Wikipedia:Reference desk/Archives/Mathematics/2011 November 16

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November 16

What is the pre-fix to indicate you have 1 & 1/2 of something?

I know in genetics, that they use terms like haploid for 1/2, and diploid for 2, and monoploid for 1; however, what theoretically might I call something that is 1.5-ploid? 완젬스 (talk) 03:19, 16 November 2011 (UTC)

I don't know much about genetics but ploidy appears to have different definitons and only allow natural numbers, and it seems those terms are only for use in genetics. More generally (but rarely used), number prefix mentions sesqui- at 1½. PrimeHunter (talk) 03:31, 16 November 2011 (UTC)

Yes! Sesqui- thank you very much. It's actually a question I had regarding what is halfway in between octofinals and double-octofinals [at this article] and it make me deathly curious what therefore I might call a "Round of 24" so by my reasoning, I can call it sesqui-octofinals, lol. This word will impress my Korean friends who are not good at English--I will make them memorize my new word on a flashcard as part of a flash-card series I'm helping my younger brother learn this "difficult" language. Apparently to him, it's a word as common as building or car is for us! 완젬스 (talk) 04:01, 16 November 2011 (UTC)

The premise here is wrong -- haploid means one copy of each chromosome, diploid means two. The term "monoploid" does not fit on this scale. Diploid is the most common condition, so an organism with 1 1/2 times the usual number of chromosomes is triploid. To my knowledge there is no such thing as an organism that is 1.5-ploid. Looie496 (talk) 17:10, 17 November 2011 (UTC)

Point symmetry - the improper rotation symmetry element

When describing the improper rotation symmetry element (in 3 dimensional Euclidean space) it is always specified that the reflection plane be perpendicular to the rotation axis. What is the reason for this restriction? What about a transformation consisting of a rotation of θ=360°/n (n = a positive integer)combined with a reflection through a plane inclined to the rotation axis by an arbitrary angle α? I am well aware of the fact that for α=0 (i.e. when the rotation axis is in the reflection plane) the combination of the above operations amounts to a simple reflection but what about 0<α<90°? Can this be a symmetry element?Shlomo Levinger (talk) 11:00, 16 November 2011 (UTC)

Any rotation (improper or otherwise) that isn't the identity or a simple reflection sends a unique line to itself. For proper rotations, this is the axis of rotation and the points are fixed by the rotation. For improper rotations, the points are flipped about the origin. It is this line that allows an improper rotation to be decomposed into a proper rotation about the line followed by a reflection that flips the line over (i.e., in a plane perpendicular to the line). For your second question, if the rotation and reflection are not correlated, then they still describe a symmetry element. In fact, any symmetry element can be described in an infinity of ways in this case. But every symmetry element has a unique normal form, in which it is given as a rotation followed (if necessary) by a reflection in the plane of that rotation. Sławomir Biały (talk) 12:28, 16 November 2011 (UTC)

Thank you Sławomir for clarifying the uniqueness of a reflection in the plane of rotation. Still, one point bothers me here. The presence of a rotation-reflection symmetry element (with the mirror plane perpendicular to the rotaion axis), Sn, of an even order (n is even) does not imply the presence of a simple plane of symmetry for a 3-dimensional object. Can this be claimed in the case of an object possessing an oblique rotation-reflection symmetry element (i.e. where the mirror plane is inclined to the rotation axis)? My "intuition" is that such a symmetry element, if at all present, will necessarily imply either the presence of a simple plane of symmetry or that of another rotation-reflection symmetry element having the mirror plane perpendicular to the rotaion axis. This would mean that such an oblique element is not an "independent" one. However, I can not yet prove this. Your comment will be appreciated.Shlomo Levinger (talk) 11:05, 17 November 2011 (UTC)

If you find the mid points between original points and where they go to that will define a subspace, here a plane unless it is just a point. That's the reflection plane. Dmcq (talk) 12:20, 17 November 2011 (UTC)

You can prove that all such improper rotations can be described in terms of a rotation in and reflection in the same plane using geometric algebra. Something like this:

Suppose your rotation is in a plane with bivector B through angle θ. Then the rotation is given by the rotor

