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[boviscopophobic]

Hope people don't mind me asking here, but I'm really quite desperate on this problem and I really would appreciate some help to acquire the marginal densities (once I have that I can do it fine, but in this particular example I don't understand how to get them):

It looks like you misread the parenthesization in the integrand -- the way you have it is pretty nasty. Try this. Also feel free to insert "for x > 0" where I have the expression for the marginal density.

EDIT for typo correction: "unnormalized exponential density with mean 1/(1 + theta * x)"

Also, a related question, for the marginal distribution, in an instance such as this, do we need to integrate over all possible values of theta as well, or is it just integrating over the random variables?

Here theta is a fixed parameter so it doesn't make sense to integrate over it. In order for that to make sense, you would need a joint distribution on (X, Y, theta).

 

[Majestad]

Any of you guys willing to help me on this short question:

I did the steps needed for implicit differentiation, but my answer came wrong. Don't even know how to put this into WolframAlpha to see what I might have missed.

 

[Jospina]

How can I integrate e^(x^2)dxdy, my textbook says it can't be integrated because the integrand is not an elementary function, but I have a problem where I'm asked to take the double definite integral of it and reverse the order of integration. Before I can do any of that I need to be able to find the indefinite integral first. Wolfram produces some weird non-real answer.

 

[kgtrep]

Any of you guys willing to help me on this short question:

It's good to check that the given point satisfies the given equation (in case there is a typo in the problem). Here it does, so let's continue.

Differentiate both sides with respect to x. Note that we can do so because the equation holds for all x and y such that x =/= 6y, and the point at which we plan to evaluate the derivative is "far away" from this line.

We get,

( (x - 6y) * dy/dx - y * (1 - 6*dy/dx) ) / (x - 6y)^2 = 9x^8.

We notice that -6y * dy/dx and + 6y * dy/dx cancel each other. That's nice.

Substitute x = 1 and y = 4/25, and solve for dy/dx evaluated at the point.

 

[kgtrep]

t

How can I integrate e^(x^2)dxdy, my textbook says it can't be integrated because the integrand is not an elementary function, but I have a problem where I'm asked to take the double definite integral of it and reverse the order of integration. Before I can do any of that I need to be able to find the indefinite integral first. Wolfram produces some weird non-real answer.

Hi there. Can you be more specific about the problem you're solving? What are the limits of the integration? Did you mean to say exp(-(x^2 + y^2)) instead of exp(x^2)?

 

[Biggest-Geek-Ever]

How can I integrate e^(x^2)dxdy, my textbook says it can't be integrated because the integrand is not an elementary function, but I have a problem where I'm asked to take the double definite integral of it and reverse the order of integration. Before I can do any of that I need to be able to find the indefinite integral first. Wolfram produces some weird non-real answer.

You do not necessarily need to know the antiderivative of e^(x^2) to such a double integral. This is a typical problem where changing the order integral will, after changing the limits appropriately and integrating with respect to y, will leave you something like ∫ x e^(x^2) dx, which does have an elementary antiderivative. To be able to help you further, we'll have to know what region you are integrating over (ie, what the limits of the double integral are).

 

[Jospina]

Here it is:

 

[Biggest-Geek-Ever]

Here it is:

You are first integrating from x = 8y (which is the same as y = 1/8x) to x = 8. Think of this as dragging a pencil from the x = 8y to the vertical line x = 8. Now, the limits of integration for the second half of the interated integral are from y = 0 to y = 1. This is then like drawing every line from x = 8y to x = 8 over the interval 0 <= y <= 1. This gives the shaded region:

This is the region bounded by the lines x = 8y, y = 0, and x = 8 This is the region that you are integrating e^(x^2) over, and is given by the iterated integral given in the problem:

To solve this problem, you must switch the order of integration to dydx, while ensuring the region you are integrating remains the same. When we integrated with respect to x first, we thought of drawing a horizontal line from x = 8y to x = 8. Now, we want to integrate with respect to y first, so we'll be thinking of drawing a vertical line. This vertical line will go from y = 0 to y = 1/8 x. To then fill up the region, we would consider drawing every line of this manner for the interval 0 <= x <= 8. So our new limits of integration will result in the iterated integral:

Let me know if you need any further explanation. A (sort of) quick way to check if you've correctly changed the limits of your iterated integral is to compare them when integrating over the constant function 1, i.e. &#8747;&#8747; 1 dxdy versus &#8747;&#8747; 1 dydx, with the proper limits of integration. In the case of this problem, we get 4, which is is course the area of the region.

