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I have four equations:
s = e^0.245t (t = time(sec) over 10 seconds) s = distance(metres)
s = 4.2ln(t + 1.2)
s = 0.2t(t-5.1)
s = 0.2t(t-5.1)(t-9.1)
and for each equation I need to find the displacement,velocity, acceleration, start postition and how long each takes to reach 10 metres away from it's starting point.
I know that:
Velocity is: ds/dt
Acceleration is :dv/dt
I have no problem finding the Velocity and acceleration:
1.
s = e^0.245t (Is this the displacement?)
Velocity = .245 e^.245t
Acceleration = (.245)^2 e^.245t
2.
s = 4.2ln(t + 1.2) (Is this the displacement?)
Velocity = 4.2* (1/t +1.2)
Acceleration = -4.2 (t+1.2)^-2
3.
s = 0.2t(t-5.1) (Is this the displacement?)
Velocity= 0.2t^2 - 1.0t (is this right?)
Acceleration = 0.4t-1.02
4.
s = 0.2t(t-5.1)(t-9.1) (Is this the displacement?)
=-(0.2t^2 - 1.02t)(t-9.1)
=0.2t^3 - 1.82t^2 - 1.02t^2 + 9.282t
=0.2t^3 - 2.84t^2 +9.282t
Is this the velocity?
Velocity =.6t^2 - 5.68t + 9.282
Is this the acceleration?
Acceleration = 1.2t-5.68
Now here's the question, how do I find the starting position that each equation starts at, and how do i find how long it took the equation to finish.
Thanks for any help!
If you need me to explain something better please ask.
Ps: Sorry for the long question.
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so you are given s=f(t) and you are asked to find the starting position and the time taken to travel 10m. I think you should review your lecture notes one more time and figure it out for yourself. Look at some example that the lecturer did and see where the starting position is, and how to work out how far something travelled.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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Suppose you have a function describing position of a particle in meters at a time:
f(t) = e^(3t^2)
Then, as you say, the first and second derivatives will be the position:
v(t) = 6 t e ^(3t^2) and
a(t) = 6 e ^ (3t^2) + 36 t ^2 e ^ (3t^2).
To see how displacement works, plug in some values:
f(0) = 1m
f(.5) = 6.351m
f(1) = 120.513m
So in the first .5 seconds, the particle is at position 6.351m, but since it started at position 1m, it has only traveled a distance of 5.351m.
I don't want to give it away, so I won't say more, but that should give you an idea of how to figure out the problem.
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I tried much in this, but I couldn't get to any thing.
I want to get 21000 and put it in an integer, when I put it in a double it give something like 1.80934987391E+301, I want all the 302 digits, so I tried parsing in a string but it parses the same thing, I heard that something like this can be done by catching the overrun exception, but I really can't do it.  Can any one help ?
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Hi,
2^1000 has more than 300 decimal digits, obviously it wont fit in an int or a long.
2^1000 = 10,715,086,071,862,673,209,484,250,490,600,018,105,614,048,117,
055,336,074,437,503,883,703,510,511,249,361,224,931,983,788,
156,958,581,275,946,729,175,531,468,251,871,452,856,923,140,
435,984,577,574,698,574,803,934,567,774,824,230,985,421,074,
605,062,371,141,877,954,182,153,046,474,983,581,941,267,398,
767,559,165,543,946,077,062,914,571,196,477,686,542,167,660,
429,831,652,624,386,837,205,668,069,376
(302 digits)
there are basically three ways to get all the digits:
1.
use very clever code, based on mathematics, to calculate one or a few digits at a time,
without ever holding them all. Hint: the last digit can easily be predicted, the powers of
two end on 2/4/8/6/2/4/8/6/...
2.
use a library that can handle "big integers"; there are some on CP, and many more on Google.
Only a very small part of such a lib is actually needed here, since 2^1000 simply gets
represented as a 1 followed by a thousand zeroes in binary. So the one thing that remains
to be done is the ToString() of that...
3.
cheat and go to http://www.newdream.net/~sage/old/numbers/pow2.htm[^]
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Luc Pattyn wrote: 2^1000 = 10,715,086,071,862,673,209,484,250,490,600,018,105,614,048,117,
055,336,074,437,503,883,703,510,511,249,361,224,931,983,788,
156,958,581,275,946,729,175,531,468,251,871,452,856,923,140,
435,984,577,574,698,574,803,934,567,774,824,230,985,421,074,
605,062,371,141,877,954,182,153,046,474,983,581,941,267,398,
767,559,165,543,946,077,062,914,571,196,477,686,542,167,660,
429,831,652,624,386,837,205,668,069,376
Can you please tell how do you calculate such a big number?
