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September 20

Planes and Spheres

I need a bit of help understanding a problem I'm doing and Wikipedia seemed like the best place.

A plane $P_{1}$ has equation

$3x-y-1=0$

and a sphere $S_{1}$ has equation

$x^{2}+y^{2}+z^{2}=7$ .

I have to determine the equation of the circle, $C_{1}$ , defined by the intersection of $P_{1}$ and $S_{1}$ .

I have tried this in a number of ways and nothing seems to help. From the equation for $P_{1}$ , I can see that $y=3x-1$ . Substituting this into the equation for $S_{1}$ gives $x^{2}+(3x-1)^{2}+z^{2}=7$ , which is simplified to

$10(x-{\frac {3}{10}})^{2}+z^{2}={\frac {69}{10}}$

This clearly isn't a circle and plotting it in my graphing application confirms it.

My next method was to rearrange the equation for $P_{1}$ to arrive at $21x-7y=7$ . Substituting this into the equation for $S_{1}$ gives $x^{2}+y^{2}+z^{2}=21x-7y$ , which eventually simplifies to

$(x-{\frac {21}{2}})^{2}+(y+{\frac {7}{2}})^{2}+z^{2}={\frac {490}{4}}\,\!$ .

This is even more confusing because somehow the intersection of plane and a sphere has become a sphere, defying all logic. Could someone please tell why these two methods have failed and point me in the right direction, ie don't explicitly tell me what to do, just give a hint? I want to try and do as much of this by myself as possible. Thanks 92.2.212.113 (talk) 15:22, 20 September 2008 (UTC)[reply]

A circle, a 1-dimensional thing, cannot be represented by 1 equation in 3-D, because each equation only constrains one dimension; so you will need 2 equations to represent your circle. --71.147.13.131 (talk) 17:17, 20 September 2008 (UTC)[reply]

Indeed. Baring degenerate cases, each equation (or "constraint") reduces the dimension by one. You've started in 3D space, so one constraint gives you a 2D subset, another gives you 1D and a third would give you a single point (0D). You can't really get much better than just giving the equations of the sphere and the plane - the question seems to be flawed. Perhaps it wants the equation of the projection of the circle on to one of the co-ordinate planes? --Tango (talk) 19:49, 20 September 2008 (UTC)[reply]

Do the two constraints come from the fact that the equation for the plane involves both x and y? 92.2.212.113 (talk) 19:54, 20 September 2008 (UTC)[reply]

Perhaps clarity will be aided with the complete question.

P_{1},S_{1},C_{1}

are the same as above.

P_{2}

is the plane with equation

x-y+1=0

,

S_{2}

is the sphere with equation

x^{2}+y^{2}+z^{2}-6y-4z+10=0

,

C_{2}

represents their intersection and L is the intersection of

P_{1}

and

P_{2}

. "Find the points where L meets each of

S_{1}

and

S_{2}

. Determine, giving your reasons, whether the circles

C_{1}

and

C_{2}

are linked." Does that help? 92.2.212.113 (talk) 20:02, 20 September 2008 (UTC)[reply]

Since a circle has an area, I see no way that it can be a 1-dimensional object. Also, if a plane intersects a sphere, not if it just touches the sphere, the points of intersection must produce a circle, see here Plane-sphere intersection. 92.2.212.113 (talk) 19:50, 20 September 2008 (UTC)[reply]

Hmm I think my confusion stems from thinking that the intersection produce a disc as well as a circumference. It's now clear that it only produces a circumference. This question comes from a well established exam though, so there must be something I've missed out that would make it work. 92.2.212.113 (talk) 19:53, 20 September 2008 (UTC)[reply]

The two constraints are the two equations, one for the plane and one for the sphere, each equation will be one constraint, regardless of how many variables are in it (as long as it isn't a vector equation, anyway - vector equations represent multiple scalar equations). I don't see anywhere in that question that asks for an equation for the circle, you just need to determine if the two circles are linked, that doesn't require having an equation for the circle. --Tango (talk) 20:42, 20 September 2008 (UTC)[reply]