$R=e^{\mathbf {B} \theta /2}$

The important thing is this is an even element of the algebra so has a scalar and bivector part. It acts on a vector to rotate it like so

$\mathbf {v} '=R\mathbf {v} R^{-1}$

where all products are the geometric product. The reflection is generated by the vector in the direction of the reflection so perpendicular to the plane of reflection, n say, as follows

$\mathbf {v} ''=-\mathbf {nv'n}$

The rotation and reflection together are then

$\mathbf {v} ''=-\mathbf {n} R\mathbf {v} R^{-1}\mathbf {n}$

The quantity nR, the product of a vector and a rotor, is an odd element of the algebra and it describes the improper rotation. What's more the product of an orthogonal vector and bivector produce a particular result. If n is perpendicular to the plane of rotation then it is orthogonal to B and so to the bivector part of the rotor. It's product with that produces a trivector/pseudoscalar. The other part of the product, with the scalar part of the rotor, produces a vector parallel to n.

More generally if your planes of rotation and reflection are not parallel the products are still composed the same way (though order matters in such a case). As n is a vector and R is a rotor nR is still an odd element of the algebra, with a vector an trivector part.

The resulting improper rotation can be read off this product. The vector part is parallel to the reflection vector, i.e. the normal vector to the reflection/rotation plane. And the angle of rotation in that plane is given by the ratio of the trivector and vector parts. If there's no vector part then you have the central inversion which has any plane of rotation/reflection. If there's no trivector part then it's a pure reflection with rotation angle zero.--JohnBlackburne^words_deeds 13:17, 17 November 2011 (UTC)

shortest connections graph

working in n dimensions, and with y nodes, i'm trying to find a sequence that will tell you how many graphs are possible if its built according to these rules.

1)each node only connects to the closest node/s 2)it is possible to go from each node to any other node following the connectors i.e. the nodes are all connected to each other 3) a graph is considored distinct if nodes connect to different nodes, but not if it's of a different shape, or has a longer line somewhere.

in 1 dimension, only 1 graph is possible. in 2, then for each extra node it goes 1,1,2,5,12... (i think) and in three 1,1,2,6,17... (again, i think). in 4, instead of 17, its 18. is this possible? has it been done before?109.148.193.179 (talk) 21:04, 16 November 2011 (UTC)

Rule 1 needs specification. If each node can ONLY connect to the closest nodes, then assume I have three nodes, A, B, and C. A and B are 1 unit of distance from each other. C is 2 units from B and 3 units from C. So, A and B connect. That is a given. C will want to connect to B, but B cannot connect to C. So, how is this handled? Is B forced to connect to C since B is closest to C? -- kainaw™ 21:11, 16 November 2011 (U

they do connect

OK, so you are thinking of some kind of recursive scheme, perhaps. Another question: You write "working in n dimensions, and with n nodes,...". Is n=n here? In other words, is the number of nodes equal to the dimension? JoergenB (talk) 19:36, 17 November 2011 (UTC)

no. I've cfhanged it now.

Derivative of the Pythagorean theorem with respect to time

What does ${\operatorname {d} b \over \operatorname {d} t}$ equal when $b={\sqrt {c^{2}-a^{2}}}$, assuming that $c$ and $a$ are functions of time and ${\operatorname {d} c \over \operatorname {d} t}=1$ and ${\operatorname {d} a \over \operatorname {d} t}=3\times {\operatorname {d} b \over \operatorname {d} t}$? Do the variables $c$ and $a$ need to be known? I've tried the chain rule, but I cannot determine how to differentiate $c^{2}-a^{2}$. --Melab±1 ☎ 22:43, 16 November 2011 (UTC)

Our article, total derivative, explains how to compute this.

