Re: Math Question
 

<font size="2" face="sans-serif, arial, verdana">You mean kind like...

[(d^2)z/(dt)^2] + [k/m*z] - [kL/m] + [C/m(dz/dt)] - [g] = [kA/m*sin(wt)]

I have a test involving this problem in about an hour and a half.

Re: Math Question
 

Yep. Fortunately I didn't have to get too deep into calc for my major..even the course I took was a pain.

Re: Math Question
 

<font size="2" face="sans-serif, arial, verdana">Geo, this is probably because you are working with functions. A function may only have one value in the range for each value in the domain.
So, in this case, the function sqrt() is defined to be positve .

Re: Math Question
 

<font size="2" face="sans-serif, arial, verdana">Geo, this is probably because you are working with functions. A function may only have one value in the range for each value in the domain.
So, in this case, the function sqrt() is defined to be positve . </font><hr /></blockquote><font size="2" face="sans-serif, arial, verdana">Ok, but isn't that circular logic? To me it sounds like you are saying that the square root of a number is always positive, because if it's negative then it's not a function. But the square root being positive is what makes it a function. How do you know it's a function, other then the fact that I typed it in function notation? Because there are times they will give you a problem in function notation and ask you if it is indeed a function.

Re: Math Question
 

It is a function by definition. As you continue in math, you find a lot of this. For example, you cannot take a derivative (calculus) of anything that is not a function. It's one those things you have to accept as defined.

Re: Math Question
 

In other words:

Take the following two sets of ordered pairs:

f: (1,1), (4,2), (9,3)
g: (1,1), (1,-1), (4,2), (4,-2), (9,3), (9,-3)

f is a function, g is not.

So, if you want to use the square root in circumstances where a function is required,it is useful to define it as f.

Re: Math Question
 

Geo said:

"Question is find the domain and range of
f(x)=sqrt(x+4) (The square root of x+4)

The domain is [-4,infinty)
The correct answer for the range is [0,infinity)

My question is, why isn't the range all real numbers? (-infinity,infinity) In class the teacher worked the problem and he said plug in any domain value for x and see what you get for f(x) is never less then zero."

If you plug in a negative number less that -4, you end up with the square root of a negative number. This is not possible with real numbers, only imaginary, i.e. sqrt(-1) by defintion equals a funny little number called i (or sometimes j).

So:

sqrt(4) = +/- 2 and
sqrt(-4)=2i where 2i=sqrt(4*(-1))=sqrt(4)*sqrt(-1)=2*i

Complex math is, for the most part useless, except when you get into analyzing differential equations and AC electrical circuits. Believe it or not, that imaginary stuff is quite useful in the real world. Well, the real of world of us engineers and other math-type geeks.

Re: Math Question
 

<font size="2" face="sans-serif, arial, verdana">Hmm... damped, driven, simple harmonic motion?

Re: Math Question
 

<font size="2" face="sans-serif, arial, verdana">Hmm... damped, driven, simple harmonic motion? </font><hr /></blockquote><font size="2" face="sans-serif, arial, verdana">No. It's obviously a complicated plot to fry my brain.


Say, do you smell smoke? http://forum.shrapnelgames.com/images/icons/shock.gif

Re: Math Question
 

<font size="2" face="sans-serif, arial, verdana">Hmm... damped, driven, simple harmonic motion? </font><hr /></blockquote><font size="2" face="sans-serif, arial, verdana">Impressive, it looks like someone else has seen a 3rd year physics course. I managed to shank this very problem on the test I took this morning.