 

[The One Who Knocks]

It looks like you misread the parenthesization in the integrand -- the way you have it is pretty nasty. Try this. Also feel free to insert "for x > 0" where I have the expression for the marginal density.

http://i.imgur.com/bQdaN1E.png[img]



Here theta is a fixed parameter so it doesn't make sense to integrate over it. In order for that to make sense, you would need a joint distribution on (X, Y, theta).[/QUOTE]


I just want to say thank you so much, I think you are correct that I misread the bracketing of the terms, I assumed (as in the picture) that it's all to the power of e, but it makes much more sense, and is far more manageable, if it is, as you've shown, b*e(^-x-y-axy). That also makes much more sense in light of the hint that we have too. Thank you so much for your help, I definitely wouldn't have copped that or to then use the expected value and normalisation to 'resolve' the integrals (it's not as bad to integrate it out like that, but using the related it has to the expected value does make it nicer undoubtedly).

Thank you and to luoapp for taking your time to help with this, it's really been driving me mad since the tenth and I definitely wouldn't have been able to catch on to my misinterpretation of the function given.

 

[kgtrep]

You are first integrating from x = 8y (which is the same as y = 1/8x) to x = 8. Think of this as dragging a pencil from the x = 8y to the vertical line x = 8. Now, the limits of integration for the second half of the interated integral are from y = 0 to y = 1. This is then like drawing every line from x = 8y to x = 8 over the interval 0 <= y <= 1. This gives the shaded region:

This is the region bounded by the lines x = 8y, y = 0, and x = 8 This is the region that you are integrating e^(x^2) over, and is given by the iterated integral given in the problem:

To solve this problem, you must switch the order of integration to dydx, while ensuring the region you are integrating remains the same. When we integrated with respect to x first, we thought of drawing a horizontal line from x = 8y to x = 8. Now, we want to integrate with respect to y first, so we'll be thinking of drawing a vertical line. This vertical line will go from y = 0 to y = 1/8 x. To then fill up the region, we would consider drawing every line of this manner for the interval 0 <= x <= 8. So our new limits of integration will result in the iterated integral:

Let me know if you need any further explanation. A (sort of) quick way to check if you've correctly changed the limits of your iterated integral is to compare them when integrating over the constant function 1, i.e. &#8747;&#8747; 1 dxdy versus &#8747;&#8747; 1 dydx, with the proper limits of integration. In the case of this problem, we get 4, which is is course the area of the region.

Just wanted to say, that was an excellent explanation you gave! I really liked your pencil analogy and checking the integration with a constant function.

 

[LightOfTruth]

Almost. A cylindrical Jacobian has an extra r in it.

For part 2, there has to be two parts to the integral. First part is r from 2 to 4, and theta from 0 to pi. Second part is r from 2 to -2/sin(theta) and theta from pi to 2*pi.

Sorry, I didn't see your reply. But yes, I think I remember that. Here it is fixed:

OK, I tried to work out how you got part 2 (see below for my interpretation of your post). Did you use x=pcos(theta) and y=psin(theta)? And then I'm assuming you set x=0 so then p=0. And then you plugged in -2 for y and solved accordingly for p again. Why can we ignore the z component here?

Also, would you agree that in context of the original problem, c (radius from the origin to the center of the torus tube) must always be 3 and a (radius of the donut tube) must always be 1?

 

[Kieli]

GAF, may I beseech thine wisdom?

Can I assume the following are true?

If I have the heat equation with u_t = c^2* u_xx for some position 0 < x < L.
I presume a solution of the form u(x,t) = X(x)T(t)

(1) Dirichlet B.C. ----> The Eigenfunctions are X_n = sin(n*pi/L * x)
(2) Neumann B.C. ----> The Eigenfunctions are X_n = cos(n*pi/L * x)
(3) Robin B.C. ----> The Eigenfunctions are X_n = cos(n*pi/L * x) + sin(n*pi/L * x)

In all three cases, the eigenvalues are lambda_n = (n*pi/L)^2

Given the initial condition u(x, 0) = f(x), to find the coefficients of the solution u(x,t) = a_0/2 + summation from zero to infinity of [ a_n * cos(n*pi/L * x) + b_n * sin(n*pi/L * x) ] * T(t)

a_n = 2/L * integral from 0 to L of f(x)cos(n*pi/L * x)

b_n = 1/L * integral from -L to L of f(x)sin(n*pi/L * x)

-------------------------------

Right now, I don't quite understand separation of variables and Fourier series enough to be able to come up with these, but if I can memorize them, then I think I can do the questions and pass the midterm.