Regards,
Arun Kumar.A
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I don't know how Luc did it, but using any large integer library will allow you to do it. Some languages support large integers, e.g. in Python (free), typing pow(2,1000) gives:
<br />
>>> pow(2,1000)<br />
1071508607186267320948425049060001810561404811705533607443750388370351051124936122493198378815695858<br />
1275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954<br />
1821530464749835819412673987675591655439460770629145711964776865421676604298316526243868372056680693<br />
76L<br />
I know that mathematica also supports large integers.
To do it in say c++, you need to write code, or use a large integer library - type large integer into the search box at the top of the page.
If you are asking how it is done, simply by using long multiplication / division. If the native int representation is 32 bits, then to represent a 1001 bit number needs 32 ints. You can write routines to do standard +,=,* and / operations on these numbers, it is much the same as in base 10 - you need to worry about carrying and borrowing. It is not that hard - I did it some years ago when I wrote some crypto routines (one of the major uses for large integer libraries these days), but libraries are readily available these days. Long division is the tricky one - I confess to resorting to Donald Knuth's book to get my algorithm right.
To work out 2^1000 is simply a number with 1 followed by 1000 '0's as Luc noted, the decimal representation is probably worked out using long division.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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Another one to look into is MIRACL[^].
"This perpetual motion machine she made is a joke. It just keeps going faster and faster. Lisa, get in here! In this house, we obey the laws of thermodynamics!" - Homer Simpson
Web - Blog - RSS - Math - LinkedIn - BM
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Hi
I have a bit of a math problem (related to graphics)
If i have three vectors and an origin representing an orthonormal system, how do i get the rotation matrix that would rotate the regular orthonormal system to this new one?
(by regular i mean at O(0,0,0) through i(1,0,0), j(0,1,0), and k(0,0,1) )
thanks for your time
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The columns of the matrix are the orthonormal vectors of the new basis. So if the transformation is
i->V1
j->V2
k->V3
where {V1,V2,V3} are an orthonormal set, then the matrix
R={V1,V2,V3}
where V1 is the first column etc, is want you want. Check by using R to transform each of the original basis vectors i,j,k and check they end up in the right place.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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Add Mathworld[^] to your bookmark. Been some time since I've done matrices :->
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You need to apply a unitary matrix transform to your coordinates (http://en.wikipedia.org/wiki/Unitary_matrix). that is, if you have a point in R3 represented by the vector v_o, then v = U v_o is a rotated coordinate. You need a unitary matrix so that no coordinate in the new basis is scaled with respect to the original basis. Concretely, if you start with an orthonormal system you want to end with an orthonormal system.
See rotation matrices on wikipedia: http://en.wikipedia.org/wiki/Rotation_matrix for more details of making the appropriate R3 rotation matrix.
good luck
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Why are you making this complicated? If he has the three new basis vectors then these are simply the columns of the transformation matrix. Simple matrix multiplication shows that the vector (1,0,0) goes to a vector equal to the first column of the matrix, (0,1,0) to the second etc. So if he has these three vectors (normalized) then he has his matrix.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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Matrix Rotation
Rotation of the plane at the origin by angle @ has the following form
x |--> R@x
where R@ is the matrix
( cos@ -sin@ )
( sin@ cos@ )
This is a 2 x 2 matrix on for a unit square (0,0), (0,1) , (1,0) and (1,1).
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I need a solution to a tedious problem please. I need to find a suitable equation that will model the Gateway Arch in St. Louis using suitable axes. I can use what is known as a catenary which is y=a/2(e^(x/a)+e^(-x/a)). Can you help me? I am desperate!!!!
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Jbones wrote: I need to find a suitable equation
Sounds like you have found it:
Jbones wrote: y=a/2(e^(x/a)+e^(-x/a))
What's the problem?
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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if you need to fit catenary parameters then use least squares.
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
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Have you tried regression
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maybe was for the OP...wasn't it?
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
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There was me talking about linear regression and trend lines 
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Frank Kerrigan wrote: Grady Booch: I told Google to their face...what you need is some serious adult supervision.
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