Need Math Help?

[GOD!]

can someone explain to me how the distance formula is related to the arc length formula?

my textbook has something demonstrating how they used the arc length formula to find the distance between (X1, Y1) and (X2, Y2), but i don't really get why.

Alright I'll try to explain it to you. If I can't sufficiently, sorry, and if I explain what you already know, sorry about that too.

If you know what the pythagorean theorem is, the distance formula should be simple. Imagine two points on a graph.

So the distance between them must be:

The distance formula (which I assume you know) uses the pythagorean theorem and uses the fact that you know both shorter legs of the triangle (as long as you know the coordinated of each point).

From here it's just c^2 = a^ + b^2.

And of course, c=d, a=x2-x1 (or y2-y1), b=y2-y1 (or x2-x1). Easy.


Now to understand integration:
In calculus (at least at the level you'll be dealing with your whole life presumably, lol), an integral is a limit of sums. Say you have this curve.

Now you know s (arc length) = S (a to b) sqrt( 1 + f'(x)^) dx

Ok, so now on differentials: differentials (e.g. dx, dy, dA, whatever) represent tiny changes in a variable. dx is a tiny change in x, dy a tiny change in y. The slope of a graph at a point = rise over run, or dy/dx, which is vertical change over horizontal change. It's important for you to know that a differential is a tiny change in direction.

Now look at that little dx hanging outside the sqrt( 1 + f'(x)^). distribute him into the equation.

s = S (a to b) sqrt (dx^2 + dy^2)

Now its starting to look like the distance formula.

All you have to know now is that the definition of an integral is a limit of sums.

If you think about it, dx and dy are tiny distances in the x and y direction, and so are the (x2-x1) and (y2-y1). The key difference here is that when you integrate, you take the sum of each of these differentials as dx and dy become infinitely small.

If your dx and dy are larger, you'll have a graph like this:

As your dx and dy become smaller, you get graphs that look like this:

And then

As you can see, you are doing the distance formula several times: the more times you do it (and the smaller the dx and dy), the more accurate your answer will be.

So the formula for arc length is just a sum of the results of the distance formula, as your dxs and dys become infinitely tiny and you use more and more pieces. The approximation becomes better and better.

I hope that explained your question. If not I just wasted a whole lot of time doing nothing

 

[Corpsecreate]

A Cylindrical Container is filled with perfectly spherical objects of the same size. The container has radius 5.5cm and height 12cm.

If the spheres each have a radius of 0.7cm, how many spheres can fit in the container?

 

[Velox]

A Cylindrical Container is filled with perfectly spherical objects of the same size. The container has radius 5.5cm and height 12cm.

If the spheres each have a radius of 0.7cm, how many spheres can fit in the container?

-7.2



adkfjadkjf

 

[Corpsecreate]

Cool so howd you figure that out?

 

[Velox]

heh, sorry, bad joke (for some reason I think negative numbers are funny)

 

[Jane]

Show that the given value(s) of c are zeros of P(x), and find all other zeos of P(x).

P(x) = x^3 - x^2 - 11x + 15, c = 3

what im having trouble with is, im looking at the answer, and it says the other two zeros are
1 + or - sqrt 6

i dont know how to get to that answer D: help anyone? preferably by tonight cuz test is tomorrow lol

 

[MintyFlesh]

Can anybody give me tips for applications of conic sections? Most specifically hyperbolas, they seem the hardest =/

For example: Three listening stations are located in (3300,0), (3300,1100), and (-3300,0) monitor an explosion. The last two stations detect the explosion 1 second and 4 seconds after the first, respectively. Determine the coordinates of the explosion. Assume sound travels at 1100 ft per sec.

I got the C value to equal 3300ft, since half the distance between the first and third listening points is equal to that. I was able to get the A value to be 2200 ft, since I multiplied the rate (1100) by the time (4 seconds, the distance from the first and third listening points) and then divided by two. From there I can get the equation of the hyperbola, but then I'm stuck.

If you could help at all, I love you

 

[terr13]

Show that the given value(s) of c are zeros of P(x), and find all other zeos of P(x).

P(x) = x^3 - x^2 - 11x + 15, c = 3

what im having trouble with is, im looking at the answer, and it says the other two zeros are
1 + or - sqrt 6

i dont know how to get to that answer D: help anyone? preferably by tonight cuz test is tomorrow lol

Since we know 3 is an answer, then we can do synthetic division and we get
x^2 +2x - 5. Then we just solve the equation x^2 +2x -5 using quadratic formula or completing the square. and we get 1+- sqrt(6).

 

[Jane]

yeah, i should have used the completing the square method

thanks :D

 

[Corpsecreate]

Can anybody give me tips for applications of conic sections? Most specifically hyperbolas, they seem the hardest =/

For example: Three listening stations are located in (3300,0), (3300,1100), and (-3300,0) monitor an explosion. The last two stations detect the explosion 1 second and 4 seconds after the first, respectively. Determine the coordinates of the explosion. Assume sound travels at 1100 ft per sec.