OK, I'm with you on the constraint issue. So is it impossible to determine an equation for the circles then? If so, I suppose a new method of approach is called for... Thanks 92.2.212.113 (talk) 21:26, 20 September 2008 (UTC)[reply]

It is impossible to determine a single equation for the circle (barring messy stuff), but it is possible to find a pair of equations for the circle: there are many such pairs, one such being the pair of equations you started with, as Tango pointed out. Try approaching your problem from some other direction (and ask for assistance if wanted). Eric. 65.96.172.100 (talk) 22:20, 20 September 2008 (UTC)[reply]

The "messy stuff" doesn't have to be all that messy:

(3x-y-1)^{2}+(x^{2}+y^{2}+z^{2}-7)^{2}=0

would do it. Of course, that is absolutely no help in solving the main problem ;) AndrewWTaylor (talk) 13:26, 22 September 2008 (UTC)[reply]

The main point of this question seems to be to provoke the answerer into providing a proper definition of that mysterious word "linked". Once you've done that, it should be obvious how to proceed with the equations. (My definition wouldn't involve the line L at all, and my method wouldn't really make use of those line/sphere intersection points, but I can see how they're related.) --tcsetattr (talk / contribs) 22:00, 20 September 2008 (UTC)[reply]

Just a note: A circle is a 1-manifold and therefore is one-dimensional in the topological sense.

Topology Expert (talk) 13:45, 21 September 2008 (UTC)[reply]

Radius vs. radian

The mean circumference of Earth is about 40041.5 km, which means 1° of arc ≈ 111.2263 km, or 1 rad of arc ≈ 6372.8 km, as 1 rad = 57.295779513...°. A radian ("rad") is the arc equivalent of radius. So why is the inverse of curvature called "the radius of curvature", and not "a radian of curvature"? Since 1° of arc, or "arcdegree", is 1/360° of the circumference, isn't 1/(2π), here 6372.8, the "arcradian", rather than "arcradius". ~Kaimbridge~ (talk) 17:06, 20 September 2008 (UTC)[reply]

The radius of curvature is so called because it is the radius of a circle with the same curvature as the given curve. It would be silly to say that a curve has x radians of curvature rather than radius of curvature x for at least two reasons. Firstly, radius of curvature is a dimensional quantity, in fact a length. Do you want to say that the surface of the earth has 6372.8 km radians of curvature? Secondly, the 'of curvature' suggests that a higher number corresponds to higher curvature, which is of course not the case. Algebraist 17:11, 20 September 2008 (UTC)[reply]

Of course not plural, anymore than you would say Earth has "6372.8 km radii of curvature"!——one radian is the arc equivalent length of a radius: In a unit circle the radius = 1, a radian = 1 and the circumferemnce = 6.2831853...radians. So if a circle's radius = 6372.8, then one radian = 6372.8 and the circumference = 40041.5. Thus Earth would have a radian of curvature equaling 6372.8. See Plane and Spherical Trigonometryand Elements of Geometry. ~Kaimbridge~ (talk) 22:47, 20 September 2008 (UTC)[reply]

What do you mean by "a radian = 1"? A radian is a radian, it doesn't equal a number. --Tango (talk) 22:54, 20 September 2008 (UTC)[reply]

See the definition of radian:

"One radian is the angle subtended at the center of a circle by an arc that is equal in length to the radius of the circle." ~Kaimbridge~ (talk) 01:25, 21 September 2008 (UTC)[reply]

Yes, I know what a radian is, but it's a measure of angle, it isn't equal to a natural number. --Tango (talk) 13:23, 21 September 2008 (UTC)[reply]

Is it accurate to use "1° of arc" like that? I thought the "of arc" was just to distinguish between other things measured in degrees (temperature, for example). "1° of arc" is still a measure of angle, not length, by my understanding. --Tango (talk) 19:44, 20 September 2008 (UTC)[reply]

Yes, see the links I gave above, and this application of it(73.208(c)(3)-(4)):

(3) Calculate the number of kilometers per degree latitude difference for the middle latitude calculated in paragraph (c)(2) as follows:

KPD[lat]=111.13209−0.56605 cos(2ML)+0.00120 cos(4ML)

