Country Boy

I'm a bit confused by the question. You say "H: R^3--->R^3" but the H given, H[(a1,a2,a3), (b1,b2,b3)] = a1b1 - 2a1b2+a2b1- a3b3 is clearly from R^3 X R^3 to R.

No, you are correct. [math]\frac{1}{\epsilon^2+ z^2}= \frac{1}{(z- i\epsilon)(z+ i\epsilon)}[/math] and so has a pole of order 1 at [math]z= i\epsilon[/math] and [math]z= -i\epsilon[/math]. As to how that affects its behavior on the real line, here is a simple example: If you were to construct its Taylor's series, about z= 0, say, you would find that the radius of convergence was [math]\epsilon[/math]. That's because, in the complex plane, the "radius of convergence" really is a "radius"! In the complex plane, the series converges as long as it doesn't hit a "problem"- which, in the case of [math]\frac{1}{\epsilon^2+ z^2}[/math] is at [math]i\epsilon[/math] and [math]-i\epsilon[/math]. Here's a nice application: Suppose you were to construct the Taylor's series for [math]\frac{1}{\epsilon^2+ x^2}[/math] about the point [math]x= x_0[/math]. What would its radius of convergence be? You don't need to actually calculate the series to answer that- it is the distance from [math]x= x_0[/math] to [math]i\epsilon[/math] and [math]-i\epsilon[/math], [math]\sqrt{x_0^2+ \epsilon^2}[/math].

The problem does not say anything about acceleration! F= ma is the force necessary to accelerate an object. Obviously, you must accelerate the object in order to get it moving but that could be a very quick acceleration which is over very quickly and does not really contribute anything to the work done. The force necessary to lift an object of mass m AT CONSTANT VELOCITY is mg, the weigth of the object. What's happening here is that A is lifting the object at 5/2 feet per minute and B is lifting it at 5 feet per minute- again, that's velocity not acceleration.

Since the problem specifically says "thrown down", I would be inclined to think the initial downward speed of the second ball is the same as the initial upward speed of the first ball, 25 m/s. I would also ignore air resistance. The acceleration for the two balls is the same: 9.8 m/s^2 downward. Taking the ground to be y= 0, at time t seconds, the first ball will have height y1(t)= -4.9t^2+ 25t. The second ball will have height y2(t)= -4.9t^2- 25t+ 15. They will pass when those heights are the same: -4.9t^2+ 25t= -4.9t^2- 25t+ 15. Since the "-4.9t^2" terms will cancel, that's a simple linear eqaution to solve for t. Once you have found t, evaluate either y1 or y2 (or both, as a check) for the height of the two balls.

You are right that it is not a circle, but it also is not a curve! It is a surface: specifically, a "paraboloid". Frankly, I don't believe it has any asymptotes! The parabola in 2 dimensions doesn't.

I'm afraid you will have to explain what you mean by a "quad plane"!

Do you understand that these are just asking for examples and definitions? Not much "thinking" is required. I'll bet if you look in your textbook, you will find all the definitions given.

They take their oxygen with them!

Looks to me like you need a trig identity! Do you know sin(x+y)= sin(x)cos(y)+ cos(x)sin(y)? If you replace y with -y that becomes sin(x-y)= sin(x)cos(y)- cos(x)sin(y) because cos(-y)= cos(y) and sin(-y)= -sin(y). Adding those two equations, 2sin(x)cos(y)= sin(x+y)+ sin(x-y) or sin(x)cos(y)= (sin(x+y)+ sin(x-y))/2. Now what does that give you for sin(5x)cos(2x)?

If you have no idea what reagents ARE, look up the definition. That should make it clear how they "work"! If you d0 know what reagents are, I can see no reason for your question.

Athiest's remark about "which statement ro oversteretch" was referring to your analogy being to a baseball in water creating "a hole for it to move through". That is also does not happen!

I can't answer your questions because it isn't clear How those images are relalted to either the sphere or pi. How are those images defined?

Hey, YOU were the one who asked "Am I mad?" Logically I could only agree with your first comment (which i did not) or with your second! I think I was typing that WHILE you were typing your response to Dave. Though I do like the "Pauline Conversion". I wondered for a moment who "Pauline" was!