(ec) By specifying that a is also a function of b, you have described a nonlinear partial differential equation. Nimur (talk) 22:56, 16 November 2011 (UTC)

That doesn't sound good. --Melab±1 ☎ 22:58, 16 November 2011 (UTC)

$a$ is a function of time. --Melab±1 ☎ 23:03, 16 November 2011 (UTC)

"but I cannot determine how to differentiate $c^{2}-a^{2}$" :
I don't understand; ${\operatorname {d} (c^{2}-a^{2}) \over \operatorname {d} t}$ is just ${\operatorname {d} (c^{2}) \over \operatorname {d} t}-{\operatorname {d} (a^{2}) \over \operatorname {d} t}$. The chain rule should work, just keep drilling down and substituting. Caveat: I'm an engineer, not a mathematician. There's a non-zero chance I'm wrong. --Floquenbeam (talk) 23:28, 16 November 2011 (UTC)

It equals one over the square root of ten. Integrate your given identities with respect to t, to find c in terms of t, and a in terms of b. Substitute the answers into the equation, rearange, and differenciate. Plasmic Physics (talk) 00:34, 17 November 2011 (UTC)

But, ${\operatorname {d} (a) \over \operatorname {d} t}$ uses ${\operatorname {d} (b) \over \operatorname {d} t}$ in its definition, which in turn has ${\operatorname {d} (a) \over \operatorname {d} t}$ in its definition. How does it work out like that? Also, $a=4$ and $b=3$ when I am looking for ${\operatorname {d} (a) \over \operatorname {d} t}$. --Melab±1 ☎ 01:38, 17 November 2011 (UTC)

If the derivative of c with respect to t is 1, then its integral is c = (t + C) , where C is the constant of integration. If the derivative of a with respect to t is thrice the derivative of b with respect to t, then its integral is a = 3b. Now substitute these two values into the original equation and rearange and solve for b, then find the derivative. Plasmic Physics (talk) 02:16, 17 November 2011 (UTC)

We have not gotten into integrals yet. --Melab±1 ☎ 02:44, 17 November 2011 (UTC)

I figured it out. ${\operatorname {d} b \over \operatorname {d} t}={1 \over 3}$ --Melab±1 ☎ 03:51, 17 November 2011 (UTC)

Wait, what? How? Plasmic Physics (talk) 04:05, 17 November 2011 (UTC)

${\operatorname {d} b \over \operatorname {d} t}={1 \over {\sqrt {10}}}$ Plasmic Physics (talk) 04:19, 17 November 2011 (UTC)

You're ignoring a constant of integration when you get that answer: a = 3b + D. Solving with this will get a non-constant derivative. D=1, t = 1 will get 1/3, which might how that answer was arrived at.--121.74.125.249 (talk) 05:32, 17 November 2011 (UTC)

There, the constant of integration is implicit. ${\operatorname {d} b \over \operatorname {d} t}$ integrated equals ${b}$, not ${b+D}$. Plasmic Physics (talk) 07:39, 17 November 2011 (UTC)

That is one solution, but not all. You're assuming that ${\operatorname {d} a \over \operatorname {d} t}=3\times {\operatorname {d} b \over \operatorname {d} t}$ implies a = 3b, but that's not necessarily true. If you let c = t and a = 3b+1, you can solve for b as a function of time and get a different solution which satisfies everything given.--121.74.125.249 (talk) 07:56, 17 November 2011 (UTC)

(edit conflict)Let ${y=ax+b}$, then ${\operatorname {d} y \over \operatorname {d} x}={a}$. The antiderivative of ${\operatorname {d} y \over \operatorname {d} x}$ does not equal ${y+c}$, because ${(ax+b)+c}$ does not equal to ${ax+b}$ unless ${c=0}$. Therefore, the antiderivative of ${\operatorname {d} y \over \operatorname {d} x}$ equals ${y}$. Plasmic Physics (talk) 07:58, 17 November 2011 (UTC)

Differentiation is a many-to-one operator, and so you cannot speak of "the antiderivative", at least not as a unique function. Anti-differentiation takes a function to a family of functions, not a single one. Try repeating the above argument starting with the equation $y+c=ax+b$.--121.74.125.249 (talk) 08:18, 17 November 2011 (UTC)

Let ${y+c=ax+b}$, then ${\operatorname {d} (y+c) \over \operatorname {d} x}={a}$. The antiderivative of ${\operatorname {d} (y+c) \over \operatorname {d} x}$ does not equal ${y+c+d}$, because ${(ax+b)+d}$ does not equal to ${ax+b}$ unless ${d=0}$. Therefore, the antiderivative of ${\operatorname {d} (y+c) \over \operatorname {d} x}$ equals ${y+c}$. Plasmic Physics (talk) 08:36, 17 November 2011 (UTC)

By definition, the antiderivative of a derivative cannot equal anything other than the original function. When the original function is known, the constant of integration is made to equal zero, for the answer to be true. Otherwise an false equality is produced, whereupon ${y=y+c}$, where ${c}$ is non-zero. Plasmic Physics (talk) 08:49, 17 November 2011 (UTC)

And yet simultaneously, if $y+c=ax+b$, then ${\operatorname {d} y \over \operatorname {d} x}=a$. So since they have the same derivative, $y+c=y$?

