Mathematics desk | ||
---|---|---|
< January 20 | << Dec | January | Feb >> | January 22 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
This is not a homework problem - I thought of it while I was playing with my cat. If you dangle a string and spin it the right way, it forms a sort of standing wave. There are points along it, usually at most 3, that are almost stationary, and the rest looks like a squished sine wave spinning around its x-axis. In trying to figure out its shape, I got stuck with an equation I couldn't solve. To start, I assume no air resistance, that the string is the same mass all the way along and completely flexible, and that it has attained its ideal shape - a stationary planar wave spinning around a vertical axis. Call the axis of rotation the x-axis, and a perpendicular line the y-axis, with the y-axis spinning with the string. Since I wasn't sure how to solve it directly, I made the string massless with point masses along it at equal intervals, and drew a free body diagram for each mass. The first mass, at the bottom end of the string, has only two forces on it: mg downwards, and F1 along the string. Every other mass, say the nth along from the bottom, has three forces: mg downwards, and Fn-1 and Fn along the string. Equating X-components, Fn,x = nmg, because no mass is moving up or down. The Y-components, then, must add up to the actual acceleration (an) of the point. Since the point moves in a circle at constant angular velocity, acceleration is towards the x-axis, and since all points move with the same angular velocity, their accelerations are proportional. That is, there's some constant a such that an = ayn for all n, where yn is the y-coordinate of the point mass. Equating Y-components, then, . In other words, each bit of the string must support the weight of everything below it, and keep everything below it moving. Since the force on each mass must be in the direction of the next mass, that is along the string, Fn,y/Fn,x = (yn+1-yn)/(xn+1-xn). Since the masses are at even intervals along the string, there's some constant d such that d2 = (yn+1-yn)2+(xn+1-xn)2 for all n. Substituting the first two equations into the third, I get for all n. Once I had the shape of the problem, I shifted it to differentials. n becomes a continuous independant variable measuring distance along the string, x and y the dependant variables. If f(n)=y, and 1 = (y')2+(x')2. I differentiate the first, so (a/g)y = (x'y"-y'x")/(x')2. Plugging in the other equation and its derivative to get rid of x, (a/g)y2(1-(y')2)3 = (y")2. Choosing b=a/g, I'm left with by2(1-(y')2)3 = (y")2 Unfortunately, I'm not sure how to solve this for y. Any ideas? Black Carrot ( talk) 00:09, 21 January 2008 (UTC)
I've hit a bit of a difficulty with the power series. When I expand it around some points, it doesn't form a wave - all the odd-degree terms are zero. Specifically, if the x term is zero, corresponding to one of the peaks of the wave, all the odd-degree terms are as well. This means that if y is analytic anywhere, it isn't for very far. Is there a way to pull some bounds on that out of the equation? Black Carrot ( talk)05:22, 21 January 2008 (UTC)
"Square" and "cube" are apparently used to describe the exponents 2 & 3 because of their relationship to geometrical forms. Are there any forms that could lend their names to higher powers? Retarius | Talk 02:47, 21 January 2008 (UTC)
I have only a very slow computer with very slow dialup access, any chance I could do mathematical computation (matlab, mathematica, etc) on a normal free shell account somewhere? I don't know where to begin.
I may start a math degree this fall. —Preceding unsigned comment added by 212.51.122.8 ( talk) 03:19, 21 January 2008 (UTC)
The measured heights are 2950cm and 35 cm, while the true values are 2945cm and 30 cm respectively. compare the absolute and relative errors —Preceding unsigned comment added by 210.212.161.105 ( talk) 06:43, 21 January 2008 (UTC)
I'm trying to find the limit as x approaches 0 of (1 - cos 4x)/x. I rewrote it as (cos 4(0) - cos 4x)/(0 - x) and took the derivative of the subtrahend in the numerator to get -4 sin 4x. Substituting 0 in for x, that gives me 0, which was indeed given as the right answer. But my question is, was I justified in rewriting
lim as x approaches 0 of (1 - cos 4x)/x
as
lim as x approaches 0 of (cos 4(0) - cos 4x)/(0 - x)
given that that seems to change the sign of the denominator? Is it acceptable in this case because x is approaching 0 and the sign before 0 doesn't matter? Thanks, anon. —Preceding unsigned comment added by 70.19.20.251 ( talk) 19:30, 21 January 2008 (UTC)
I'm trying to evaluate ∫02(2x3 + 3)dx. Here's what I have so far:
But that's not one of the choices given; it says here that the answer is 14. Why is this? Thanks, anon. —Preceding unsigned comment added by 70.19.20.251 ( talk) 20:47, 21 January 2008 (UTC)
There are two cones with intersecting axes. They don't necessarily have the same slope. That is, one may be pointier than the other (I don't know the way you're supposed to say it). I assume the intersection between them is a conic section. How do I find the plane the intersection is on? Specifically, I need a vector at right angles to it. — Daniel 21:49, 21 January 2008 (UTC)
I forgot to add: the cones are the same width where their axes intersect.