 

[kgtrep]

GAF, may I beseech thine wisdom?

Can I assume the following are true?

If I have the heat equation with u_t = c^2* u_xx for some position 0 < x < L.
I presume a solution of the form u(x,t) = X(x)T(t)

(1) Dirichlet B.C. ----> The Eigenfunctions are X_n = sin(n*pi/L * x)
(2) Neumann B.C. ----> The Eigenfunctions are X_n = cos(n*pi/L * x)
(3) Robin B.C. ----> The Eigenfunctions are X_n = cos(n*pi/L * x) + sin(n*pi/L * x)

In all three cases, the eigenvalues are lambda_n = (n*pi/L)^2

Given the initial condition u(x, 0) = f(x), to find the coefficients of the solution u(x,t) = a_0/2 + summation from zero to infinity of [ a_n * cos(n*pi/L * x) + b_n * sin(n*pi/L * x) ] * T(t)

a_n = 2/L * integral from 0 to L of f(x)cos(n*pi/L * x)

b_n = 1/L * integral from -L to L of f(x)sin(n*pi/L * x)

-------------------------------

Right now, I don't quite understand separation of variables and Fourier series enough to be able to come up with these, but if I can memorize them, then I think I can do the questions and pass the midterm.

I think you have the right idea. We should assume that everything is nice enough such that we can use Fourier series and separable functions. (I'm not sure what the exact necessary conditions are.)


Math people like the set of sines and the set of cosines. The functions in each set are mutually orthogonal in L^2 norm (the integral that you took) and are orthogonal with any function in the other set.

That is, for any integers m and n (different from m),

\int_{0}^{2*pi} sin(m*x) * sin(n*x) dx = 0
\int_{0}^{2*pi} cos(m*x) * cos(n*x) dx = 0
\int_{0}^{2*pi} sin(m*x) * cos(n*x) dx = 0

Together with the constant function 1, the sines or the cosines form a basis for L^2 space (where the nice functions live). Hence, we can write a smooth function u = u(x) as a linear combination of the sine or cosine basis functions:

u = a0 * 1 + a1 * sin(1*x) + a2 * sin(2*x) + ...

We then multiply both sides of the equation by 1, or sin(1*x), or sin(2*x), and so on, and integrate both sides from 0 to 2*pi. This gives us the corresponding coefficient a0, or a1, or a2, and so on.


For the heat equation, where u is supposed to depend on both position x and time t, we probably don't have a clue about what u looks like.

So instead, we are going to approximate u as a sum of separable functions. Separable functions are nice because they may simplify a partial differential equation into two ordinary differential equations (one in x and one in t).

And of course, the idea is that, the more terms that we consider in the sum of separable functions, the better our approximation will be.

 

[ibyea]

Sorry, I didn't see your reply. But yes, I think I remember that. Here it is fixed:

OK, I tried to work out how you got part 2 (see below for my interpretation of your post). Did you use x=pcos(theta) and y=psin(theta)? And then I'm assuming you set x=0 so then p=0. And then you plugged in -2 for y and solved accordingly for p again. Why can we ignore the z component here?

Also, would you agree that in context of the original problem, c (radius from the origin to the center of the torus tube) must always be 3 and a (radius of the donut tube) must always be 1?

Sorry, I didn't mention the z part, which is the same. For the second term, the order of integration is then z, r, and then theta. Also, yes, c=3 and a=1.

 

[Kieli]

snip

Thanks a lot! I definitely appreciate the intuitive explanation of separation of variables and the need for Fourier series.

I don't quite have the mathematical maturity to understand what L^2 space means (I have a very, very rudimentary understanding of spaces of vectors or functions).

----------------

May I piggy-back and ask one more question?

Suppose you wish to find the power series solution to a non-linear second order ODE at a regular singular point, x_0.

After finding the indicial equation, we obtain two roots.

There are 3 scenarios:

1: The difference between the two is not zero and not a positive integer --> this case is easy, no problemo

2: The two roots are the same.

3: The difference between the two roots is a positive integer.

