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The Math Help Thread

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M.D said:
I have this equation that

The only instruction is to "solve the system of equations"

y=-x^2+6x
y=x+4

What exactly am I supposed to do here? Find the intersection of the parabola and the line?
Click to expand...

since you have more than one equation (one being y = -x^2 + 6x and the other being y = x+4), you have a "system of equations".

The solution to the system of equations can be thought of a few different ways:
- The first equation is a parabola and the second equation is a line. The points on their curves represent solutions to their respective equation. The solution to the system is the point where they intersect.
- The solution is also x value that satisfies both equations when plugged in. Graphically, this means the same thing as the above remark.
- The solution is also the y value that satisfies both equations when plugged in.

You can solve the system different ways. Graphically, you could plot the parabola and the line and find the point where they meet. That x value and y value where they meet is your solution--a coordinate location where the solution to each equation is the exact same. You can also solve the system analytically, i.e. using algebra.

To do that, you can arrange each equation to "y = something" format (already done for you) then you can set each equation equal to each other. The logic behind this is that since, you're looking for the solution, the y value you find by doing this will be the same for each system. Therefore y1 = y2 = y, you can set the equations equal to each other.

x + 4 = -x^2 + 6x

x^2 - 5x + 4 = 0

You can use the quadratic equation or factoring to find that this system actually has two solutions, or places where the parabola meets the line: at x = 1 and x = 4. You can plug in these x-values and use algebra to find the respective y-value they are associated with.

You will find that the system has two solutions, one at x = 1, y = 5 and one at x = 4, y = 8 or (1,5) and (4,8).

Picture:
OnSJ6e2.gif


You could also have arranged one equation into "x = something" format, and substituted that expression into the x value for the other equation.
 
M.D said:
I have this equation that

The only instruction is to "solve the system of equations"

y=-x^2+6x
y=x+4

What exactly am I supposed to do here? Find the intersection of the parabola and the line?
Click to expand...
Exactly that, but do it algebraically. The parabola and the line meet where their y-coordinates are the same, so solve x+4 = -x^2 + 6x.
 
Wow, calculus is rough

Give 4 terms (that are not 0), of the MacLaurin series of f(x)=e^(sin(x))

Now, im wondering if anyone know of an easy trick out of that. I calculated the 3rd derivative of this and it was a a pain in the ass, im looking at the work for the 4th one and i can't believe they would ask for this. There must be an easier way out right?

The answer is 1 + x + (x^2)/2 - (x^4)/8 ...
 
Buggy Loop said:
Wow, calculus is rough

Give 4 terms (that are not 0), of the MacLaurin series of f(x)=e^(sin(x))

Now, im wondering if anyone know of an easy trick out of that. I calculated the 3rd derivative of this and it was a a pain in the ass, im looking at the work for the 4th one and i can't believe they would ask for this. There must be an easier way out right?

The answer is 1 + x + (x^2)/2 - (x^4)/8 ...
Click to expand...

IIRC you take the maclaurin series of e^x and insert the maclaurin series of sinx in that.
 
Leezard said:
IIRC you take the maclaurin series of e^x and insert the maclaurin series of sinx in that.
Click to expand...

Ah i forgot to say that they say that this is an error. SUM (n=0 to infinity) of ((sin(x))^n)/n! is not a MacLaurin series, i guess its because you aren't supposed to have things like, cos, e, sin, arctan, ln in a power series since the whole goal of it is to be doable with basic arithmetics.

This is coming from Stewart's book (in french). But i can't find an answer really. I cheated and checked the 4th derivative from wolfram alpha and if i use it for the sum then i have the right answer but wow, 4th derivative of this is INSANE.
 
Buggy Loop said:
Ah i forgot to say that they say that this is an error. SUM (n=0 to infinity) of ((sin(x))^n)/n! is not a MacLaurin series, i guess its because you aren't supposed to have things like, cos, e, sin, arctan, ln in a power series since the whole goal of it is to be doable with basic arithmetics.

This is coming from Stewart's book (in french). But i can't find an answer really. I cheated and checked the 4th derivative from wolfram alpha and if i use it for the sum then i have the right answer but wow, 4th derivative of this is INSANE.
Click to expand...

Leezard is correct, you can use the series of both functions and insert them into each other. See the "first example" on Wikipedia for log(cos(x))

If it's true that exp(x) = sum(0 to inf) (x^n)/n!, then it's also true that exp(sin(x)) = sum(0 to inf) (sin(x)^n)/n!. Maybe it's not a MacLaurin series anymore, but that's just a name (which I didn't even know until today, haha)

Now you just take the series expansion of sin(x) around 0 and insert it into the series of exp(sin(x)).

