krameriffic
Member
A challenge problem. Give yourself a pat on the back if you can do it.
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The Math Help Thread
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A challenge problem. Give yourself a pat on the back if you can do it.
hemtae
Member
SaskBoy said:
I'm not entirely sure this is right but,
when you derive the u equation, you get du/dt = sec^2(t)
so du = sec^2(t)*dt
using what I think is your initial conditions, you get dr = sec^2(t)*dt
integrate both sides and you get r = tan(t)+C
At the very least, if I was wrong I hope this helped
cpp_is_king
Member
SaskBoy said:
I'm not sure there's enough information here. Are you sure that's not dr/du(pi/4) = ...? It's really dr/du(1)? Or is it just dr/du, but not evaluated at the point u=1?
Sorry for my dumb questions, but can someone help me to understand this?:
http://www.neogaf.com/forum/showpost.php?p=27856707&postcount=14978
No solutions please, I want to solve it by myself, but I just cant understand the main concept
http://www.neogaf.com/forum/showpost.php?p=27856707&postcount=14978
No solutions please, I want to solve it by myself, but I just cant understand the main concept
mt1200 said:
In (a) you show that the whole function family [y=...] is a solution to the differential equation, meaning that any function of that family is a solution or, clearly speaking, no matter what you declare C as (say C=1 or C=127 or C=e) the resulting function is always a solution to the differential equation.
In (c) and (d) you are supposed to find a solution (= a certain function of the family) that also fulfills additional parameters (y(1)=2, y(2)=1).
If that's not clear enough:
y(1)=2 means that for x=1 y should be 2, so you end up with 2 = (ln 1 + C) / 1. Isolate C and insert it back into the initial term of the function [y=...] and you end up with the specific function of the family that is asked for.
When you take that equation, rearrange, and integrate to solve for y, every solution takes on that general form. Since you're going from y' to y, you'll wind up with an unknown constant, whose identity you can't get unless you're given more information about the original y.mt1200 said:
Zeona said:
Thanks, regarding what you said, does that mean that the solution for C is
Y = (Lnx + 2)/X ??
But how did you know that in Y(1) = 2, the x is 1 ?, from what I understand Y(1) = 2 means that "Y is 2 at the instant 1, or second 1"
Orayn said:
I havent touched integration for solving DE's yet, as I said, i'm on page 571 of the Stewart's book, 6th edition, and I wont advance until I complete all the exercises and I fully understand the DE concept.
TimesEunuch
Member
SaskBoy said:
You don't need to know (and from the information given there is no way to know) r(t). You just need to use the chain rule on the function r(u(t)): for a particular value t=t0 it says that dr/dt(t0) = dr/du(u(t0)) * du/dt(t0). Evaluating at t0=pi/4 (so u(t0)=tan(pi/4)=1) will give you dr/dt(pi/4)=dr/du(1)*sec^2(pi/4) = 2*dr/du(1).
Just beware that Leibniz notation (dr/du and so on) is somewhat ambiguous for expressing results in calculus. What you're really doing is considering the function r o u, where "o" denotes composition of functions. Then the precise expression of the chain rule is that (r o u)'(t0) = r'(u(t0)) * u'(t0), where ' denotes derivative.
Hello, I'm having a little trouble solving this 2nd order ODE:
y'' + 4y = 2cos(2x) + sin(x)
Now, I have found the CF solution:
y=C1*cos(2x) + C2*sin(2x) (C1 and C2 constants)
But my trouble comes with finding the PI solution. I'm not sure what to try...
y=x(A1*cos(2x) +A2*sin(2x)) + B1*sin(x) + B2*cos(x) (A1, A2, B1, B2 constants to be determined)
Is my first instinct but it seems a bit overcomplicated.
Thanks.
y'' + 4y = 2cos(2x) + sin(x)
Now, I have found the CF solution:
y=C1*cos(2x) + C2*sin(2x) (C1 and C2 constants)
But my trouble comes with finding the PI solution. I'm not sure what to try...
y=x(A1*cos(2x) +A2*sin(2x)) + B1*sin(x) + B2*cos(x) (A1, A2, B1, B2 constants to be determined)
Is my first instinct but it seems a bit overcomplicated.