In all three cases, finding one solution using the larger of the two roots is easy.

I'm having trouble understanding how to find the second linearly independent solution.

Suppose y_1 is the solution obtained using the bigger root.

Then in case 2:
y_2 = y_1 * ln(x - x_0) + sigma from n = 1 to infinity of b_n(x-x_0)^(n+r) where b_n are coefficients of the power series, r is the only root of the indicial equation.

I wonder how do you find b_n?

In my notes, I see that b_n is simply the derivative with respect to r of a_n (which is the sequence of coefficients in the solution, y_1).

How do you take the derivative of a_n? Does that mean taking the derivative of the recurrence relation?

E.g. a_(n+2) = a_(n+1)/(n+r-1) is a recurrence relation. So we just take the derivative of this with respect to r?

How about case 3?

 

[MisterLuffy]

Math gaffers, I have two problems I need help on, and a third one that I might post later:

At a NASA research center free-fall research is performed by dropping experiment packages from the top of an evacuated shaft 146 m high. Free-fall imitates the so-called microgravity environment of a satellite in orbit. Hint: In this problem, all the motion is downward, so make the top y=0, and have the positive y-axis pointing downward.

1) What is the maximum time interval for free fall if an experiment package were to fall the entire 146 m? Answer: 5.46 s
2) Actual NASA specifications allow for a 5.27 s drop time interval. How far do the packages drop in 5.27 s? Answer: 136.35 m
3) What is their speed at 5.27 s? Answer: 51.72 m/s
4) NASA does not want the experiment package to slam into the ground and get smashed to pieces. They fill the bottom of the shaft with lightweight foam to break the fall. What constant acceleration would be required to gently stop an experiment package in the distance remaining in the shaft after its 5.27 s fall?

For part 4, I used this formula: a = (Vf)^2 - (Vi)^2 / 2D where D should be the distance of the dropped package. So, a = (0 m/s)^2 - (51.72 m/s)^2 / 2(136.35) = -9.81 m/s^2 but the answer is wrong. I may have either used the wrong formula or plugged in the numbers incorrectly.

The height of a helicopter above the ground is given by h = 2.95t^3, where h is in meters and t is in seconds. After 1.90 s, the helicopter releases a small mailbag. Assume the upward direction is positive and the downward direction is negative.

1) What is the velocity of the mailbag when it is released?
V = 8.85t^2 = 8.85(1.9)^2 = 31.95 m/s
2) What maximum height from the ground does the mailbag reach?
h = 2.95(1.9)^2 = 20.23 m (wrong answer)
3) What is the velocity of the mailbag when it hits the ground?
4) How long after its release does the mailbag reach the ground?

I got part 2 wrong, and yes I've been using the internet to help solve these problems. Will the answer for part 1 be the same for part 3? Not sure what to do at this point.

Can anyone help me please?

 

[ibyea]

I got part 2 wrong, and yes I've been using the internet to help solve these problems. Will the answer for part 1 be the same for part 3? Not sure what to do at this point.

Can anyone help me please?

You got part 2 wrong because you squared the time instead of cubing it.

 

[ibyea]

Math gaffers, I have two problems I need help on, and a third one that I might post later:



For part 4, I used this formula: a = (Vf)^2 - (Vi)^2 / 2D where D should be the distance of the dropped package. So, a = (0 m/s)^2 - (51.72 m/s)^2 / 2(136.35) = -9.81 m/s^2 but the answer is wrong. I may have either used the wrong formula or plugged in the numbers incorrectly.

Wouldn't the answer be 0 since you don't want your package to increase velocity anymore?

 

[kgtrep]

Then in case 2:
y_2 = y_1 * ln(x - x_0) + sigma from n = 1 to infinity of b_n(x-x_0)^(n+r) where b_n are coefficients of the power series, r is the only root of the indicial equation.

I wonder how do you find b_n?

In my notes, I see that b_n is simply the derivative with respect to r of a_n (which is the sequence of coefficients in the solution, y_1).

How do you take the derivative of a_n? Does that mean taking the derivative of the recurrence relation?

E.g. a_(n+2) = a_(n+1)/(n+r-1) is a recurrence relation. So we just take the derivative of this with respect to r?

How about case 3?

I think you would still substitute the power series into the differential equation and find a recurrence relations for the b's. It has been a while since I studied solving differential equations analytically, so I may be wrong here.