(4) Calculate the number of kilometers per degree longitude difference for the middle latitude calculated in paragraph (c)(2) as follows:

KPD[lon]=111.41513 cos(ML)−0.09455 cos(3ML)+0.00012 cos(5ML)

Here, KPD[lat] is "M" and KPD[lon] is "N(×cos(ML)", the principal radii of curvature in degree form of their series expansion, and the latitude and longitude differences are angular distances (i.e., "degrees of arc"). ~Kaimbridge~ (talk) 22:47, 20 September 2008 (UTC)[reply]

That quote doesn't include the phrase "degree of arc", so how does it answer my question? --Tango (talk) 22:54, 20 September 2008 (UTC)[reply]

It says "per degree latitude difference" and "per degree longitude difference". As I said above, the latitude and longitude differences are "degrees of arc", so one degree of latitude difference equals one degree of arc.

Let's say ΔLat = 20°: Using the previously cited mean values, distance = 2224.526 km = 20°× 111.2263 = .3490658 rad × 6372.8. What would you call 111.2263 and 6372.8? I would say 111.2263 is the value of an arcdegree and 6372.8 is the arcradius, though it would seem it should be arcradian. ~Kaimbridge~ (talk) 01:25, 21 September 2008 (UTC)[reply]

I've never seen the terms used that way before, although I now see Nautical mile uses them that way. I would have just said "the arc length corresponding to one degree" or something. Even with those definitions, "radian of curvature" wouldn't make sense, a radian is still a measure of angle, you could say "radian of arc of curvature" since a radian of arc is, by definition, the radius, but I don't see much point. --Tango (talk) 11:15, 21 September 2008 (UTC)[reply]

Okay, then that brings the question——he-he——full circle: Should it be "arcradius" or "arcradian".

Consider example 2, here:

Example 2. An angle of .75 radians means that the arc is three fourths of the radius. s = .75r

So letting r = 6372.8, we have

4779.6 =.75·6372.8 = 42.9719°·111.2263,

thus an arcdegree = 111.2263, but what is 6372.8, the arcradius or an arcradian? Googling around, I find several references to arcradius, though mostly referencing height and width calculations of circles, but none for arcradian (or "arc radian"), with regard to what I'm talking about. On one hand, it would seem arcradian belongs with arcdegree, arcminute and arcsecond, not arcradius, but on the other we are talking about the arc equivalent of the plane radius, which would suggest arcradius. ~Kaimbridge~ (talk) 17:48, 21 September 2008 (UTC)[reply]

Using your definitions, "arc-radian" would mean the same as "radius". I have no idea what "arc-radius" would mean. --Tango (talk) 18:01, 21 September 2008 (UTC)[reply]

Okay, I think I've figured it out.

The radius of an arc(i.e., "arcradius") of a circle can be found, given the height, "H", and width, "W", of an arc segment:

Circle: radius of arc = plane radius = (4H^2 + W^2)/H;

The arc length, "AL", of a circle equals the angular distance/central angle, "AD", times the (arc)radius, "r":

AL = AD rad × r = AD° × (π/180×r);

So the above example would be,

4779.6 =.75 rad × 6372.8 = 42.9719° × 111.2263,

where ".75" is the radian quantity ("arc" is intrinsic to the term "radian", itself), "6372.8" is the length of a radian ("arcradius"), "42.9719°" is the arcdegree quantity and "111.2263" is the length of an arcdegree.

Thus, for a unit circle,

1 = unit radius = unit arcradius = radian = 57..295779513... arcdegrees.