Thanks.
cpp_is_king
Member
MLH said:
When f(x) contains M sin(Nx) or M cos(Nx), you want to try a PI of the form A sin(Nx) + B cos(Nx)
2 cos(2x) --> A sin(2x) + B cos(2x)
sin(x) --> C sin(x) + D cos(x)
So you should try something of the form A sin(2x) + B cos(2x) + C sin(x) + D cos(x)
Similar to what you came up with, but why did you multiply the first part by an x surrounding the 2 terms?
Anyway, suppose your PI is called Y(x)
Y(x) = A sin(2x) + B cos(2x) + C sin(x) + D cos(x)
Y'(x) = 2A cos(2x) - 2B sin(2x) + C cos(x) - D sin(x)
Y''(x) = -4A sin(2x) - 4B cos(2x) - C sin(x) - D cos(x)
Substituting these into the original DE, we get:
Y''(x) + 4 Y'(x) = 2cos(2x) + sin(x)
[-4A sin(2x) - 4B cos(2x) - C sin(x) - D cos(x)] + 4 [2A cos(2x) - 2B sin(2x) + C cos(x) - D sin(x)] = 2cos(2x) + sin(x)
Let's simplify now.
-4A sin(2x) - 4B cos(2x) - C sin(x) - D cos(x) + 8A cos(2x) - 8B sin(2x) + 4C cos(x) - 4D sin(x) = 2cos(2x) + sin(x)
(8A - 4B)cos(2x) - (4A + 8B)sin(2x) - (C + D)sin(x) + (C - D)cos(x) = 2cos(2x) + sin(x)
(8A - 4B - 2)cos(2x) - (4A + 8B)sin(2x) - (C + D - 1)sin(x) + (C-D)cos(x) = 0
So now we can get some simultaneous equations:
System 1
4A - 2B - 1 = 0
4A + 8B = 0 (A = -2B)
4(-2B) - 2B = 1
-10B = 1
B = -1/10
A = 1/5
System 2
C + D - 1 = 0
C - D = 0 (C = D)
2D = 1
D = 1/2
C = 1/2
A = 1/5
B = -1/10
C = 1/2
D = 1/2
Thus, PI = sin(2x)/5 - cos(2x)/10 + sin(x)/2 + cos(x)/2
Hopefully I didn't make an error somewhere, but does this get you on the right track?
Thanks for that, maybe my notes are wrong. They say when finding the particular solution of the given inhomogeneous equation we use that
if f(x)=Acos(ax) or f(x)=Asin(ax) (A and a are known) (In my case f(x)=2cos(2x) + sin(x))
we try Y(x)=B1cos(ax)+B2sin(ax)
unless the roots of the auxiliary equation are +or- ia, (auxiliary equation is a^2 + 4=0 in my case, therefore roots are +or- 2i)
where we try Y(x)=x(B1cos(ax)+B2sin(ax)
which was why I multiplied by 'x'.
But what you have done does seem to work
if f(x)=Acos(ax) or f(x)=Asin(ax) (A and a are known) (In my case f(x)=2cos(2x) + sin(x))
we try Y(x)=B1cos(ax)+B2sin(ax)
unless the roots of the auxiliary equation are +or- ia, (auxiliary equation is a^2 + 4=0 in my case, therefore roots are +or- 2i)
where we try Y(x)=x(B1cos(ax)+B2sin(ax)
which was why I multiplied by 'x'.
But what you have done does seem to work
cpp_is_king
Member
MLH said:
It's been long enough that I actually don't recall one way or another whether that note you've written down is correct or not.
I also didn't check my answer, if you check it and it works out correctly, then I guess you can double check with your prof about the inconsistency with your notes. If my answer doesn't work, then maybe you can proceed the same way that I did but with the x in there and see if you end up with something that is still solvable for A, B, C, and D.
It's also possible that you got lucky and either method leads to the same solution.
I need some help with Conic Sections - missed class and... well, clueless.
(x-3)^2 + (y+2)^2 = 16
Identify the conic section
center
Radius
(x-5)^2/36 + (y+5)^2/64 = 1
Identify the conic section - either horizontal or vertical?