What makes this question difficult and weird is that they want the first four non-zero terms, so it's hard to judge how many terms of either function you need to arrive at exactly four. Normally the goal is to get a series up to a certain power of x. I'd try with the expansion of exp(x) up to ^4 and the expansion of sin(x) up to ^3 (which is the second term). Then you just have to collect the terms with powers of x.

Your method makes sure you get exactly four, but requires much more work.
 
Buggy Loop said:
Ah i forgot to say that they say that this is an error. SUM (n=0 to infinity) of ((sin(x))^n)/n! is not a MacLaurin series, i guess its because you aren't supposed to have things like, cos, e, sin, arctan, ln in a power series since the whole goal of it is to be doable with basic arithmetics.

This is coming from Stewart's book (in french). But i can't find an answer really. I cheated and checked the 4th derivative from wolfram alpha and if i use it for the sum then i have the right answer but wow, 4th derivative of this is INSANE.
Click to expand...

No, inserting the maclaurin series of sin(x) into the maclaurin series of e^x is the trick. You don't insert sin(x) into the maclaurin series. See this site for an example:
http://calculus.seas.upenn.edu/?n=Main.ComputingTaylorSeries

edit: YianGaruga seems to have explained it more properly.
 
Thank you for that link, some nice info in there. And good explanation YianGaruga.

I'll go try it this way now.

I just hope to god this type of question doesn't end up in the exam i have on tuesday -_-
 
Linear algebra question :)

Would someone mind helping me determine if my problem is a subspace of R^2 or not? Or at least put me on the right track if I'm thinking about this problem incorrectly.

I understand the rules of a subspace:

1.) The zero vector is in W.
2.) (vector x + vector y) is in W when vector x and vector y are in W.
3.) (a * vector x) is in W when x is in W and when 'a' is any nonzero scalar.

My problem is applying these rules to a problem such as this:


Code: 
vector x = x1
           x2


W = { x: |x1| + |x2| = 0}

So this is what I'm thinking for each rule:

1.) Both x1 and x2 could be 0, which means the zero vector exists. So this is true.
2.) If both x1 and x2 are zero, then |x1| + |x2| = 0. This would also be true.
3.) If both x1 and x2 are zero, then multiplying vector x by any nonzero scalar would result in x1 = 0 and x2 = 0, which would satisfy the condition. Then this would be true as well and it would be a subspace.

I'm not sure if I'm thinking through this problem correctly or not. If any of you guys could help me out I'd really appreciate. If anything is unclear about what I've stated please let me know and I'll try to clarify.
 
Piercedveil said:
Linear algebra question :)

Would someone mind helping me determine if my problem is a subspace of R^2 or not? Or at least put me on the right track if I'm thinking about this problem incorrectly.

I understand the rules of a subspace:

1.) The zero vector is in W.
2.) (vector x + vector y) is in W when vector x and vector y are in W.
3.) (a * vector x) is in W when x is in W and when 'a' is any nonzero scalar.

My problem is applying these rules to a problem such as this:


Code: 
vector x = x1
           x2


W = { x: |x1| + |x2| = 0}

So this is what I'm thinking for each rule:

1.) Both x1 and x2 could be 0, which means the zero vector exists. So this is true.
2.) If both x1 and x2 are zero, then |x1| + |x2| = 0. This would also be true.
3.) If both x1 and x2 are zero, then multiplying vector x by any nonzero scalar would result in x1 = 0 and x2 = 0, which would satisfy the condition. Then this would be true as well and it would be a subspace.

I'm not sure if I'm thinking through this problem correctly or not. If any of you guys could help me out I'd really appreciate. If anything is unclear about what I've stated please let me know and I'll try to clarify.
Click to expand...

You need to be more careful. Make sure to differentiate x1 and x2, the components of the vector x, and x y, two distinct vectors of your set.
In particular x=(x1,x2) and y=(y1,y2)

What you need to show is that if |x1|+|x2|=0 and |y1|+|y2|=0, then x+y=(x1+y1,x2+y2) also verifies the condition, i.e. |x1+y1|+|x2+y2|=0.
Same thing for 3; if |x1|+|x2| is =0, you need to show |ax1|+|ax2|=0

A hint, though, is that it's probably easier to characterize directly what the set W is. It's a very simple set. As a bonus, try to see if the answer is different for say W={(x1,x2): |x1|+|x2|=1}
 
harriet the spy said:
You need to be more careful. Make sure to differentiate x1 and x2, the components of the vector x, and x y, two distinct vectors of your set.
In particular x=(x1,x2) and y=(y1,y2)

What you need to show is that if |x1|+|x2|=0 and |y1|+|y2|=0, then x+y=(x1+y1,x2+y2) also verifies the condition, i.e. |x1+y1|+|x2+y2|=0.
Same thing for 3; if |x1|+|x2| is =0, you need to show |ax1|+|ax2|=0

A hint, though, is that it's probably easier to characterize directly what the set W is. It's a very simple set. As a bonus, try to see if the answer is different for say W={(x1,x2): |x1|+|x2|=1}
Click to expand...