After a little experimentation, it looks like if the cones have the same slope, the normal of the plane they intersect on is the cross product of their axes. The plane doesn't go through the intersecting axes as I thought it would, so I need to find a point on the plane. If the cones don't have the same slope, they appear to intersect on a hyperbola stretched infinitely in the third dimension. For what I'm doing, it would probably be best just to approximate it as a plane by averaging the slopes and taking the first case. Can anyone tell me how to find the point I need for the first case? — Daniel 00:39, 22 January 2008 (UTC)
Never mind. I solved it myself. I'm apparently better at solving the problem then explaining it. — Daniel 18:27, 22 January 2008 (UTC)
I need to find the absolute minimum value of f(x) = 2x3 - 3x2 - 12x, which is defined on the interval [-3, 2]. Relative extrema are marked by sign changes in the derivative, which is f'(x) = 6x2 - 6x - 12 = 6(x2 + x + 2) = 6(x - 2)(x + 1). So the possibilities are -3, -1, and 2. f(-3) = -45, f(-1) = 7, and f(2) = -20, so the answer is -45, isn't it? —Preceding unsigned comment added by 70.19.20.251 ( talk) 22:07, 21 January 2008 (UTC)
I'm looking for the critical numbers of f(x)= x√(16 - x2). I took the derivative: x•(1/2)(16 - x2)1/2 + (16 - x2)1/2. I factored out 16 - x2, giving me (16 - x2)[(x/2) + 16 - x2]. So my critical numbers are ±4 and ... what? I tried running the contents of the other parenthesis through the quadratic formula, but I got the gobbledygook 1/4 ± √(257)/8 ... —Preceding unsigned comment added by 70.19.20.251 ( talk) 22:33, 21 January 2008 (UTC)
You neglected the chain rule and you left 1/2 as the exponent where you needed (1/2) − 1. Michael Hardy ( talk) 23:23, 21 January 2008 (UTC)
What is 16,777,216 to the 2,079,600th power? Tamashiihiroka ( talk) 23:33, 21 January 2008 (UTC)
Mathematics desk | ||
---|---|---|
< January 20 | << Dec | January | Feb >> | January 22 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
This is not a homework problem - I thought of it while I was playing with my cat. If you dangle a string and spin it the right way, it forms a sort of standing wave. There are points along it, usually at most 3, that are almost stationary, and the rest looks like a squished sine wave spinning around its x-axis. In trying to figure out its shape, I got stuck with an equation I couldn't solve. To start, I assume no air resistance, that the string is the same mass all the way along and completely flexible, and that it has attained its ideal shape - a stationary planar wave spinning around a vertical axis. Call the axis of rotation the x-axis, and a perpendicular line the y-axis, with the y-axis spinning with the string. Since I wasn't sure how to solve it directly, I made the string massless with point masses along it at equal intervals, and drew a free body diagram for each mass. The first mass, at the bottom end of the string, has only two forces on it: mg downwards, and F1 along the string. Every other mass, say the nth along from the bottom, has three forces: mg downwards, and Fn-1 and Fn along the string. Equating X-components, Fn,x = nmg, because no mass is moving up or down. The Y-components, then, must add up to the actual acceleration (an) of the point. Since the point moves in a circle at constant angular velocity, acceleration is towards the x-axis, and since all points move with the same angular velocity, their accelerations are proportional. That is, there's some constant a such that an = ayn for all n, where yn is the y-coordinate of the point mass. Equating Y-components, then, . In other words, each bit of the string must support the weight of everything below it, and keep everything below it moving. Since the force on each mass must be in the direction of the next mass, that is along the string, Fn,y/Fn,x = (yn+1-yn)/(xn+1-xn). Since the masses are at even intervals along the string, there's some constant d such that d2 = (yn+1-yn)2+(xn+1-xn)2 for all n. Substituting the first two equations into the third, I get for all n. Once I had the shape of the problem, I shifted it to differentials. n becomes a continuous independant variable measuring distance along the string, x and y the dependant variables. If f(n)=y, and 1 = (y')2+(x')2. I differentiate the first, so (a/g)y = (x'y"-y'x")/(x')2. Plugging in the other equation and its derivative to get rid of x, (a/g)y2(1-(y')2)3 = (y")2. Choosing b=a/g, I'm left with by2(1-(y')2)3 = (y")2 Unfortunately, I'm not sure how to solve this for y. Any ideas? Black Carrot ( talk) 00:09, 21 January 2008 (UTC)
I've hit a bit of a difficulty with the power series. When I expand it around some points, it doesn't form a wave - all the odd-degree terms are zero. Specifically, if the x term is zero, corresponding to one of the peaks of the wave, all the odd-degree terms are as well. This means that if y is analytic anywhere, it isn't for very far. Is there a way to pull some bounds on that out of the equation? Black Carrot ( talk)05:22, 21 January 2008 (UTC)
"Square" and "cube" are apparently used to describe the exponents 2 & 3 because of their relationship to geometrical forms. Are there any forms that could lend their names to higher powers? Retarius | Talk 02:47, 21 January 2008 (UTC)
I have only a very slow computer with very slow dialup access, any chance I could do mathematical computation (matlab, mathematica, etc) on a normal free shell account somewhere? I don't know where to begin.