Center
Vertices
co-vertices
foci
any help addressing and solving these problems would be appreciated - have some other ones, but would look like a jackass posting.
(x-3)^2 + (y+2)^2 = 16
Identify the conic section
center
Radius
(x-5)^2/36 + (y+5)^2/64 = 1
Identify the conic section - either horizontal or vertical?
Center
Vertices
co-vertices
foci
any help addressing and solving these problems would be appreciated - have some other ones, but would look like a jackass posting.
hemtae
Member
MikeTyson said:
To help with the identification, you should probably read up on the general forms of the different conic sections, ellipses, circles, hyperbolas, and parabolas.
For the first one, you can tell that its a circle since it matches the x^2+y^2 = r^2 form. The center would be 3,-2 and the radius is 4
I can't help you much on the second one since I've forgotten the whole vertices and focus thing, but I can tell you that its an ellipse and since the major axis is parallel to the y-axis(since the denominator of the two terms is the radii) it is vertical. You can also come up with the center the same way as in a circle so it would be 5,-5
i'm doing an introductory stats course and this question has me stumped, i'm not sure where to begin with it. i am 24 and haven't done maths since i was 15 so i'm sure this is incredibly easy for most of you, however i am totally lost. any help would be greatly appreciated!
The mean level of statistics anxiety in the population of second year psychology students is 18 and the population standard deviation is 12 (this characteristic is normally distributed in the population). What is the probability of randomly selecting a sample of 30 scores that have a mean above 24?.
The mean level of statistics anxiety in the population of second year psychology students is 18 and the population standard deviation is 12 (this characteristic is normally distributed in the population). What is the probability of randomly selecting a sample of 30 scores that have a mean above 24?.
hemtae
Member
Jake. said:
It's been awhile for me too but I get .31%
I don't remember what the concept is called (Central Limit Theorem keeps popping up in my head but I don't think thats correct), but basically you need to get the normal model for your sample.
The mean for the normal curve would be the mean of the population while the standard deviation would be the standard deviation of the population divided by the square root of the sample size
close to the edge
Member
Hey, I need some help with stats in R, I need to do a two sample t-test (just the default t.test(x,y)) and I am unsure about how to interpret the results.
What does t, df and p-value mean? I've learned different variable names in my class, I'm slightly confused. :/
Code:
Welch Two Sample t-test
data: x and y
t = -0.8208, df = 196.687, p-value = 0.4127
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.3795959 0.1564744
sample estimates:
mean of x mean of y
0.9590466 1.0706074What does t, df and p-value mean? I've learned different variable names in my class, I'm slightly confused. :/
hemtae
Member
close to the edge said:
df is degrees of freedom
the p-value is the probability of the alternative hypothesis happening given that the null hypothesis is true
the t-value in this case, I think would be the difference in means
Honestly, the easiest way to do these questions is just to sub in values extremely close to the number in question. For 3-, try 2.999999, see what happens. For 3+, try 3.00000001.Heysoos said:Sorry for the dumb question, but can someone please help me understand this problem? For some reason, I'm just not getting it. Every other limit problem has been really easy, but for some reason I'm completely frozen here.
For -3, you'll need to try both -2.9999999 and -3.0000001 to see if they converge.
This is ignoring basic/advanced techniques like L'Hopital's rule, but it's almost always reliable.
Feep said:
Sorry, but I'm not quite sure if I understood you. Do you mean like, for example the first one:
lim x--> -3- (2(-2.9999999)+4)/((-2.9999999)^2+6(-2.9999999)+9) and (2(-3.0000001)+4)/((-3.0000001)^2+6(-3.0000001)+9) and using that to determine if it's negative/positive infinity?
ChocolateCupcakes
Member
Or if you have a graphing calculator type it in and see if it's positive or negative. Probably the cheapest way to do it though to be honest.Heysoos said:
harriet the spy
Member
Heysoos said:
You need to factor the polynomial in the denominator. Assuming you know how to do that, the answer is
x^2+6x+9=(x+3)^2
What happens to to that when x gets close to -3?