I got the C value to equal 3300ft, since half the distance between the first and third listening points is equal to that. I was able to get the A value to be 2200 ft, since I multiplied the rate (1100) by the time (4 seconds, the distance from the first and third listening points) and then divided by two. From there I can get the equation of the hyperbola, but then I'm stuck.

If you could help at all, I love you

Super tricky question this one is. I managed to get it but I didnt use any hyperbola or anything like that. Just some critical thinking and a little bit of use of the Distance between 2 points formula.

The only way that point B can hear the explosion 1s after point A and have point C hear it 4s after point A is if the x coordinate of the explosion is at 3300. Its a little hard to realise why thats the case but once you get past that, its simple. You have your x coordinate and you need to solve an equation (distance formula) involving one variable - y.

Anyway, I end up getting the explosion happening at the point (3300, -2750).

 

[MintyFlesh]

Hmm, well apparently, you create the equation based off of what I said in the previous post, and then "assume" the explosion is underneath the point (3300,1100), with the point (3300,y). From there you just substitute those into the equation of the hyperbola, in order to find y.

 

[joshthelegodude]

my skype username is joshthelegodude. pm me or call me WITH YOUR SKYPE if you need help

 

[Corpsecreate]

or post your question here...

 

[Kason Birdman]

Hey guys I got a grade 11 physics question for ya.


Stu Dent, driving with an expired license tag is travelling at 10.0 m/s (36km/h) along a street. A policeman, standing beside his motorcycle, takes 5.00 s to finish his donut, and then gives chase, accelerating at 2.00 m/s^2 (the squared is on the seconds.. I suppose you all know that though lol)

Determine
a) the time required to catch the car
b) the distance the officer travels while overtaking Stu Dent.


like wtf how does I solve this? lol. I assume I need to solve the system by setting up 2 equations.. but my physics work is not usually like this.. so I dont know how to go about this or how my teacher wants me to do it..

 

[terr13]

eh, I haven't done physics in awhile, so let's give this a shot. We know s = vt+(at^2)/2. s is position, v is initial velocity, t is time, a is acceleration. We are trying to solve for s, or for t. The equation for Stu would be
s = 10(t+5), since he gets 5 seconds head start, and no acceleration.
The police officer would be
s = t^2, since the two cancels out, and he has 0 initial velocity.

Then we have t^2-10t-50 = 0 by combining the equations, solving for t we get 5(1+-sqrt(3)). One of them is negative time, which we discard and we're left with 5+5sqrt(3), which is the time in seconds. Then we solve for s, using either equation and we get about 183 meters.

 

[Kason Birdman]

wow, that makes a lot of sense.. I get it now =)

thanks a lot!

 

[Merkuri]

This is due tonight so I would really appreciate the help.

Edit: I figured out the answer to the last question and all that's left is this.

I need to get the scale in my map down to 1 cm = ? meters. I have it down as fat as 12.5 centimeters = 1 meters, but how do I find out the meter equivalent?(it seems like it should be fairly simple but mind is too burned out to think math) please help me.

 

[Merkuri]

^^ bumped so people see my edit.

 

[ballin4life]

If 12.5 cm = 1m, then 1cm = 1/(12.5) m = .08 m

Right?

 

[Corpsecreate]

no wai man i dun beleeve u

 

[Druggedfox]

I need this before noon tomorrow, so I don't really expect a response; but hey, maybe somebody will pass by this in time? Even if its past the deadline, if you know how to solve it I'd still love to hear an answer =D

Given an n x n matrix A with real entries such that A^2 = -I. Prove the following:

(a) A is nonsingular.
(b) n is even.
(c) A has no real eigenvalues.
(d) det A = 1.

 

[ballin4life]

Here's my best shot:

(a) If nonsingular just means that matrix has an inverse, then obviously A^3 is A's inverse. (AxA^3 = A^4 = (-I)^2 = I). Note also that A^3 = -A.

(b) A is a matrix will all real coefficients, and therefore det(A) is real. Now, we also know that det(A^2) = det(-I) implies that det(A)det(A) = det(-I). Now, if n is even, then det(-I) = 1 and we get det(A) = 1 or det(A) = -1 as possible solutions. If n is odd, then det(-I) = -1 and we get det(A) = i or det(A) = -i as possible solutions, but we know that det(A) is real, so we can rule out the case where n is odd. Therefore n is even.

(c) Suppose y is a eigenvector and c is a corresponding eigenvalue. If Ay = cy, then AAy = cAy, and then we have -Iy = cAy, ie (-1/c)y = Ay. By symmetry we get (-1/c)y = cy which implies -1/c = c, ie c^2 = -1. So any eigenvalue must be nonreal. (note also that 0 can't be an eigenvalue since A is nonsingular, so we don't have to worry about c=0).

(d) From part c we know that there are two eigenvalues i and -i. Since the matrix has only real values, the characteristic polynomial has real coefficients, meaning that in order for it to have roots of i and -i, it must be (x^2+1)^(n/2), which equals ((x+i)(x-i))^(n/2). So there are n/2 roots of i, and n/2 roots of -i. When we multiply out all these eigenvalues we get (-ii)^(n/2) = 1^(n/2) = 1 (and of course the determinant is equal to all the eigenvalues multiplied together).