I was reading Earl Coddington's "Introduction to Ordinary Differential Equations" to see what he had to say, and think this is what you're looking for.

 

[MisterLuffy]

You got part 2 wrong because you squared the time instead of cubing it.

Oh it was a mistake on my part. I mean 2.95(1.9)^3 = 20.23 and it's still wrong

Wouldn't the answer be 0 since you don't want your package to increase velocity anymore?

I tried that, still gives me the wrong answer.

 

[ibyea]

Oh it was a mistake on my part. I mean 2.95(1.9)^3 = 20.23 and it's still wrong



I tried that, still gives me the wrong answer.

Okay, for the helicopter question, you have to take account of the fact that since the helicopter is going up, the package released has an upward initial velocity as stated in part a).

For the elevator shaft question, for d, plug in (146-136.35) for the initial equation you wrote, the reason being you want to stop the package in the span of that distance.

 

[MisterLuffy]

Okay, for the helicopter question, you have to take account of the fact that since the helicopter is going up, the package released has an upward initial velocity as stated in part a).

For the elevator shaft question, for d, plug in (146-136.35) for the initial equation you wrote, the reason being you want to stop the package in the span of that distance.

For d, if its - (51.72 m/s)^2 / 2(146-136.35) = -139.32 m/s^2, right?

 

[ibyea]

For d, if its - (51.72 m/s)^2 / 2(146-136.35) = -139.32 m/s^2, right?

Yeah that's what I was thinking.

 

[MisterLuffy]

Yeah that's what I was thinking.

Unfortunately, still got the wrong answer. It should be -138.59 m/s^2 but its still wrong. But thanks for your help. I'll leave that problem alone.

 

[luoapp]

Unfortunately, still got the wrong answer. It should be -138.59 m/s^2 but its still wrong. But thanks for your help. I'll leave that problem alone.

That's the net acceleration, which is the "acceleration required" minus g. I think the answer is -138.59 - g = -148.8

And the helicopter Q2 is 72.30m?

 

[Pau]

So turns out that every instructor's office hours are during another class. So going to any given office hour means skipping another class, which means being behind in that one, which means I might have to go to those office hours and continue the cycle. Basically, no office hours for me. There's a math tutoring center but they don't have any tutoring for my courses. I've spoken to my instructors and they can't meet any other time.

I feel completely and utterly fucked. No one I know in person has every done this kind of math or gotten this far. I've done well in all of my previous courses, getting an A+ in every math class until now (except Probability Theory, where I got a B+). So I don't think I'm exactly bad at this stuff, but I do feel a lot more confident knowing that help is around when I need it. I'm considering looking for a tutor, but it sounds incredibly expensive in NYC.

I'm taking Stochastic Processes, Introduction to Mathematical Probability, and a graduate course in Probability Theory that's really the exact same course I took last year. A lot of my issues are probably just Calc III stuff that I'm rusty on, but others are new things.

One specific question:

I need to find the moment generating function, and then use it to find the mean and variance. Is there a simpler way to integrate the following? Or is it just integration by parts again and again? And I just realized that I can find the mean and variance without using the moment generating function...

 

[MisterLuffy]

That's the net acceleration, which is the "acceleration required" minus g. I think the answer is -138.59 - g = -148.8

And the helicopter Q2 is 72.30m?

For the helicopter, it's 72.31m but yeah its correct. In this problem, g wasn't given but we know its 9.81m/s^2 right? I calculated and it gave me -148.4. I tried both of this and the answer you've given me, its still wrong.

I'm worried about my upcoming exam, I can easily miss something like that for sure. And most of all, I may not know what formulas or how to do the problems.

 

[luoapp]

For the helicopter, it's 72.31m but yet its correct. In this problem, g wasn't given but we know its 9.81m/s^2 right? I calculated and it gave me -148.4. I tried both of this and the answer you've given me, its still wrong.
.

I couldn't see what was wrong, so I recalcuated using g=9.81, and got 146.54.

 

[Reven]

Hey everyone I'm looking for some help on this problem, i'm its something simple I forgot from algebra and my textbook is practically useless. Thanks in advance.

 

[Therion]

Hey everyone I'm looking for some help on this problem, i'm its something simple I forgot from algebra and my textbook is practically useless. Thanks in advance.

Note that each of the three individual pieces of the function are guaranteed to be continuous on their given domains (the first would be continuous for all x!=2 and the others continuous everywhere). Therefore you only need to concern yourself with the places where the pieces join together, at x=2 and x=3.