Well, if it's just a tiny bit less than -3, (x+3) is very small negative, but squared becomes very small positive.
The numerator becomes 2(-3)+4=-2, which is negative.
-2/(0+) is -infinity, so the limit for -3^- is -infinity.
The same reasoning applies if x is just a tiny bit more than 3.
Which means that the limit for -3^+ is also -infinity.
Since both of these limits agree, the limit for -3 is -infinity as well (if they disagreed it would not exist)
Can someone help me understand the basis of a homogeneous system. I understand the basic concepts of a basis such as (1 0 0), (0 1 0), (0 0 1) spans R3 and is linear independent so it is a basis. However I'm given a problem like
x -y +z +w =0
2x +y -3z +2w =0
and I'm lost.
I put the above system into reduce row echelon to get
1 0 -2/3 1
0 1 -5/3 0
but I don't know what to do from here.
x -y +z +w =0
2x +y -3z +2w =0
and I'm lost.
I put the above system into reduce row echelon to get
1 0 -2/3 1
0 1 -5/3 0
but I don't know what to do from here.
Ace 8095 said:
To solve a system of linearly independent equations you need the same number of equations as variables. So we have two free parameters in this system that could be anything. If you call these u and v, so z=u and w=v, then x=2/3u-v and y=5/3u. We can write this as the vector (2/3u-v,5/3u,u,v) or u(2/3,5/3,1,0)+v(-1,0,0,1). Any linear combination of these vectors will solve the problem, so it is a basis as they are linearly independent. This is the null space of the matrix, if you were just given the matrix without the =0s.
So I am asked to calculate the outward fluxes of a vector field through the four faces of a tetrahedron.
I got a negative answer. Does this mean the outward "flow" through the face is actually going inward?
Edit: I think I made a mistake so it might not be negative, but my question still stands.
I got a negative answer. Does this mean the outward "flow" through the face is actually going inward?
Edit: I think I made a mistake so it might not be negative, but my question still stands.
Jaladinozozo
Banned
can someone help me out with this problem?
(-4y-9)^2+11=21
when i work this out the solutions I get are -9+sqrt(10)/4, and -9-sqrt(10)/4, however my webwork application keeps telling me this is wrong....
(-4y-9)^2+11=21
when i work this out the solutions I get are -9+sqrt(10)/4, and -9-sqrt(10)/4, however my webwork application keeps telling me this is wrong....
Jaladinozozo
Banned
lol nevermind i just got it right by playing around with how i enter the solution...stupid webwork. I had to make the 4 a negative instead of putting it infront of the 9...but I thought you cant have negatives on the denominator of an equation?
Jaladinozozo
Banned
ok I need to solve this problem by completing the square and i am completely stuck
its 7/9x^2 + 9/10x + (-11/90) = 0
any help is greatly appreciated....
its 7/9x^2 + 9/10x + (-11/90) = 0
any help is greatly appreciated....
Jaladinozozo
Banned
bump
Jaladinozozo said:
First, take the 7/9 outside of some brackets/parenthesis. So you have 7/9 [x^2 + (9/10 divided by 7/9)x + (-11/90 divided by 7/9)] and then keep going as you normally would to complete the square. Then when you get to the end, multiply by 7/9 to get rid of the square brackets/parenthesis.
Jaladinozozo
Banned
Zoibie said:
uhh this sounds like a lot of fractions lol...im like still lost, so once i have that do i do 9/10/(7/9), get that answer, than divide that by 2 and square it?
Jaladinozozo
Banned
im still kind of lost and i dont know if this looks right, im getting huge numbers where my equation so far looks like 7/9(x^2+81/70x+6561/19600) = 11/70 + 6561/19600
I believe those are the same numbers I got when I tried it. At least 6561/19600 is definitely right. It's just an ugly problem.Jaladinozozo said:
Edit: Although you need to carry the 7/9 through to the stuff on the right side as well (or divide it out of the problem entirely, as I mentioned before).
Jaladinozozo
Banned
thanks for the help, but i think im just gonna guess on a problem like this on my final, way too much freakin work, they even want the answer to the nearest hundredth...
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