# The Math Help thread :)



## pbandjam (Sep 24, 2011)

I'm in Calculus 2 right now and am really struggling. I thought maybe we could have a math help thread to help with problems, post ideas and discuss concepts. Anything math related can be posted 

So post away! :yes


----------



## Sacrieur (Jan 14, 2013)

I can assist with any problems you have.


----------



## Princu (Jun 10, 2013)

Does it have to be only Calculus related or Algebra and trignometry can too kick in?


----------



## Kekai (May 31, 2013)

This website was an invaluable asset for acing the Calc II final:

http://www.wolframalpha.com


----------



## Sacrieur (Jan 14, 2013)

Princu said:


> Does it have to be only Calculus related or Algebra and trignometry can too kick in?


I should be able to handle any mathematical question.


----------



## nullptr (Sep 21, 2012)

Kekai said:


> This website was an invaluable asset for acing the Calc II final:
> 
> http://www.wolframalpha.com


That's my secret to passing calculus


----------



## Vee87 (Jul 16, 2012)

I need help with multiple integration


----------



## pbandjam (Sep 24, 2011)

Vee87 said:


> I need help with multiple integration


Could you either type your problem or post a picture of it?


----------



## Vee87 (Jul 16, 2012)

pbandjam said:


> Could you either type your problem or post a picture of it?


Find the volume of the region that lies inside z = x^2 + y^2, and
below z = 16

I know r = 4, so i guess my limits would be 0 <= r <= 4 ? I need help with determining the limits of integration for theta! :|


----------



## Tensor (Mar 9, 2013)

Vee87 said:


> Find the volume of the region that lies inside z = x^2 + y^2, and
> below z = 16
> 
> I know r = 4, so i guess my limits would be 0 <= r <= 4 ? I need help with determining the limits of integration for theta! :|


So you're integrating










where D is a disk of radius 4. The area element dA in polar coordinates is r dr dθ and z = x² + y² = r² (independent of θ), so you end up with


----------



## Princu (Jun 10, 2013)

Alright Geniuses.
Solve this problem for me.

1.Let P denotes the number of ways of selecting 3 people out of n people sitting in a row if no two of them are consecutive and Q is the corresponding figure when they are in a circle.If (P-Q)=6.Find n.


----------



## Ape in space (May 31, 2010)

Princu said:


> Alright Geniuses.
> Solve this problem for me.
> 
> 1.Let P denotes the number of ways of selecting 3 people out of n people sitting in a row if no two of them are consecutive and Q is the corresponding figure when they are in a circle.If (P-Q)=6.Find n.


Every valid way of selecting 3 from the circle will also be a valid way of selecting 3 from a row, if you imagine breaking the circle at a specific place and making it into a row. But not every valid way of selecting from the row will be a valid way of selecting from the circle, because it could happen that you have selected one from each end of the row, and so when you fold it into a circle those two ends will be next to each other. So (P-Q) will be the number of valid ways of selecting from the row that are not valid when you fold it into a circle - i.e. the number of ways of selecting from the row such that you select one from each end.

To find this, you select the ones at the ends, and then ask how many ways are there of picking the third person. Since the third person can't be sitting next to the two already picked, and you obviously can't re-select the two you've already picked, that leaves n-4 possible choices.

So P - Q = n - 4.
We're given that P - Q = 6, so that means n - 4 = 6, so *n = 10*.


----------



## pbandjam (Sep 24, 2011)

Allright guys I need help :/

Set up, but do not evaluate, an integral for the area of the surface obtained by rotating the curve about the given i) x axis ii) y axis

y = tan x; 0 <= x <= pi/3

So I was able to do the x axis part.

i) y = tan x, y' = sec^2(x)

1 + (sec^2(x))^2

L = 2pi ∫0 to pi/3 tanx √ 1 + sec^4(x)


Now how do I do the y axis?


----------



## Princu (Jun 10, 2013)

Ape in space said:


> Every valid way of selecting 3 from the circle will also be a valid way of selecting 3 from a row, if you imagine breaking the circle at a specific place and making it into a row. But not every way of selecting from the row will be a valid way of selecting from the circle, because it could happen that you have selected one from each end of the row, and so when you fold it into a circle those two ends will be next to each other. So (P-Q) will be the number of ways of selecting from the row that are not valid when you fold it into a circle - i.e. the number of ways of selecting from the row such that you select one from each end.
> 
> To find this, you select the ones at the ends, and then ask how many ways are there of picking the third person. Since the third person can't be sitting next to the two already picked, and you obviously can't re-select the two you've already picked, that leaves n-4 possible choices.
> 
> ...


That's the sickest explanation for this question.Too good that you did it without making any hassle.My teachers did it by some other way which I did not got.This was the best.
Thank you Sir.Are you a teacher or something?


----------



## Vee87 (Jul 16, 2012)

how did you make the cool graphics?