A function f is continuous at x=c if the limit as x->c of f(x) exists and is equal to f(c). At x=2, for example, you have f(2) = a*2^2-b*2+3 = 4a-2b+3. Because of the continuity of ax^2-bx+3, you already know that the limit from the right matches this value. So you need to check the limit from the left. This is determined by the (x^2-4)/(x-2) piece of the function. You can find the limit as x->2 of (x^2-4)/(x-2) by factoring and simplifying, and this must equal 4a-2b+3.

Do the same thing at x=3, and you have a system of two linear equations in two variables which you can solve to find the values of a and b.

 

[Reven]

Thanks so much for the help, I swear the hardest part of calculus is actually the algebra.

 

[LightOfTruth]

Sorry, I didn't mention the z part, which is the same. For the second term, the order of integration is then z, r, and then theta. Also, yes, c=3 and a=1.

Yup I caught those and I think I actually got it. Thanks buddy!

 

[Kieli]

Thanks so much for the help, I swear the hardest part of calculus is actually the algebra.

Absolutely. The concepts are surprisingly intuitive, with the exception of solving integrals and some proofs of properties of sequences and series (which isn't so straightforward).

 

[XenodudeX]

Hey guys, I'm having a little trouble with this problem: Solve the polynomial equation. In order to obtain the first root, use synthetic division to test the possible rational roots.


2x^4 + 21x^3 + 89x^2 + 171x + 117 = 0

I'm trying to test the possible rational roots using synthetic division, but i've used -2,-21,-89,-171,-117 and they all end up with no zero as the remainder. What am I doing wrong?

edit: I just figured it out. I wasn't bring downing the 2. Haha..

edit 2 : I now have 2x^3+15x^2+44x+39 and now I some how have to factor this. How?

 

[mewmew42]

If a gold leaf is 10^-7 metres thick, and an atom of gold has a diameter of 2.88 X 10^-11 metres, calculate how many atoms thick gold leaf is?

Please help

 

[TUSR]

If a gold leaf is 10^-7 metres thick, and an atom of gold has a diameter of 2.88 X 10^-11 metres, calculate how many atoms thick gold leaf is?

Please help

10^-7 metres thick/2.88 X 10^-11 metres

?

 

[Stumpokapow]

Easiest way to solve elementary applied math problems is to look at what units you have:

meters per sheet
meters per atom

And then look at what units you want:
atoms per sheet

meters/sheet
------------------
meters/atom

= atoms per sheet

 

[ibyea]

Hey guys, I'm having a little trouble with this problem: Solve the polynomial equation. In order to obtain the first root, use synthetic division to test the possible rational roots.


2x^4 + 21x^3 + 89x^2 + 171x + 117 = 0

I'm trying to test the possible rational roots using synthetic division, but i've used -2,-21,-89,-171,-117 and they all end up with no zero as the remainder. What am I doing wrong?

edit: I just figured it out. I wasn't bring downing the 2. Haha..

edit 2 : I now have 2x^3+15x^2+44x+39 and now I some how have to factor this. How?

For the factoring, notice that for the binomial, you have 8 choices, 2x and x for the first term, and 1,3,13.39 for the second. Now, for the first term, just pick one. I personally picked 2x. If so the other equation must have an 1*x^2 term since no other term will add up to it after the multiplication since you are multiplying the highest term of one polynomial with the highest term of another one. So you have (2*x + a)*(x^2 + bx + c).
Now, you have 4 choices for the binomail, which are 1,3,13, and 39. 1 and 39 won't do, since 39 is too high and 1 makes constant c too high. So you have two choices, 3 and 13. Turns out 3 is the right answer after testing it out.

 

[Timedog]

How hard are partial differential equations texts/ material compared to ordinary?

 

[Biggest-Geek-Ever]

How hard are partial differential equations texts/ material compared to ordinary?

In my experience, most intro PDE courses will cover 5-6 methods of solving such equations, which basically boil down to turning them into ODEs and then . Using Fourier Series, Fourier Transforms, Separation of Variables, Sturm–Liouville theory, and Green's Function are probably the minimum in any quarter-based class.