Okay, I see what you're saying about the two distinct vectors. I have a few more questions as well just to make sure what I'm doing is correct.

1.) There is nothing specifying what values x1 and x2 are allowed to be, so I am able to just pick their values in this, correct? That is how I got my zero vector in W, for example.

2.) Can I change the value as I work through the problem if needed once I establish that the zero vector exists? (if it's a case like the example I provided.) In this case, I can just say x1 = 0, x2 = 0, y1 = 0, and y2 = 0, to make my condition(s) true to | (0) | + | (0) | = 0? And then do the same thing to make sure scalar multiplication would work.

Thanks for your help. We touched on this in class and did a few examples but we didn't get in to specifics too much.

And what you said about if it was | x1 | + | x2 | = 1, this would not be a subspace because the zero vector wouldn't exist in W. Is that right?
 
I have some probability-related questions if anyone wants to help me out:

1) Suppose you observe 5,200 heads in 10,000 tosses with a coin you suspect to be biased. Find a 95% confidence interval for p, the true probability that the coin produces heads.

Now, I know how to find a confidence interval for, say, the number of times a fair coin produces heads, but I have no idea how to find a confidence interval for a probability.

2) Consider a "best of 2n-1" type playoff series, where to teams A and B play each other repeatedly until one has n wins (no ties allowed). Such a series lasts a minimum of n games and a maximum of 2n-1 games. Assume the probability that A wins a given game against B is p. For general n and p, find an exact formula for the probability that the series lasts a full 2n-1 games. (Hint: re-interpret this probability as a probability involving the n-th win, then apply an appropriate binomial formula)
 
Incandenza said:
I have some probability-related questions if anyone wants to help me out:

1) Suppose you observe 5,200 heads in 10,000 tosses with a coin you suspect to be biased. Find a 95% confidence interval for p, the true probability that the coin produces heads.

Now, I know how to find a confidence interval for, say, the number of times a fair coin produces heads, but I have no idea how to find a confidence interval for a probability.
Click to expand...
This is going to depend on your priors, taken literally. I suspect that what's really being asked is just "What is the lowest value of p that will produce at least this many heads 95% of the time and what is the highest value of p will produce at most this many heads 95% of the time?"

2) Consider a "best of 2n-1" type playoff series, where to teams A and B play each other repeatedly until one has n wins (no ties allowed). Such a series lasts a minimum of n games and a maximum of 2n-1 games. Assume the probability that A wins a given game against B is p. For general n and p, find an exact formula for the probability that the series lasts a full 2n-1 games. (Hint: re-interpret this probability as a probability involving the n-th win, then apply an appropriate binomial formula)
Click to expand...

What is the probability that 2n-2 games occur with both sides winning exactly n-1 games?
 
Gotchaye said:
This is going to depend on your priors, taken literally. I suspect that what's really being asked is just "What is the lowest value of p that will produce at least this many heads 95% of the time and what is the highest value of p will produce at most this many heads 95% of the time?"



What is the probability that 2n-2 games occur with both sides winning exactly n-1 games?
Click to expand...

Did the first one the way you described and it came out nicely to 0.51<p<0.53. Seems good enough to me. And then for the second one I ended up with (2n-2 C n-1) p^(n-1) (1-p)^(n-1), which gives the correct probability for a 7-game series so I figure that's correct too. Thanks!
 
if anyone wants something to do can you check my math problems

find volume of solid obtained by rotating given region about specified line

1. y = sqrt(25 - x^2) ; y = 0; x = 2; x = 4, about x-axis

i got approximately 98.46

2. x = y - y^2 ; x = 0; about y-axis

i got pi/30
 
It's really sad that I have to ask this, but could anyone give me any tips for this integral?

Integral of the following function with respect to x: x^10 * sqrt(x^7 + 3).

I can't do a simple sub, nor does it seem like I could use integration by parts (maybe I could, but I dunno how many iterations I'll need).

I totally forgot what partial fraction decomposition is, but I don't think it'll be helpful here.
 
Kieli said:
It's really sad that I have to ask this, but could anyone give me any tips for this integral?

Integral of the following function with respect to x: x^10 * sqrt(x^7 + 3).