 

[Corpsecreate]

Man I got 91% in my Linear Algebra class and I cant follow ballin4life's answer. Why must my brain forget things!?

 

[Druggedfox]

Wow, thank you so much ballin4life. That all makes complete sense now, and you even gave an amazingly timely response. Bah you made it seem so easy, now I'm regretting not seeing the solution XD

Many thanks

 

[ballin4life]

Lol I was embarrassed because I took like 20 minutes to do these.... I'm rusty on linear algebra.

I think part d might have an easier solution, because mine relies on the fact that if (x-i)^(n/2) is a factor of the polynomial, then there must also be a factor of (x+i)^(n/2). This becomes clear because the coefficients of the resulting polynomial have to be real, and having factors of (x+i) is the only way to "cancel out" the factors of (x-i), but still it seems a bit overly complicated so I think there has to be a simpler way.

A better way to explain it might be that if i is the root of a polynomial with real coefficients, then -i has to be a root as well (pretty easy to show this by looking at the general form of a polynomial). Then we can just divide our polynomial by (x-i)(x+i) = x^2 + 1 and then if i is a root, -i must be a root as well and so on to show that for each root of i there has to be a root of -i so they have the same power when you factor the polynomial.

 

[Druggedfox]

We'll probably go through the problems at some point, so if an easier solution arises I'll be sure to let you know.

Embarrassed that it took you 20 minutes? Hahaha, at least you could solve it. I was sitting there at 3 am like "... I should be able to do this... nope, nothing."

 

[Corpsecreate]

Actually looking back at it now has my brain flowing with information again. I just dont remember how to get the characteristic polynomial.

 

[ballin4life]

Whenever I'm not sure about something in math I just go look it up on wikipedia haha.

Characteristic polynomial of a matrix A is det(A-xI) where x is our polynomial variable. The roots of the characteristic polynomial are the eigenvalues of the matrix.

 

[Druggedfox]

xI-A also works if it makes the matrix look prettier

Um for 2x2 just use the standard formula for determinants, for anything more, put it into upper triangular form and multiply the entries of the main diagonal.

Is it ringing a bell back from linear algebra days? XD

 

[[oni]LoKo]

I always use Paul's Notes when i need a quick refresher
http://tutorial.math.lamar.edu/

I don't always like Wikipedia because it tends to confuse me more @___@

 

[ballin4life]

Wikipedia is more complete.

Although sometimes that is a bad thing because every theorem is stated in like full generality, and all the consequences and properties are listed, which may be more than you need.

 

[[oni]LoKo]

Ya, in that sense it is why i prefer Paul's Notes as a quick refresher since it provides steps, examples, and references. Even has a cheat sheets that i loooooooooooovvvvee so much~

 

[Velox]

Wow, I didn't know Paul's notes were so widespread, I thought they were just a Houston thing. Yeah, they're pretty good references (I especially used it for basic Linear Algebra.)

 

[MintyFlesh]

Dumb question...

what is the domain of f(x) = tan(x)?

I understand that it is all reals excluding the odd multiples of pi/2, but how can I express that?

 

[theeboredone]

I figure it would be...

-pi/2 < x < pi/2

Or

-90 degrees < x < 90 degrees

right? I haven't seen this stuff in years, but I'm pretty sure that's right.

 

[GOD!]

Uhh I can't use latex on this computer

But its something like D : {x ϵ R : x =/= Pie + Pie/2*n }

Or the domain of the function is all x such that x is a real number that does not equal to pie + pie/2*(every integer)

Idk this could be horribly wrong but I'm pretty sure the notation is right

 

[ballin4life]

GOD! is close but I think he means Pi*(every integer) + Pi/2 so

{x ϵ R : x != (Pi)*n + Pi/2, n ϵ Z }

 

[Jane]

Halp :<

Find all solutions of the systems of equations

x^2 + y^2 = 9
x^2 - y^2 = 1

 

[Druggedfox]

x^2 + y^2 = 9
x^2 - y^2 = 1

Rearrange either equation to solve for one variable in terms of another. I arbitrarily select the second equation:

x^2 = 1 + y^2

Substitute into the first equation to get:

(1+y^2) + y^2 = 9

2y^2 +1 = 9

2y^2 = 8

y^2 = 4

y= 2, -2

Substitute that into either equation, once again I arbitrarily choose the second one:

x^2 - 4 = 1 (Either value of y gives the same result of 4)

x^2 = 5

x= Sqrt5, -Sqrt5

 

[GOD!]

GOD! is close but I think he means Pi*(every integer) + Pi/2 so

{x ϵ R : x != (Pi)*n + Pi/2, n ϵ Z }

lol fuuuuuuuu forgot how to define set of integers

lol as my math prof says, more high level math exposure = inability to do lower level math

yeah and that typo **** it.