That's interesting- I would have thought just the opposite! The "Handbook" and reference to "Methods" would make me think it was NOT theoretical but devoted to applications.

The first part of this is true- since the radius is perpendicular to the direction of motion, the radius will not change. But you do NOT gety a contradiction" that R= r and R< r. What happens is that C= 2 pi r is no longer true- the geometry is no longer Euclidean. Absolutely not. If you mean measuring rods moving with the circumference of the disk, from the point of view of a person in the "lab frame", in which the disk is spinning, both circumference and measuring rods, moving at the same rate, will be reduced by same amount and so the same number of rods will cover the circumference. From the point of view of a person ON the circumference of the disk, the neither rods nor circumference will have changed. From the point of view of a person on the disk but closer to the center, both circumference and rods will have reduced by the same amount but less than observed from the laboratory frame- he will see a non-Euclidean geometry but closer to Euclidean geometry than the laboratory frame. Of course all that change in "geometry" is because rotation is an acceleration, producing "fictitious forces".

Apparently you are mad! If V and W are not the SAME vector space, then there is NO "identity operator". Yes, the set of linear transformations from V->W is a (dim(v))(dim(W)) dimensional vector space with addition of vectors defined by (L_1+ L_2)(v)= L_1(v)+ L_2(v). In particular, the 0 transformation must satisfy (L_1+ 0_L)v= L_1(v)+ 0_L(v)= L_1(v). In other words, the 0 transformation must be the transformation that takes every vector in V into the 0 vector in W. Of course, if we look at linear transformations V->V then we can form a RING in which the "multiplication" is composition of transformations. (We cannot do that with V->W with V and W different because then L_1(L_2(v)) would be meaningless- L_2(v) is in W, not V, so we cannot apply L_1 to it.) In THAT case the "multiplicative" indentity is the identity transformation.

I'm going to assume that that last condition is L_1+ L_2= 0_L, the "0" transformation that maps every vector in v to the 0 vector in W. Take a look at L_2(v)= -(L_1(v)).

The crucial point is the front wheel. As I am sure you know, if you tilt a spinning wheel, downward to your right, say, the wheel will tend to turn to the right. That means it turns to "catch' the fall.

Okay, a bijective function HAS an inverse. But none of that addresses the question of how you determine whether or not two given functions ARE inverse to one another! Fortunately that has been pretty well handled already.

Notice that Phil has written that entirely in terms of ORDINARY derivatives. That's perfectly correct- you just have to remember that since all of the partial derivatives are with respect to x, that "y" is just a constant.

"That sux"

You can memorize "Please excuse my dear aunt Sally". That helps you remember "PEMDAS": first do whatever is in the Parentheses, then any Exponents, then multiplications, then any divisions, then any additions, then any subtractions. For your original question, 2 - (-5)+ 3^2, first do the exponent: 3^2= 9, the additions 2+ 9= 11, finally, the subtraction: 11-(-5) is really the same as 11+ 5= 16.

So basically, then it is your ATTITUDE that is the problem rather than the chain rule.

I would say that a quadratic equation is an equation whose highest power IS 2 (not "no higher than 2- that would include linear equations.).

I hope your teacher DIDN'T say precisely that! [math]\sqrt{25}= 5[/math], not -5. What is true is that the equation, [math]x^2= a[/math] has TWO solutions. One is [math]\sqrt{a}[/math], the other is [math]-\sqrt{a}[/math]. The reason we need to write the "-" (or sometimes [math]\pm[/math]) is BECAUSE [math]\sqrt{a}[/math] does NOT include the negative. You might then, on the basis of other information (whether a negative number is physically possible if it is a physic problem, the fact that "distance" is always positive) decide that one of those solutions does not actually work in that particular problem. No. [math]25^{\frac{1}{2}}[/math] or [math]\sqrt{25}[/math] are defined as POSITIVE 5. In general, for any positive number, a, [math]a^{\frac{1}{2}}[/math] or [math]\sqrt{a}[/math] are defined as the POSITIVE number whose square is a.