Of course, with an ellipse it is more complicated, as the arcradius equals the meridional radius of curvature, not the plane radius, and then only at a given point ("φ")——"prorated" or "per capita" (i.e., if that point value was spread out over the whole length of the radian)? ~Kaimbridge~ (talk) 18:39, 24 September 2008 (UTC)[reply]

Integration

sin(Tan^-1) is not a function? —Preceding unsigned comment added by Saadmunir (talk • contribs) 20:22, 20 September 2008 (UTC)[reply]

Not in the common sense, no. It's a multivalued function though, with all input values (except 0) having two output values. (I don't see what this has to do with integration.) -- Jao (talk) 20:32, 20 September 2008 (UTC)[reply]

If you restict the inverse tan to the principal branch, as is commonly done, it's fine, though. --Tango (talk) 20:34, 20 September 2008 (UTC)[reply]

Isn’t the symbol tan^-1(x) by definition only defined on

({\frac {\pi }{2}},{\frac {\pi }{2}})

? GromXXVII (talk) 13:07, 21 September 2008 (UTC)[reply]

It's often (although not universally) defined to take values only on that interval, is that what you mean? It's defined everywhere. --Tango (talk) 13:22, 21 September 2008 (UTC)[reply]

Err yeah that’s what I meant. GromXXVII (talk) 13:35, 21 September 2008 (UTC)[reply]

Note also that sin(tan⁻¹(x)) = x/√(1 + x²). Michael Hardy (talk) 05:09, 21 September 2008 (UTC)[reply]

functions

what is parent function and daughter function? —Preceding unsigned comment added by Saadmunir (talk • contribs) 20:24, 20 September 2008 (UTC)[reply]

Looking at Google I see parent function is as I suspected yet another of those ?#@[+! terms invented by people in education. This one meaning basic functions like exp or sin like those on a calculator. I didn't see anything about daughter function. Aged parents often believe it is to look after them in their dotage but daughters tend to have other ideas about it. I can't see where the maths comes into that though. :) Dmcq (talk) 21:31, 20 September 2008 (UTC)[reply]

Child function, perhaps? (Being gender-sensitive). I did find [1] which is a horrible way of thinking about things, I think. I believe "parent functions" refer to a "general" function, such as

f(x)=x^{2}

, while "child functions" refer to thinks like

f(x+1)=(x+1)^{2}

or

f(x+3)=(x+3)^{2}

, where f is defined previously. The closest I learnt to this concept would be "family of curves". I think it mostly refers to translation although I can't find any definitions, so to restrict this to translation would probably be wrong. x42bn6 Talk Mess 21:01, 21 September 2008 (UTC)[reply]

A family of mathematical objects is usually a collection of objects where each is on an equal footing, I wouldn't distinguish between children and parents. --Tango (talk) 21:06, 21 September 2008 (UTC)[reply]

The link x42bn6 gives doesn't look like it's using "parent" and "daughter" as rigidly defined technical terms, it looks like it's just using them to structure the discussion. Like a "family" of sets, is a set of sets, but using different words for them temporarily makes it easier to describe their relationship. The words themselves aren't essential and are discarded when they're no longer convenient. Black Carrot (talk) 13:50, 26 September 2008 (UTC)[reply]

Algebra

-6 = -6

9-15 = 4-10

9-15+ 25/4 = 4-10+ 25/4

(3-5/2)² = (2-5/2)²

3-5/2 = 2-5/2

3 = 2

what's wrong in it? —Preceding unsigned comment added by Saadmunir (talk • contribs) 20:34, 20 September 2008 (UTC)[reply]

When you take the square-root, there are two possibilities. You can have them both with the same sign, as you've done, or you can have them with different signs, which gives 3-5/2=-2+5/2, which is true. See Extraneous and missing solutions. --Tango (talk) 20:38, 20 September 2008 (UTC)[reply]

More directly,

{\sqrt {x^{2}}}=|x|

. Dragons flight (talk) 21:22, 20 September 2008 (UTC)[reply]

Or, just look at line four, then simplify it *before* taking the square root: (1/2)² = (-1/2)². From there, if you can conclude that +1/2 = -1/2, then maybe 3 does equal 2! --DaHorsesMouth (talk) 22:51, 20 September 2008 (UTC)[reply]

It all boils down to this:

Even if x^2 = y^2

you CANNOT conclude that x = y (without additional information or additional constraints)

122.107.140.79 (talk) —Preceding undated comment was added at 00:57, 21 September 2008 (UTC).[reply]

See also Invalid proof. This is one of the classic tricks in deliberately invalid proofs. PrimeHunter (talk) 01:30, 21 September 2008 (UTC)[reply]