I can't do a simple sub, nor does it seem like I could use integration by parts (maybe I could, but I dunno how many iterations I'll need).

I totally forgot what partial fraction decomposition is, but I don't think it'll be helpful here.
Click to expand...

I don't think a nice closed for expression exist for that one, so you should probably just solve it numerically if you have a specific integration range.
 
Kieli said:
It's really sad that I have to ask this, but could anyone give me any tips for this integral?

Integral of the following function with respect to x: x^10 * sqrt(x^7 + 3).

I can't do a simple sub, nor does it seem like I could use integration by parts (maybe I could, but I dunno how many iterations I'll need).

I totally forgot what partial fraction decomposition is, but I don't think it'll be helpful here.
Click to expand...
Distribute the x^10 into the radical for: sqrt( x^107 + 3*x^100)

Work through the integration mentally, and with scratch paper.

The integrand is now ( x^107 + 3*x^100)^(1/2)

What is that the derivative of? Take the derivative of ( x^107 + 3*x^100)^(3/2)

(3/2)*(107*x^106 + 300x^99)*( x^107 + 3*x^100)^(1/2) notice the integrand in bold?


Thus, ( [(3/2)*(107*x^106 + 300x^99)]^(-1) ) *( x^107 + 3*x^100)^(3/2) + C = int( x^10 + sqrt(x^10 + 3) )

Ugly, but do-able by hand.
 
campfireweekend said:
I'm having trouble with this problem. I know I'm supposed to integrate to get velocity position vectors but I can't seem to determine what the constants are that I get from integration
Click to expand...

After you integrate to get the velocity equation, you set t = 0 and the velocity equal to the initial values it gives you (I think its <2,3,0>) and then solve for the constant of integration.
 
Could someone help me? Calc 2

&#8747;squareroot(-x^2+2x+8dx)

(express radicand as the difference of squares)



I have the answer but do not understand the process

(9/2)arcsin((x-1)/(3))-(1/2)(x-1)squareroot(-x^2+2x+8)+C

Could someone please explain? :D
 
Tater Tot said:
Could someone help me? Calc 2

&#8747;squareroot(-x^2+2x+8dx)

(express radicand as the difference of squares)



I have the answer but do not understand the process

(9/2)arcsin((x-1)/(3))-(1/2)(x-1)squareroot(-x^2+2x+8)+C

Could someone please explain? :D
Click to expand...
I think that the radicand as the difference of squares refers to factoring the polynomial.
(-1)x^2 + 2x + 8 = (4-x)(2+x)

(4-x) = (2^2 - (sqrt(x))^2) "difference of squares"

(4-x) = (2 - sqrt(x))(2 + sqrt(x))

so the integral with the difference of squares is:

int( sqrt( (2 - sqrt(x))(2 + sqrt(x))(2+x) ) dx

I think you might be able to solve it with trig+ u-v substitution and get a better answer than what you have. Is it from a book or output from a symbolic solver? It just looks like a funky maple/mathematica output since you are coming from calc II. It doesn't seem like something from a table of integrals either. I could be wrong.

edit:
given: int( sqrt( (x-4)(x+2) ) dx

let: x+2 = u+6 so dx = du
and (-1)(x+2) = (-1)(u+6)
(-x-2) + 6 = (-u-6) + 6
(4-x) = -u

sub u for x: int( sqrt( (-u)(u+6) ) du
int( sqrt (-u^2 + 6u) ) du

6u - u^2 = (sqrt(6u) - u)(sqrt(6u) + u)

I think the difference of square would call for only two terms in the radicand,
int( sqrt( (sqrt(6u) - u)(sqrt(6u) + u) ))du
that seems a bit more palpable
 
Partial Gamification said:
Distribute the x^10 into the radical for: sqrt( x^107 + 3*x^100)

Work through the integration mentally, and with scratch paper.

The integrand is now ( x^107 + 3*x^100)^(1/2)

What is that the derivative of? Take the derivative of ( x^107 + 3*x^100)^(3/2)

(3/2)*(107*x^106 + 300x^99)*( x^107 + 3*x^100)^(1/2) notice the integrand in bold?


Thus, ( [(3/2)*(107*x^106 + 300x^99)]^(-1) ) *( x^107 + 3*x^100)^(3/2) + C = int( x^10 + sqrt(x^10 + 3) )

Ugly, but do-able by hand.
Click to expand...

I don't think that works.... Quotient rule would not get you anything near the original integrand.
 
Can we still use the Green's theorem if a curve defining a bounded region is not given in the positive orientation?

Also, I still don't quite understand what an open region is (does it not include the boundary?), and what the difference between a connected and simply-connected region is (aren't they the same thing?).
 