I may start a math degree this fall. —Preceding unsigned comment added by 212.51.122.8 ( talk) 03:19, 21 January 2008 (UTC)
The measured heights are 2950cm and 35 cm, while the true values are 2945cm and 30 cm respectively. compare the absolute and relative errors —Preceding unsigned comment added by 210.212.161.105 ( talk) 06:43, 21 January 2008 (UTC)
I'm trying to find the limit as x approaches 0 of (1 - cos 4x)/x. I rewrote it as (cos 4(0) - cos 4x)/(0 - x) and took the derivative of the subtrahend in the numerator to get -4 sin 4x. Substituting 0 in for x, that gives me 0, which was indeed given as the right answer. But my question is, was I justified in rewriting
lim as x approaches 0 of (1 - cos 4x)/x
as
lim as x approaches 0 of (cos 4(0) - cos 4x)/(0 - x)
given that that seems to change the sign of the denominator? Is it acceptable in this case because x is approaching 0 and the sign before 0 doesn't matter? Thanks, anon. —Preceding unsigned comment added by 70.19.20.251 ( talk) 19:30, 21 January 2008 (UTC)
I'm trying to evaluate ∫02(2x3 + 3)dx. Here's what I have so far:
But that's not one of the choices given; it says here that the answer is 14. Why is this? Thanks, anon. —Preceding unsigned comment added by 70.19.20.251 ( talk) 20:47, 21 January 2008 (UTC)
There are two cones with intersecting axes. They don't necessarily have the same slope. That is, one may be pointier than the other (I don't know the way you're supposed to say it). I assume the intersection between them is a conic section. How do I find the plane the intersection is on? Specifically, I need a vector at right angles to it. — Daniel 21:49, 21 January 2008 (UTC)
I forgot to add: the cones are the same width where their axes intersect.
After a little experimentation, it looks like if the cones have the same slope, the normal of the plane they intersect on is the cross product of their axes. The plane doesn't go through the intersecting axes as I thought it would, so I need to find a point on the plane. If the cones don't have the same slope, they appear to intersect on a hyperbola stretched infinitely in the third dimension. For what I'm doing, it would probably be best just to approximate it as a plane by averaging the slopes and taking the first case. Can anyone tell me how to find the point I need for the first case? — Daniel 00:39, 22 January 2008 (UTC)
Never mind. I solved it myself. I'm apparently better at solving the problem then explaining it. — Daniel 18:27, 22 January 2008 (UTC)
I need to find the absolute minimum value of f(x) = 2x3 - 3x2 - 12x, which is defined on the interval [-3, 2]. Relative extrema are marked by sign changes in the derivative, which is f'(x) = 6x2 - 6x - 12 = 6(x2 + x + 2) = 6(x - 2)(x + 1). So the possibilities are -3, -1, and 2. f(-3) = -45, f(-1) = 7, and f(2) = -20, so the answer is -45, isn't it? —Preceding unsigned comment added by 70.19.20.251 ( talk) 22:07, 21 January 2008 (UTC)
I'm looking for the critical numbers of f(x)= x√(16 - x2). I took the derivative: x•(1/2)(16 - x2)1/2 + (16 - x2)1/2. I factored out 16 - x2, giving me (16 - x2)[(x/2) + 16 - x2]. So my critical numbers are ±4 and ... what? I tried running the contents of the other parenthesis through the quadratic formula, but I got the gobbledygook 1/4 ± √(257)/8 ... —Preceding unsigned comment added by 70.19.20.251 ( talk) 22:33, 21 January 2008 (UTC)
You neglected the chain rule and you left 1/2 as the exponent where you needed (1/2) − 1. Michael Hardy ( talk) 23:23, 21 January 2008 (UTC)
What is 16,777,216 to the 2,079,600th power? Tamashiihiroka ( talk) 23:33, 21 January 2008 (UTC)