In terms of difficulty vs ODEs, I'd say it's a pretty noticeable step-up. Just objectively speaking, there's typically a lot more setup and work that goes into solving a PDE, and it can seem pretty daunting. If you did well in an ODE course, I'm sure you can do well in PDEs, but you should expect to have to put a good amount more thought and work to do as well... but that's in general true for any math course that builds upon prior material.

 

[Pau]

My Introduction to Mathematical Probability homework is kicking my ass. Can't seem to do anything right. Worse, I don't know where I'm going wrong. Some questions:

Adding those two together gives me 0, which is wrong. The answer is 1/4. What am I doing wrong?

For this last one, I'm supposed to calculate the correlation coefficient. They already give me cov(x,y) as = -2/75

Problem is when I try to find the means of X and Y, I get a different answer than what they give. (Which is E[X] = E[Y] = 2/5; I get 8/5) The correlation coefficient should be 2/3. I was going to also post a picture of my work, but my phone decided that it couldn't send emails anymore...

 

[Gotchaye]

My Introduction to Mathematical Probability homework is kicking my ass. Can't seem to do anything right. Worse, I don't know where I'm going wrong. Some questions:



Adding those two together gives me 0, which is wrong. The answer is 1/4. What am I doing wrong?

It looks like you're on the right track here. You want to integrate f(x)f(x-y) for x>y and f(x)f(y-x) for y>x. That you got -1/6 for the second part should throw up a red flag - waiting times should always be positive. I think the first part is right. But when you do the integration for y from 0 to 1 and x from 0 to y you're first integrating by y again - you're doing exactly what you did the first time just with a negative sign. Actually the first intermediate step should be -6/4 x^4 y + 6/3 x^3 y^2 evaluated at 0 and y. Instead of switching the order of integration, you could also have gotten the part for y>x by integrating y from x to 1.

For this last one, I'm supposed to calculate the correlation coefficient. They already give me cov(x,y) as = -2/75

Problem is when I try to find the means of X and Y, I get a different answer than what they give. (Which is E[X] = E[Y] = 2/5; I get 8/5) The correlation coefficient should be 2/3. I was going to also post a picture of my work, but my phone decided that it couldn't send emails anymore...

E[X] = 8/5 is obviously wrong because the pdf is only non-zero for 0<x<1. You want to integrate x*f(x,y) over all x and y. f(x,y)=24xy on a triangle bounded by x=0, y=0, and x+y=1. So the integral you set up to get E[X] is 24 x^2 y dy dx for 0<y<1-x and 0<x<1. https://www.wolframalpha.com/input/?i=integrate+24x^2+y+dy+dx+for+0%3Cy%3C1-x+and+0%3Cx%3C1 . Edit: Taking another look at this I suspect you've gotten this far but that when you evaluate 12 x^2 y^2 at y = 0 and (1-x) you're not expanding (1-x)^2 correctly. You're using 1-x^2 instead of 1-2x+x^2. Or at least I get 8/5 doing that.

 

[MisterLuffy]

I have a question: A fish swimming in a horizontal plane has velocity vi = (4.00 i + 5.00 j) m/s at a point in the ocean where the position relative to a certain rock is ri = (-10.0 i - 4.00 j) m. After the fish swims with constant acceleration for 22.0 s, its velocity is v = (19.0 i - 5.00 j) m/s.

For part a, I got a_x = .68 m/s^2 and a_y = -.45 m/s^2

But for part b, if the fish maintains constant acceleration, where is it at t = 31.0 s, with respect to the rock?
x = 440.74 m using this equation x = x_0 + v_{0x}t + (1/2)a_xt^2
I used the same equation for y, and got -65.225m but it's giving me the wrong answer. Anyone can help me?

 

[kgtrep]

I have a question: A fish swimming in a horizontal plane has velocity vi = (4.00 i + 5.00 j) m/s at a point in the ocean where the position relative to a certain rock is ri = (-10.0 i - 4.00 j) m. After the fish swims with constant acceleration for 22.0 s, its velocity is v = (19.0 i - 5.00 j) m/s.

For part a, I got a_x = .68 m/s^2 and a_y = -.45 m/s^2

But for part b, if the fish maintains constant acceleration, where is it at t = 31.0 s, with respect to the rock?
x = 440.74 m using this equation x = x_0 + v_{0x}t + (1/2)a_xt^2
I used the same equation for y, and got -65.225m but it's giving me the wrong answer. Anyone can help me?