Again, you mistakenly believe the antiderivative of a function to be a unique function. The antiderivative of a function is a family of functions. Once again, a = 3b+1 can give you consistent solutions to every requirement in the original problem.--121.74.125.249 (talk) 09:07, 17 November 2011 (UTC)

Does it not seem strange to you that your logic indicates that ${y=y+c}$? It's like saying ${2=1}$, in non-mathematical terms: I'm the only person in the room, and you are also the only person in the same room. You can believe me when I say that that the antiderivative of a derivative of a function, is equal to the function and only the function, I didn't get a B+ in university-level Calculus for nothing. Plasmic Physics (talk) 09:20, 17 November 2011 (UTC)

Try this, you are a forensic detective, you find a foot print and you want to describe the person who made the foot print. All you can say is that the foot print was made by a person with a specific size foot, but the person can be any height.

Next, you find a foot print and you're told that a person who is 1.84 m tall made the foot print. Now, you can determine that the foot print was made by a person with a specific size foot, and a specific height. You can no longer say that the person can be any height, because you already know that the person is 1.84 m tall. Only a person who is 1.84 m tall could have made that particular foot print. Plasmic Physics (talk) 09:43, 17 November 2011 (UTC)

Likewise, you are given ${b}$. Now you can determine that the derivative is of a function with a rate of change and a specific constant. You can no longer say that the function can have any constant, because you already know that the constant is ${b}$. Plasmic Physics (talk) 09:48, 17 November 2011 (UTC)

You're still wrong. Watch this function: f(t) = 2×t + 4 has a derivative k(t) = f' (t) = 2. You say that having k(t) alone you know its antiderivative is f(t) = 2×t + 4. 'Because you already know the constant is 4.'
However, the function g(t) = 2×t + 7 has exactly the same derivative k(t) while having different constant term. Now what is the antiderivative of k(t) — is it f(t) or g(t)...? --CiaPan (talk) 07:22, 18 November 2011 (UTC)

(ec)The equations are P_i = 0 where the polynomials P_i are

P₁ = a²+b²−c²

P₂ = dc−dt

P₃ = da−3db

differentiate:

P₄ = dP₁/2 = ada+bdb−cdc

eliminate da:

P₅ = P₄−aP₃ = 3adb+bdb−cdc

eliminate a:

P₆ = (3adb−(bdb−cdc))P₅ = 9a²db²−(bdb−cdc)² = 9a²db²−b²db²−c²dc²+2bcdbdc

P₇ = 9db²P₁ = 9db²a²+9db²b²−9db²c²

P₈ = P₇−P₆ = 9db²b²−9db²c²+b²db²+c²dc²−2bcdbdc = (10b²−9c²)db²−2bcdbdc+c²dc²

The equation P₈ = 0 is a nonlinear differential equation

$(10b^{2}-9c^{2})\left({\frac {db}{dc}}\right)^{2}-2bc{\frac {db}{dc}}+c^{2}=0$

or, setting c=t:

${\frac {db}{dt}}={\frac {bt}{10b^{2}-9t^{2}}}\pm {\sqrt {\left({\frac {bt}{10b^{2}-9t^{2}}}\right)^{2}-{\frac {t^{2}}{10b^{2}-9t^{2}}}}}$

Bo Jacoby (talk) 10:30, 17 November 2011 (UTC).

You failed to eliminate ${b}$ from the equation. Plasmic Physics (talk) 10:32, 17 November 2011 (UTC)

I don't think you can eliminate $b$ without reintroducing $a$ . Not every differential equation can be solved by elementary integration. Bo Jacoby (talk) 10:56, 17 November 2011 (UTC).

