OK, read my edit of the prev post for a slightly easier way, but as played, to have no solutions, you need to ensure that (k+3)-(10/k)=0 but 4-(8/k)=/=0.

(k+3)=10/k <=> k^2+3k-10=0 <=> k=-5 or k=2 (I guessed it instantly using the inverse Vieta's theorem).

If k=2, then there are infinitely many solutions because 4-8/k=0.

If k=-5, there are no solutions because 4-8/k=5.6=/=0.

Edit: oh, and of course you can't divide by k if it's 0, so you have to consider k=0 as a special case. That's why I recommended to swap the 2nd and 3rd rows.

To get to my second matrix though I had to subtract (2/k) to one of the coefficients, am I allowed to do this since k might be 0? What happens if k = 0?

EDIT: Too slow, yeah thats what I was wondering also. Is there ever only one solution? Would that be possible only in 2-dimension since in R^3 we are dealing with planes? Thank you very much!

If k is neither -5 nor 2, there's a single solution. In general, a system of linear equations has a single solution for some (and, equivalently, for any) right-hand side iff [=if and only if] its left-hand-side matrix is square and

1) its rank equals its number of rows/columns, or, equivalently,
2) its determinant is nonzero, or, equivalently,
3) it can be transformed into a unit matrix via the Gauss (row reduction) method.

That holds for any number of dimensions of the matrix.

It's worth noting that (computer science rant alert!):

1. In computers, determinants of big square matrices are almost always calculated via the Gauss method (because the determinant of an upper triangular square matrix is the product of its diagonal elements) as the Gauss method is cubic in time (takes time roughly proportional to the cubed number of rows), while the method of calculating the determinant outlined in its definition is extremely slow as the number of rows grows.

2. Not only is swapping rows in the Gauss method a convenient option, but any computer scientist will tell you that, when using it to solve a linear system on a computer, it's imperative to swap rows so that the principal row (the one which you're subtracting from the other with coefficients) has the number with the biggest available absolute value in the corresponding principal column, for the biggest numerical stability. That's because division by a number with a smaller abs. value introduces a bigger abs. computational error.

2a. It's of course not an issue when you do hw with 2x2-4x4 matrices on paper, swap rows as you wish then; in fact, on paper, it's easier to add or subtract the row with a 1 or -1 in the principal column - it saves the time and effort needed for division by coefficients.

this is the last question on the assignment:

find the average speed over the time interval 1,5 of a particle whose position at time t is s(t)=t^3-6t^2

i think its a trick question b/c i tried doing 1/4 * integral from 1 to 5 of s(t) dt and it wasnt the right answer. I'm not good at physics or understanding math so I'm a bit confused. Do i need to find the second derivative of pos then take the integral of that? or that would be the same as the first derivative of pos.

The speed is the derivative of the position. So to get the average speed you should take 1/4th of the integral of ds(t)/dt over [1,5] but, by the fundamental theorem of Calculus this is just 1/4*(s(5)-s(1))= -5

teh back of the book says the answer is "17/2 m/s"

It wouldn't be even a question if the particle was moving in one direction all the time.

Here, however, it changes the direction once.

To see this, note that s'(t)=3t^2-12t. Hence the critical points are found from the equation 3t^2-12t=0 <=> t=0 or t=4. They also happen to be local extrema of the position function because the derivative changes its sign at both points (as s(t) is twice differentiable here, we can verify that by observing that the second derivative is nonzero at both points), in other words, the particle changes the direction of movement at both t=0 (a local maximum, outside of our interval of interest) and t=4 (a local minimum, inside it).

So the distance covered by the particle (путь частицы in Russian) is [s(1)-s(4)] + [s(5)-s(4)] = (-5-(-32)) + (-25-(-32)) = 27+7 = 34 and the average speed is 34/4=17/2.

In general, the average speed is the ratio of the distance to the elapsed time, and the distance is int |v(t)| dt, where v(t)=s'(t) is the (signed) velocity.

By the main theorem of calculus, int_{s1}^{s2} v(t) dt = s(t2) - s(t1); but when calculating the average speed, we need to account for change of the direction and break the integration interval down into subintervals where the particle maintains the same direction, and int |v(t)| dt = |s(t2)-s(t1)| + |s(t3)-s(t2)| +..., where t1 is the starting moment and t2, t3,... are local extrema of s(t), i.e. moments when the direction changes.

yeah i had the definition of speed wrong. It's the absolute value of the derivative of the position

anyone know of a logic calculator online that i can check my values for discrete math against? for instance i could plug in "if p then q, if r then q, therefore, if r then p" and it spits out valid / invalid.