Your equation is correct, i.e. when the acceleration is constant, we have (by integration)

x(t) = x(0) + t * v(0) + 1/2 * t^2 * a

The x, v, and a above are all vectors, that is, the equation holds true in both x- and y-components (and z, if in 3D).

Your answers are being marked incorrectly because of round-off errors. Use a = [15/22; -10/22] m/s^2 instead, and you will find that the fish is now at x(31) = [441.6136; -67.4091] m.

 

[MisterLuffy]

Your equation is correct, i.e. when the acceleration is constant, we have (by integration)

x(t) = x(0) + t * v(0) + 1/2 * t^2 * a

The x, v, and a above are all vectors, that is, the equation holds true in both x- and y-components (and z, if in 3D).

Your answers are being marked incorrectly because of round-off errors. Use a = [15/22; -10/22] m/s^2 instead, and you will find that the fish is now at x(31) = [441.6136; -67.4091] m.

Thank you. I got to make sure to keep them in fractions the next time I have this round-off problems.

 

[Pau]

It looks like you're on the right track here. You want to integrate f(x)f(x-y) for x>y and f(x)f(y-x) for y>x. That you got -1/6 for the second part should throw up a red flag - waiting times should always be positive. I think the first part is right. But when you do the integration for y from 0 to 1 and x from 0 to y you're first integrating by y again - you're doing exactly what you did the first time just with a negative sign. Actually the first intermediate step should be -6/4 x^4 y + 6/3 x^3 y^2 evaluated at 0 and y. Instead of switching the order of integration, you could also have gotten the part for y>x by integrating y from x to 1.



E[X] = 8/5 is obviously wrong because the pdf is only non-zero for 0<x<1. You want to integrate x*f(x,y) over all x and y. f(x,y)=24xy on a triangle bounded by x=0, y=0, and x+y=1. So the integral you set up to get E[X] is 24 x^2 y dy dx for 0<y<1-x and 0<x<1. https://www.wolframalpha.com/input/?i=integrate+24x^2+y+dy+dx+for+0%3Cy%3C1-x+and+0%3Cx%3C1 . Edit: Taking another look at this I suspect you've gotten this far but that when you evaluate 12 x^2 y^2 at y = 0 and (1-x) you're not expanding (1-x)^2 correctly. You're using 1-x^2 instead of 1-2x+x^2. Or at least I get 8/5 doing that.

Okay, right. So it was integrating over y instead of x for the first step of the second part of the absolute value.

And got the second one as well! Thank you, I was using 1-x^2.

I need to figure out how to stop from doing stupid mistakes like that. :/

 

[MisterLuffy]

I have another problem I need help on: A stone at the end of a sling is whirled in a vertical circle of radius 1.40 m at a constant speed v0 = 1.20 m/s. The center of the string is 1.50 m above the ground. (using the figure I linked below)

Here's the diagram.

For part a, I need to find the range of the stone if it is released when the sling is inclined at 30 degrees with the horizontal at A.

For part b, its the same as part a but with the horizontal at B.

Can anyone help me and explain to me?

 

[Gotchaye]

I have another problem I need help on: A stone at the end of a sling is whirled in a vertical circle of radius 1.40 m at a constant speed v0 = 1.20 m/s. The center of the string is 1.50 m above the ground. (using the figure I linked below)

Here's the diagram.

For part a, I need to find the range of the stone if it is released when the sling is inclined at 30 degrees with the horizontal at A.

For part b, its the same as part a but with the horizontal at B.

Can anyone help me and explain to me?

This is fundamentally just a "I throw a ball from a height H at a velocity (Vx, Vy), how far does it go?" problem.

The trick here is figuring out how high the stone is when it's released and what direction it's going. You know its speed is 1.2 m/s. Its initial height will be somewhere between 0.1 and 2.9 meters (center of string +/- length of string). I guess you might also want to consider the horizontal distance the stone starts at too. But basically you want to work out the position of the stone at A and its velocity at A and then do the "I throw a ball..." problem for those conditions. Then repeat for B.

 

[luoapp]

I have another problem I need help on: A stone at the end of a sling is whirled in a vertical circle of radius 1.40 m at a constant speed v0 = 1.20 m/s. The center of the string is 1.50 m above the ground. (using the figure I linked below)

Constant speed isn't possible with a sling, isn't it?

 

[cpp_is_king]

Constant speed isn't possible with a sling, isn't it?

If you had variable force applied to perfectly counteract gravity at every instant