_{x}(h)={\frac {dg}{V_{y}(h)}}={\sqrt {\frac {e(h)g}{2h}}}$ . From here it's trivial: compute $V^{2}(h)=V_{x}^{2}(h)+V_{y}^{2}(h)=g\left({\frac {e(h)}{2h}}+2h\right)$ and find ${\frac {dV}{dh}}=0$ , giving ${\frac {he'(h)-e(h)}{2h^{2}}}+2=0$ . Lots of algebra later, a cubic in h appears, but it has the (obviously specious) root $h=0$ , so you get one more quadratic and you're done with h; I get $h=R\left(1+{\frac {3{\sqrt {2}}}{4}}\right)$ . From there one can trivially calculate the initial velocity and angle, of course: evaluate $e(h)$ and $V_{y}$ , then $V_{x}$ from those and convert to polar coordinates. The important thing to note here is that since g dropped out, the problem is scale-invariant, and we must have a $h\propto R$ with a single proportionality constant for all problems (and all g). (Aside: is there a way to tell a priori that the problem was scale-invariant?) Finally, we note that for $h=2R$ (the single-point tangency case), we have $d={\sqrt {e}}=2R$ , which is the minimum tangent parabola (as required by the continuity of $e(h)$ ; it can't suddenly pick up a suboptimal parabola just because the problem is underdetermined at that one point); this means that we need not consider the one-tangency case separately, and this solution is general and optimal. Hope this helps. --Tardis 22:49, 31 July 2006 (UTC)[reply]

Apart from an obvious typo (it should read

1+{\frac {3{\sqrt {2}}}{4}}

) this agrees with my computation.--gwaihir 23:00, 31 July 2006 (UTC)[reply]

Thanks — fixed. I think I just grabbed the wrong root of the final quadratic, although I don't have my notes to check now. --Tardis 17:01, 1 August 2006 (UTC)[reply]

For comparison to the apex skimming jump, here's what the slow jump looks like. --KSmrq^T 21:19, 1 August 2006 (UTC)[reply]

Probably the simplest solution. Consider a parabola $y=a-bx^{2}$ , and a circle $x^{2}+y^{2}=2ry$ . Find the points of intersection by eliminating $x^{2}$ , that is solving a simple quadratic equation for y, that is

y^{2}-(2r+1/b)y+a/b=0

Its determinant is $D=(2r+1/b)^{2}-4a/b$ . Now the condition D<0 means no points of intersection. D>0 means 3 or 4 points of intersection when the parabola goes through the circle or possibly touching the circle at the apex from inside the circle. D=0 is the condition which we want, that is at most 2 intersections when the parabola touches the circle at two symmetric points or the apex of the circle. Finally the problem becomes to minimize the initial velocity

V_{0}^{2}(a,b)=g\left(2a+{\frac {1}{2b}}\right)\to \min

under the condition

a={\frac {b}{4}}(2r+1/b)^{2}

.

(Igny 16:16, 4 August 2006 (UTC))[reply]

homework help

please can u guys help me out wid this log it is mind boggling

\log _{\sqrt {3}}x+\log _{3}x+\log _{1/3}x=4

then find x i know u dont like hw but please i need help

Tossing it in my calculator, it claims something within ten significant digits of 9.0 solves it. The rules of logarithms should help with this. Break each one into two parts using the change of base formula, then distribute, then solve. Black Carrot 06:16, 29 July 2006 (UTC)[reply]

Here it is:

(answer removed, because poster should follow directions at top of this page and Do [their] own homework.)

—Mets501 (talk) 11:17, 29 July 2006 (UTC)[reply]

Mets501, please never again do homework for someone. Question them, guide them, teach them; that's all fine. But posting answers is not acceptable. Not only does it support cheating, it encourages more abuse. Your understanding and cooperation will be appreciated. --KSmrq^T 01:33, 30 July 2006 (UTC)[reply]

Yes, you're right. My apologies. —Mets501 (talk) 17:05, 30 July 2006 (UTC)[reply]

Make use of the formula

\log _{a}x={\frac {\log _{3}x}{\log _{3}a}}

. The equation will become much simpler. Conscious 09:14, 30 July 2006 (UTC)[reply]

New mathematical operation?