Since $a=\int {\frac {da}{dt}}\,dt\!$

$a=\int 3{\frac {db}{dt}}\,dt\!=3b$

Therefore $b={\sqrt {c^{2}-(3b)^{2}}}={\frac {c}{\sqrt {10}}}$

Therefore ${\frac {db}{dt}}={\frac {1}{\sqrt {10}}}{\frac {dc}{dt}}={\frac {1}{\sqrt {10}}}$

Simple, and true. Plasmic Physics (talk) 11:28, 17 November 2011 (UTC)

Nice! c²=a²+b²=(3b)²+b²=10b². But a=3b+K is also a possibility for K≠0. Then c²=a²+b²=(3b+K)²+b²=10b²+K²+6Kb. Bo Jacoby (talk) 13:40, 17 November 2011 (UTC).

You are making the same mistake as 121.74.125.249 above, when integrating the derivative of a known function, the constant of integration must always equal to zero for the integral to equal the known function, otherwise an stuation is created where $y+k=y$ which is false for all $y$. Let $y=1$, and $k=1$, then $2=1$. See the point? Only when the original function is unknown can we introduce $k$. Plasmic Physics (talk) 20:35, 17 November 2011 (UTC)

This is still wrong. Suppose I have two functions, $f(x)=x$ and $g(x)=3x+1$. I don't tell you what the functions are, but I do tell you that ${\operatorname {d} g \over \operatorname {d} x}=3\times {\operatorname {d} f \over \operatorname {d} x}$. I have not lied to you. If you conclude that $g(x)=3f(x)$, you're wrong.--130.195.2.100 (talk) 00:57, 18 November 2011 (UTC)

Well, the problem was that the sides of a triangle increased in such a way that ${\operatorname {d} c \over \operatorname {d} t}=1$ and ${\operatorname {d} a \over \operatorname {d} t}=3\times {\operatorname {d} a \over \operatorname {d} t}$. I used related rates. I'll post my solution in a bit. --Melab±1 ☎ 16:10, 17 November 2011 (UTC)

$a^{2}+b^{2}=c^{2}$, so differentiating implicitly, $2a{\frac {da}{dt}}+2b{\frac {db}{dt}}=2c{\frac {dc}{dt}}$.

At the desired value of t, ${\frac {dc}{dt}}=1$ and ${\frac {da}{dt}}=3{\frac {db}{dt}}$ so

${\frac {db}{dt}}(6a+2b)=2c$

Also at the desired value of t, $a=4,b=3$ so $c=5$ because $a^{2}+b^{2}=c^{2}$.

${\frac {db}{dt}}(24+6)=10$

${\frac {db}{dt}}={\frac {1}{3}}$

No integrating required. Readro (talk) 16:26, 17 November 2011 (UTC)

My solution works like this:

$c={\sqrt {a^{2}+b^{2}}}=(a^{2}+b^{2})^{1 \over 2}$

${\operatorname {d} c \over \operatorname {d} t}={1 \over 2}\times (a^{2}+b^{2})^{-{1 \over 2}}\times ([2\times a]\times {\operatorname {d} a \over \operatorname {d} t}+[2\times b]\times {\operatorname {d} b \over \operatorname {d} t})$

${\operatorname {d} c \over \operatorname {d} t}={1 \over 2}\times (a^{2}+b^{2})^{-{1 \over 2}}\times ([2\times a]\times (3\times {\operatorname {d} b \over \operatorname {d} t})+[2\times b]\times {\operatorname {d} b \over \operatorname {d} t})$

$1={1 \over 2}\times (4^{2}+3^{2})^{-{1 \over 2}}\times ([2\times 4]\times (3\times {\operatorname {d} b \over \operatorname {d} t})+[2\times 3]\times {\operatorname {d} b \over \operatorname {d} t})$

$1={1 \over 10}\times (30\times {\operatorname {d} b \over \operatorname {d} t})$

$10=30\times {\operatorname {d} b \over \operatorname {d} t}$

${\operatorname {d} b \over \operatorname {d} t}={1 \over 3}$

--Melab±1 ☎ 21:25, 17 November 2011 (UTC)

Where are you getting $a=4,b=3$? That wasn't in the original description of the problem.--130.195.2.100 (talk) 00:57, 18 November 2011 (UTC)