This only gets you 99% of the way there.

http://turner.faculty.swau.edu/mathe...library/truth/

if the problem is "find area right of x = y^2 + 4y -22 and left of x=3y+8" what would you guys do? would you keep it as-is or switch the x's and y's and hope you do it right so that the correct one is ur top one. Assume u cant use a calculator for this, only pencil and paper (to get the right answer to the question).


edit: what i would do is switch every x to y and every y to x. then find the zeros of the two equations, then pick a value between the two zeros and plug it into each equation to find out which is on top and which is on bottom, then its like solving a normal problem for me. Basically my strategy is to take something that looks alien (x = y^2+... and "right to left") and turn it into something i understand (y=x^2, area below top one and above bottom one). What I'm wondering is if you guys think this is okay for me to do or if you think I'm hurting myself by doing this

Why cant i just do that integral i put in the bottom to solve this problem? (my answer is not the answer in back of book)

edit: found this somewhere:
"The radius r at any height y is given by r = sqrt[R^2-(R-y)^2]" how / where does this come from?

"To get the area though a plane that does not go through the center, you need to know the offset distance of the plane from the center of the sphere. Calling this distance d, the area is A = pi * (R^2 - d^2) [Positive values only. If the answer comes out negative, the offset distance is greater than the radius of the sphere, so the plane does not intersect the sphere] This can be derrived from a simple 2-D diagram."

how to find this simple 2d diagram?

To get the volume you can sum (integrate) the volumes of little disks with height dy starting from y=-R to y=h-R. The volume of a disk is given by pi*r^2*dy where r is the radius of the disk, this is equal to the horizontal distance from the y axis to the edge of the sphere. Which is also equal to the r in the following diagram



Do you see why this is true and can you find this r?

I see why this is true but i didnt see how you could find r. I asked someone in my clac2 class today and they explained it to me. They said that the distance from the center of the big circle to point (h,r) is R. This means you can use the similar triangles thing (b/h = b/h) to solve for r.

This is what I think i should do:
First, find x'(t). Then, set x'(t) = 2 m/s and solve it for t (this should give me the time at which the velocity is == 2m/s. So now I know the time when vx = 2.00m/s but the question wants the particle's position. So then I would take this time and plug it back into the position function to get the answer. What do you guys think? (obv, this didnt work xD)

Did you get -0.25m? What you said sounds ok. What did you get and what is the answer to see if they have a typo or they mean something else (ie if they mean speed vs velocity etc)?

yeah the way I did it and the answer you came up with were both correct. apparently the mistake i made had to do with significant figures

Can someone tell me why part (b) is wrong? Isn't it the downward force minus the projection?

And this one, if you can:

"Consider a regular tetrahedron with vertices
(0, 0, 0), (k, k, 0), (k, 0, k), and (0, k, k),
where k is a positive real number."

Find the angle between the line segments from the centroid (k/2, k/2, k/2) to two vertices. This is the bond angle for a molecule such as CH4 or PbCl4, where the structure of the molecule is a tetrahedron. (Round your answer to one decimal place.)

I'm not seeing the path to the solution.

Isnt it like 43000*cos(10deg )if the force calculated is parallel to the incline leaving as only Normal component (vertical to incline slope) the weight component.

Try dot (interior) products again of properly selected vectors. Use vertices A,B,C,D etc. Same as the other problem before.

What Masque said. But to answer specifically why what you are doing is wrong... you are doing x + y = d (and therefore y = d - x). You want to be doing the Pythagorean x^2 + y^2 = d^2 (and therefore y = sqrt[d^2 - x^2]). Vector magnitudes add like lengths in geometry.

Math noob.

Use the method of completing the square to prove that x^2+4x+6 > 0.
So I get to (x+2)^2 > -2. How to proceed?

I would simply show the minimum of that equation is greater than -2. The minimum should be obvious and easily justifiable.

I mean it's even simpler than that since anything squared is at least 0 which means it's greater than -2.

That's a really silly question/problem(on the teachers part, not your q in this thread)

Actually you get A(x)= x^2+4x+6=(x+2)^2+2 with that method that is by definition A(x)>=2 (as (x+2)^2>=0) so the initial inequality is safe.

I have a data set that I want to run a regression, but i dont expect it to behave entirely like a function (i.e. I expect to have duplicate X values return different Y values). In other words, I expect for a given X value a range of Y values will be acceptable.

My data set contains a large amount of X=0 values as my x increases i expect the range of values to trend upwards.

What are the common issues that will cause my t and p values to misstate the significance of the variable I am regressing?

Let me put some numbers down to illustrate my "ideal" data set:

200 observations of x = 0 that produce y from 5 to 10

100 observations of x = 1 that produce y from 6 to 11

50 observations of x =2 that produce y from 7 to 12

10 observations of x =3 that produce y from 8 to 13

The data starts out range bound and then the range increases by 1 for each x value. Also my observations decrease as X increases.

My actual data set is way noisier. The above ranges might be more like where I would expect my data to cluster rather than the actual ranges observed.

When I ran my regression I got statistically significant X variables, but I wanted to make sure it was OK to have the bulk of my data be around X=0.

Also, does anyone have a good book on stats that's more technical than something like Signal and the Noise but less technical than a textbook? Thanks.

x^2 + 4x + 6 = (x+2)^2 + 2 >= 0 + 2 > 0

Where the bolded >= is true because (x+2)^2 [anything squared, really] >= 0