Is there any name for a mathematical operation where you raise a number to its own exponent that number of times? For example operation on 2 would yield 2^(2^2) = 2^4 = 16 and operation on 3 would yield 3^(3^(3^3)) = 3^(3^27) = 3^7625597484987 (I won't bother calculating that). How is this similar to the Ackerman function? And how would it be possible for me to invert the function (find out what number generates say 28).

Try our entry on Tetration. I think it's what you are describing. (ESkog)^(Talk) 05:47, 29 July 2006 (UTC)[reply]

Trinomial

Could anyone factor this for me? x³-5x²+18x-15 Black Carrot 06:18, 29 July 2006 (UTC)[reply]

Irreducible over the rationals. McKay 06:51, 29 July 2006 (UTC)[reply]

If you'll allow for rounded decimals, the roots are 1.954 ± 3.149i and 1.092, so you could write that as a product of 3 terms. But probably that's not what you were after... digfarenough (talk) 15:17, 29 July 2006 (UTC)[reply]

If you're interested in a few more digits of the real root: 1.0923893610465733519. --Lambiam ^Talk 16:15, 29 July 2006 (UTC)[reply]

Why go halfway? The exact real root is

{\frac {1}{3}}\left(5-29{\sqrt[{3}]{\frac {2}{-155+9{\sqrt {1501}}}}}+{\sqrt[{3}]{\frac {-155+9{\sqrt {1501}}}{2}}}\right).\,\!

Find a friend at Category:free computer algebra systems, such as Maxima.

And, by the way, this is not a trinomial; it has four terms, not three. It has degree three, so it can be called a cubic.

You can plot the zero crossing with a Bézier curve whose control points are (−141,−3), (−116,0), (107,3), (129,6) --KSmrq^T 17:25, 29 July 2006 (UTC)[reply]

Integrals and Floor Functions

Is $\int _{0}^{4}\lfloor x\rfloor dx$ a valid integral? Does it matter that $\lfloor x\rfloor$ isn't continuous for integer values of x ?. Intuitively it seems valid because the area under the curve can be calculated, but strictly speaking I'm not so sure. Thanks for your help, --Codeblue87 19:08, 29 July 2006 (UTC)[reply]

Unless you're using some extremely strict definition of an integral, a finite number of discontinuity points isn't a problem (as long as the function is bounded - otherwise it can get slightly tricky). So yes, the integral exists, and is equal, of course, to 6. -- Meni Rosenfeld (talk) 19:24, 29 July 2006 (UTC)[reply]

Thanks! I'm writing a test that involves this concept and I just want to make sure nothing is disputable.--Codeblue87 19:28, 29 July 2006 (UTC)[reply]

Who is this test for? You would probably do well to check what is the exact definition taught to the students. This is definitely not a question for a multiple-choice exam given to students with unknown background. -- Meni Rosenfeld (talk) 19:31, 29 July 2006 (UTC)[reply]

The test doesn't specifically ask the competitors(the test is for a math competition) if the aforementioned integral is valid, but rather to evaluate an example. I'll show you the actual question on your talk page. --Codeblue87 20:10, 29 July 2006 (UTC)[reply]

That reminds me a lot of how we were taught to use calculus in Physics. We were give simple piecewise functions and told to draw their integrals and calculate the area under them. The integral for this would be a horizontal line, then three increasingly steep diagonal lines. Black Carrot 19:43, 29 July 2006 (UTC)[reply]

I've replied at my talk page. -- Meni Rosenfeld (talk) 20:54, 29 July 2006 (UTC)[reply]

Wouldn't this be the same thing as

\sum _{n=0}^{3}n

? -Maelin 02:56, 4 August 2006 (UTC)[reply]

It's equal to it, and the evaluation of the integral involves calculating this sum, but I'm not sure about saying they are "the same thing". If nothing else, the question had meaning when asked about the integral, but would have been meaningless if asked about the sum. So they certainly are different entities. -- Meni Rosenfeld (talk) 04:54, 4 August 2006 (UTC)[reply]

Retinitis Pigmentosa Gene Therapy

Moved to the Science Refence Desk.