I know they weren't, but I didn't want to leave all of the work to someone else. --Melab±1 ☎ 02:56, 18 November 2011 (UTC)

That seems correct, but so does mine. How can there be two correct answers? Plasmic Physics (talk) 20:10, 17 November 2011 (UTC)

Yes, that also seems correct. Still, how can there be two correct answers? Plasmic Physics (talk) 22:15, 17 November 2011 (UTC)

Because your answer assumes the constant $C$ in $a=3b+C$ is zero. Apparently there was additional information in the statement of the problem that allows one to conclude that $C=-5$.--130.195.2.100 (talk) 00:57, 18 November 2011 (UTC)

Actually, you are wrong. $a$ does not equal $3b$, instead the derivative of $a$ equals 3 times the derivative of $b$. --Melab±1 ☎ 02:42, 18 November 2011 (UTC)

If ${\frac {da}{dt}}=3{\frac {db}{dt}}$, then $\int {\frac {da}{dt}}\,dt\!=\int 3{\frac {db}{dt}}\,dt\!$.

$\int {\frac {da}{dt}}\,dt\!=a$

$\int 3{\frac {db}{dt}}\,dt\!=3\int {\frac {db}{dt}}\,dt\!=3b$.

Therefore $a=3b$, showing that I am indead correct. In short, the $a=3b$, because $\int {\frac {da}{dt}}\,dt\!=\int 3{\frac {db}{dt}}\,dt\!$. Plasmic Physics (talk) 06:52, 18 November 2011 (UTC)

C must be zero for $b=b+C$ to be true. Note that C ≠ c, one is a constant of integration, the other is a function of $t$. Plasmic Physics (talk) 01:14, 18 November 2011 (UTC)

I'll use prime for a derivative, We have:
1. c' = 1
2. a' = 3×b'
3. a² + b² = c².

From 1. we have
4. c = t + K (for some constant K)

From 2.
5. a = 3×b + L (for some constant L)

Now from 3 and 5:
6. 10×b² + 6L×b + L² = (t + K)²

Solve the quadratic equation 6. with respect to b
7. b = [ −3L ± sqrt(10(t + K)² − L²) ] / 10

and finally differentiate b(t).

For L=0, i.e. a=3b, equation 7. simplifies to
b = |t + K| / √10,
so assuming positive t and K
b = (t + K) / √10 and finally b = 1 / √10 — but L=0 is not that obvious to me, so the general answer would look much more ugly...
CiaPan (talk) 07:10, 18 November 2011 (UTC)

The only case when $L$ would equal a non-zero number, would be if $\int {\frac {da}{dt}}\,d{\frac {db}{dt}}\!$. Plasmic Physics (talk) 07:44, 18 November 2011 (UTC)

Plasmic Physics, you are mistaken. Consider the case a=4, b=3, c=5. Then L=a-3b=-5. This triangle can in dt seconds expand to c=5+dt, a=4+da=4+3db, b=3+db, satisfying 0=c²-a²-b²=25+10dt-16-24db-9-6db=10dt-30db, so in this special case db/dt=1/3. Bo Jacoby (talk) 08:16, 18 November 2011 (UTC).

If that is the case then you are suggesting that $b=b-5$. Go ahead and solve for $b$. Plasmic Physics (talk) 08:33, 18 November 2011 (UTC)

No, I am merely suggesting that if a=3b-5 then da/dt=3db/dt. Bo Jacoby (talk) 09:48, 18aNovember 2011 (UTC).

General solution is

$a={\frac {3t}{\sqrt {10}}}+3k$

$b={\frac {t}{\sqrt {10}}}+k$

$c=t+{\sqrt {10}}k$

where k is an arbitrary constant representing the value of b when t = 0. Gandalf61 (talk) 12:11, 18 November 2011 (UTC)

That is a solution, but not the general solution, because it always statisfies the condition that a=3b, which is not generally true for right-angled triangles. Bo Jacoby (talk) 14:16, 18 November 2011 (UTC).