----------



## Princu (Jun 10, 2013)

Please don't let this thread die..
1.Total no of outcomes when N identical dices are rolled?.
Please Help


----------



## nullptr (Sep 21, 2012)

Princu said:


> Thank you Sir.Are you a teacher or something?


He be a physics master.



Princu said:


> Please don't let this thread die..
> 1.Total no of outcomes when N identical dices are rolled?.
> Please Help


Oh this is an easy question, the number of possibilities for a die is S^N
where S is the number of sides and N is well you know. If it's 6 sided then 6^N/


----------



## s0mebody (Mar 30, 2013)

Hey guys, I need help with my calculus too.

Apparently, I need to find x from this equation:
4x^3 - 6x^2 + 1 = 0

I have no idea what to do. Been staring at this for hours. Hope someone can help me.


----------



## Remnant of Dawn (Feb 22, 2012)

s0mebody said:


> Hey guys, I need help with my calculus too.
> 
> Apparently, I need to find x from this equation:
> 4x^3 - 6x^2 + 1 = 0
> ...


I mean I don't see any way to do this with calculus, unless you were using Newton's Method or something. This seems to be just an annoying case of factoring. In which case I would just try all the possible rational roots (plus or minus 1, 1/2, 1/4), find one of them (1/2 in this case), and use it to realize that (2x - 1) must be a factor of the original equation. Then you can use polynomial long division to factor the original equation into (2x-1)(2x^2-2x-1) and use the quadratic formula to get the other 2 roots of 0.5(1-sqrt(3)) and 0.5(1 + sqrt(3)).

I feel like there should be a better way, but I can't think of one.


----------



## nullptr (Sep 21, 2012)

s0mebody said:


> Hey guys, I need help with my calculus too.
> 
> Apparently, I need to find x from this equation:
> 4x^3 - 6x^2 + 1 = 0
> ...


I tried factoring it for a while and failed, and because I was too lazy to plug it into the quadratic equation I just plugged it into wolfram alpha

http://www.wolframalpha.com/input/?i=find+x+in+4x^3-6x^2+1=0


----------



## lostinlife (Jun 2, 2010)

Remnant of Dawn said:


> I mean I don't see any way to do this with calculus, unless you were using Newton's Method or something. This seems to be just an annoying case of factoring. In which case I would just try all the possible rational roots (plus or minus 1, 1/2, 1/4), find one of them (1/2 in this case), and use it to realize that (2x - 1) must be a factor of the original equation. Then you can use polynomial long division to factor the original equation into (2x-1)(2x^2-2x-1) and use the quadratic formula to get the other 2 roots of 0.5(1-sqrt(3)) and 0.5(1 + sqrt(3)).
> 
> I feel like there should be a better way, but I can't think of one.


Rational roots theorem is the best way. If you use Newton's method, you would still have to use a calculator to locate a "best" x0 to start from as well as do the actual computations. If you want to factor, it's going to be a method from algebra or precalculus. Calculus really doesn't cover factoring methods.


----------



## lostinlife (Jun 2, 2010)

pbandjam said:


> Allright guys I need help :/
> 
> Set up, but do not evaluate, an integral for the area of the surface obtained by rotating the curve about the given i) x axis ii) y axis
> 
> ...


Surface area formula for rotation about the y-axis is:
2pi∫x*ds for [a, b]

ds can be either:
sqrt(1 + (dy/dx)^2) dx for x = a to x = b
or
sqrt(1 + (dx/dy)^2) dy for y = a to y = b

Here you are given a function of x and x bounds, so sqrt(1 + (dy/dx)^2) is the easiest to use.

2pi∫x*sqrt(1 + (secx)^4) dx for [0, pi/3]


----------



## Princu (Jun 10, 2013)

galacticsenator said:


> He be a physics master.
> 
> Oh this is an easy question, the number of possibilities for a die is S^N
> where S is the number of sides and N is well you know. If it's 6 sided then 6^N/


Sorry Sir but you misunderstood the question.Here the dices are identical not distinct hence the total no of outcomes will decrease drastically..
Suppose there are 3 identical dice,then the outcomes-
(1,2,3) is same as (3,2,1) or (1,3,2).Hope you understand what I am trying to day.Think over it for some time and then tell.


----------



## s0mebody (Mar 30, 2013)

Remnant of Dawn said:


> I mean I don't see any way to do this with calculus, unless you were using Newton's Method or something. This seems to be just an annoying case of factoring. In which case I would just try all the possible rational roots (plus or minus 1, 1/2, 1/4), find one of them (1/2 in this case), and use it to realize that (2x - 1) must be a factor of the original equation. Then you can use polynomial long division to factor the original equation into (2x-1)(2x^2-2x-1) and use the quadratic formula to get the other 2 roots of 0.5(1-sqrt(3)) and 0.5(1 + sqrt(3)).
> 
> I feel like there should be a better way, but I can't think of one.