Kieli said:
Can we still use the Green's theorem if a curve defining a bounded region is not given in the positive orientation?

Also, I still don't quite understand what an open region is (does it not include the boundary?), and what the difference between a connected and simply-connected region is (aren't they the same thing?).
Click to expand...

If you have a negative orientation, Green's theorem should just give you the negative of what it would for a positive orientation. You can see this by reparametrizing to get the same curve with positive orientation, but it's probably more work than it's worth.

And yes, in this case open essentially just means without the boundary included. You can also think of a simply connected set as having no holes. So a large disk with a smaller one cut out from its center would be connected, as it's all one piece, but not simply connected since it has a hole. (Note all the things in this paragraph are usually defined in a much more general way, but if all you are concerned with is Green's theorem, this should be enough for you.)
 
RELAYER said:
why
Click to expand...

The numerator and the denominator have the form [ a-b ] / [ a^2 - b^2 ]. This means you can simplify the fraction if you multiply by the conjugate of a-b, since (a-b)(a+b) = a^2 - b^2.

In this case, you would use 3+&#8730;x. You multiply by [3+&#8730;x] / [ 3+&#8730;x ] = 1 instead of just 3+&#8730;x because multiplying by 1 doesn't change the value of the fraction, leaving you with 1 / [ 3+&#8730;x ] in the end.
 
I need help on this word problem for a c# class... not sure why I can't figure it out.

"The company has instituted a bonus program to give its employees an incentive to sell more. For every dollar the store makes in a four-week period, the employees receive 2 percent of sales. The amount of bonus each employee receives is based on the percentage of hours he or she worked during the bonus period (a total of 160 hours).

The screen will allow the user to enter the employee's name, the total hours worked, and the amount of the store's total sales.

The Calculate button will determine the bonus earned by this employee."

This is all the information i'm given. So basically the user inputs the hours they worked and the total amount the store made, with a constant bonus rate percentage of 2% being applied to the stores total sales (?). From this information, I need to find out the amount of bonus the employee makes.
 
Unless I misunderstood the question, you just need to find the total bonuses by multiplying the total sales by 0.02, then determine the employee's portion of the bonuses by multiplying it by the hours worked divided by total hours.
 
I'm rotating the region enclosed by

y = x; y = 0, x = 1

about y axis.

by disc method

How would I set that integral up

Just pi integral from 0 to 1 of 1 - pi integral from 0 to 1 of y dy?


actually that doesn't seem to work because it gives me a negative area

edit: figured out what was wrong
antiderivative of 1 is y
 
RELAYER said:
I'm rotating the region enclosed by

y = x; y = 0, x = 1

about y axis.

by disc method

How would I set that integral up

Just pi integral from 0 to 1 of 1 - pi integral from 0 to 1 of y dy?


actually that doesn't seem to work because it gives me a negative area
Click to expand...
A graph allways helps me.
00_shell6fubw.png

So each disc's[ reallly, each ring, its like a stack of records with a cone-shape missing from the middle] area is pi*radius^2 and that's what you are integrating.
 
yea me too the book even supplied a graph i guess i just had a d'oh moment

the thing that threw me off was one of the curves being a constant and having to integrate that.
i left it as 1 instead of actually integrating it

although now that i look at my work i'm not sure why it makes a difference since the boundaries are from 0 to 1...


could someone check the problem for me?
I got 2pi/3
 
RELAYER said:
I did this

pi*int(1^2)dy - pi*int(y^2)dy

pi(y - y^3/3)
pi (1 - 1/3)
pi (3/3 - 1/3)

2pi/3

where did i mess up
Click to expand...

That's what I did, since the outer radius is 1 and the inner radius is y.

Doing it by the shell method gives me 2pi/3 as well so I'm assuming it is the correct answer.
 
RELAYER said:
I did this

pi*int(1^2)dy - pi*int(y^2)dy

pi(y - y^3/3) <<<didn't take the difference-of-both-radii squared>>>
pi (1 - 1/3)
pi (3/3 - 1/3)

2pi/3

where did i mess up
Click to expand...
the radius^2
The radius of each disc (area of circle).
pi*(1-y)^2 as the integrand. You have the difference of the radii but you need to square the whole quantity, (1-y)^2 = 1 - 2y + y^2

pi* int( 1 - 2y + y^2 [dy]) from 0 to 1

integrated:
pi*[ y - y^2 + (1/3)y^3]_{0,1}

pi*[1-1+(1/3)(1) - 0]

pi*[1/3]

edit: did I set it up wrong isn't it the integral of the area of the discs [rings]?
 
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