I didn't start by assuming a=3b - this just fell out of the solution. I assumed that c = t + c0 for some constant c0, and then that a and b were polynomial functions of t. It is fairly easy to see that a and b must in fact be linear functions of t, so a = 3rt + a0 and b = rt + b0. Equating coefficients of t^2 in c^2 = a^2 + b^2 then gives 10r^2 = 1 and hence the value of r. Equating the coefficients of t gives an expression for c0 in terms of a0 and b0. Using c0^2 = a0^2 + b0^2 (which we know is true by setting t to 0) to eliminate c0 gives a quadratic in a0/b0. It just so happens that this quadratic has two equal roots, which are both 3, so a0 = 3b0, and so a = 3b for all values of t.

If you think there is a more general solution, can you show us what it is ? Gandalf61 (talk) 14:56, 18 November 2011 (UTC)

You have shown that the solution for a and b is not polynomial functions unless a=3b. The differential equation above can perhaps not be solved by elementary functions. Bo Jacoby (talk) 17:16, 18 November 2011 (UTC).

Even if we drop the assumption that a and b are polynomial functions of t, we can still show that a = 3b. We know that c = t + c0 because dc/dt=1. And because da/dt = 3 db/dt for all t, we know that a(t) = 3b(t) + k for some constant k; this is true whatever type of functions a(t) and b(t) may be. We have

$c(t)^{2}=a(t)^{2}+b(t)^{2}$

for all t, so we must have

$c(-c_{0})^{2}=a(-c_{0})^{2}+b(-c_{0})^{2}$

but c(-c0) = 0, so a(-c0) = b(-c0) = 0. But a(-c0) = 3b(-c0) + k, therefore k = 0, therefore a(t) = 3b(t) for all t. Gandalf61 (talk) 20:50, 18 November 2011 (UTC)

If you assume these are intended to hold for all t. Often questions like this are only for certain t, for example non-negative t. In particular, since the OP was told to investigate when b = 3 and a = 4, that must be the case.--121.74.125.249 (talk) 21:37, 18 November 2011 (UTC)

As has been shown, if we assume a and b are polynomial functions of t for t > 0 then a = 3b and there is no solution with b=3 and a=4 at time 0. This is because the solution set has only one degree of freedom - you can specify a value for a or b at time 0 (or, indeed, for c) but not for both a and b at the same time. If we go beyond polynomial functions then we are into the very speculative territory of non-elementary functions that have to satisfy the functional equation given by Bo below. I very much doubt that a student who does not know how to calculate a total derivative (which is where we came in) would be expected to venture into such uncharted territory. It is much more likely that they have misunderstood or mis-stated the problem. Gandalf61 (talk) 10:21, 19 November 2011 (UTC)

Related rates problems are a standard feature of a calculus sequence; indeed, that's how the OP went on to solve it. There's no need to determine the functions to solve such a problem.--121.74.125.249 (talk) 11:02, 19 November 2011 (UTC)

The OP appears to believe that db/dt = 1/3, which is definitely incorrect. If db/dt is constant then both a and b are linear functions of t, and it is immediately clear by differentiating c^2 = a^2 + b^2 that 10(db/dt)^2 = 1 and so db/dt = 1/sqrt(10). Deriving a, b and c as explicit functions of t in this case shows that the solution set only has one degree of freedom, and there is no solution in which b=3 and a=4 at the same time. If db/dt is not constant then we are into Bo's territory of non-elementary functions that satisfy a quadratic functional equation. Gandalf61 (talk) 09:13, 20 November 2011 (UTC)

The unknown sides of the triangle a,b,c satisfy a²+b²=c², and at time t=0 they have the known values a₀,b₀,c₀, satisfying a₀²+b₀²=c₀². Set c=c₀+t satisfying dc=dt, and set a=a₀+3x and b=b₀+x satisfying da=3db. We have (a₀+3x)²+(b₀+x)²=(c₀+t)². Or 10x²+(6a₀+2b₀)x=2c₀t+t². Solve this second degree equation and substitute the solution x into the expressions for a and b. Bo Jacoby (talk) 00:17, 19 November 2011 (UTC). For 0<t<<c₀ the square terms are neglegible and (6a₀+2b₀)x=2c₀t, such that db/dt=dx/dt=2c₀/(6a₀+2b₀). For c₀<<t the linear terms are neglegible and 10x²=t², such that db/dt=dx/dt=1/sqrt(10). Bo Jacoby (talk) 08:19, 19 November 2011 (UTC).