This is what I need! The problem is actually finding absolute extrema and the equation I gave was at the step of setting f'(x) = 0. I needed the values of x to know the critical numbers. Thank you for helping!

Thanks galactic senator or that site too! That would be helpful in the future.


----------



## Remnant of Dawn (Feb 22, 2012)

Princu said:


> Please don't let this thread die..
> 1.Total no of outcomes when N identical dices are rolled?.
> Please Help


I think it's (n + 5) choose n. If you imagine each of the possible values of the dice as a bin, then you are putting n indistinguishable dice rolls into 6 distinguishable bins. Then there's a formula for this which I can't seem to remember, but if you imagine you have 5 dividers splitting up the 6 bins, and n dice rolls to be placed into these bins, then you have 5 dividers and n rolls that's n + 5 items, and by finding all possible ways to pick the n rolls out of these items (and thus let the remaining ones be dividers), you're finding all the ways of partitioning n rolls into 6 bins. This is (n + 5) choose n or (n + 5) choose 5.

I guess more generally it would be (n + s - 1) choose n, where s is the number of sides on the dice. This might not be right though.

Also, this is a problem I ran across while studying that has had me stumped for hours. If anyone could help me with it I would be eternally grateful.

Let R be a symmetric relation. Show that R^ is symmetric for all positive integers n.


----------



## nullptr (Sep 21, 2012)

Princu said:


> Sorry Sir but you misunderstood the question.Here the dices are identical not distinct hence the total no of outcomes will decrease drastically..
> Suppose there are 3 identical dice,then the outcomes-
> (1,2,3) is same as (3,2,1) or (1,3,2).Hope you understand what I am trying to day.Think over it for some time and then tell.


Ah sorry I guess I did misread the question.


----------



## Vee87 (Jul 16, 2012)

Can't figure out how to find the constants in this differential equation

y'' + y' - 2y = 0

y1 (0) = 1
y'1 (0) = 0

y2 (0) = 0
y'2 (0) = 1


never seen the initial values separated like that, my c values are -1 and 1/2. But it seems like no one can agree on one set way to work this problem :/


----------



## lostinlife (Jun 2, 2010)

Vee87 said:


> Can't figure out how to find the constants in this differential equation
> 
> y'' + y' - 2y = 0
> 
> ...


characteristic equation: r^2 + r - 2 = 0
(r - 1)(r +2) = 0
r = 1, -2

general solution: y(t) = a*e^(t) + b*e^(-2t)
derivative of general solution: y ' (t) = ae^(t) - 2be^(-2t)

I think you mean you need to solve for a and b given each y(t) and y ' (t) case. So you should get 2 equations at the end.

y1: 
y1(0) = a*e^(0) + b*e^(-2*0) = 1
a + b = 1
y1 ' (0) = a*e^(0) - 2b*e^(-2*0) = 0
a - 2b = 0
a = 2/3, b = 1/3
specific solution: y(t) = (2/3)e^(t) + (1/3)e^(-2t)

y2: 
y2(0) = a*e^(0) + b*e^(-2*0) = 0
a + b = 0
y2 ' (0) = a*e^(0) - 2b*e^(-2*0) = 1
a - 2b = 1
a = 1/3, b = -1/3
specific solution: y(t) = (1/3)e^(t) + (-1/3)e^(-2t)


----------



## Vee87 (Jul 16, 2012)

lostinlife said:


> characteristic equation: r^2 + r - 2 = 0
> (r - 1)(r +2) = 0
> r = 1, -2
> 
> ...


Thanks! Can't believe he put a problem like this on the test, we have never done one like this in class. This kind of problem isn't even in the chapters practice :|


----------



## Slumknox (Feb 25, 2013)

I know this integration problem is simple, but I cannot figure out how to derive at 2/3 (the answer).


----------



## lostinlife (Jun 2, 2010)

Slumknox said:


> I know this integration problem is simple, but I cannot figure out how to derive at 2/3 (the answer).


U-sub will work since it's function within a function.
u = x + 2
du = dx

integral[2/(x + 2)^2 dx] becomes
integral[2/u^2 du]
= integral[2u^(-2) du]
= -2u^(-1)
= -2(x + 2)^(-1)
= -2/(x + 2)

Plugging in bounds [0, 4]:
= -2/(4 + 2) - (-2/(0 + 2))
= -2/6 + 2/2
= -1/3 + 1
= 2/3


----------



## Slumknox (Feb 25, 2013)

lostinlife said:


> U-sub will work since it's function within a function.
> u = x + 2
> du = dx
> 
> ...


Completely gaped sub method. Thank you so much!


----------



## lostinlife (Jun 2, 2010)

Slumknox said:


> Completely gaped sub method. Thank you so much!


No problem  Rule of thumb is if it's not straight x^(n+1) / (n+1) rule, try u-sub next. If that doesn't work, try int by parts. Other methods that it might be if those don't work are partial fractions or trig substitution.


----------

