Currently working on a math problem that requires maple to be solved.
The math problem is
The volume V pf the paralellpiped is given by the following scalar triple product.
V = |(P5->P8 X P5->P6) * P5->P1|
P1 = ( [-17/12] , [11/36], [-65/36])
P5 = ([-11/12], [-7/36], [-47/36])
P6 = ([-2/3],[-1/9],[-8/9])
P8 = ([-5/4],[-1/12],[-17/12])`
answer should be 4 cubic units.
Several times trying it several different ways gives errors or nothing.
with(Student:-MultivariateCalculus):
P1 := <-17/12, 11/36, -65/36>:
P5 := <-11/12, -7/36, -47/36>:
P6 := <-2/3, -1/9, -8/9>:
P8 := <-5/4, -1/12, -17/12>:
abs(TripleScalarProduct(P1-P5, P8-P5, P6-P5));
#1/18
#Executed in Maple 2018
Related
If f(u,v) = vsinu+v^2, u(x,y)=tan^−1(y/x), v=sqrt(x^2 + y^2)
Calculate using chain rule and directly substituting for u(x,y), v(x,y).
Your syntax is not altogether clear to me, but I am guessing you are trying to achieve something like this:
uveqs := [u(x,y)=arctan(y/x), v(x,y)=sqrt(x^2+y^2)]:
f := v(x,y)*sin(u(x,y))+v(x,y)^2:
Differentiate f with respect to x (noticing the effect of the chain-rule),
K := diff(f,x):
diff(v(x,y),x) * sin(u(x,y))
+ v(x,y) * diff(u(x,y),x) * cos(u(x,y))
+ 2*v(x,y) * diff(v(x,y),x)
Now substitute for u(x,y) and v(x,u), and simplify,
S1 := eval(K,uveqs):
simplify(S1);
2 x
Alternatively, first substitute for u(x,y) and v(x,y) into f itself,
fxy := eval(f,uveqs):
(x^2+y^2)^(1/2)*y/x/(1+y^2/x^2)^(1/2)+x^2+y^2
And then differentiate with respect to x, and simplify,
S2 := diff(fxy,x):
simplify(S2);
2 x
If you have trouble understanding the mechanisms, you might also compare these two pairs of operations,
diff( v(x,y), x );
eval( %, uveqs );
and,
eval( v(x,y), uveqs );
diff( %, x );
I have a task:
Let’s say a bank has a cypher key K (just a long random string). That
bank wishes to split it into two pieces p1 and p2 so that both are
required for decryption. The p1 is then given to one executive, and p2
- to another, so both must contribute their pieces for decryption to proceed.
To accomplish that, the bank generates random k1 and sets k′1←k⊕k1. Note
that k1⊕k′1=k. The bank gives k1 to one executive and k′1 to another.
Both must be present for decryption to proceed since, by itself, each
piece contains no information about the secret key k.
Now, suppose the bank wants to split k into three pieces p1,p2,p3 so
that any two of the pieces enable decryption using k. This ensures
that even if one executive is out sick, decryption can still succeed,
but no employee can decrypt a message alone. To do so, the bank
generates two random pairs (k1,k′1) and (k2,k′2) as in the previous
paragraph so that k1⊕k′1=k2⊕k′2=k. How should the bank assign pieces
so that any two pieces enable decryption using k, but no single piece
can decrypt?
What is the answer to the question above?
p1 = (k1, k2), p2 = (k`1, k2), p3 = (k`2);
p1 = (k1, k2), p2 = (k`1, k`2), p3 = (k`2);
p1 = (k1, k2), p2 = (k1, k2), p3 = (k`2);
p1 = (k1, k2), p2 = (k1), p3 = (k`2);
p1 = (k1, k2), p2 = (k2, k`2), p3 = (k`2);
Explain me please, how does it work.
The first is the answer:
p1 = (k1, k2), p2 = (k`1, k2), p3 = (k`2);
because each possible pair can generate k:
p1 & p2: k1 ⊕ k'1 = k
p1 & p3: k2 ⊕ k'2 = k
p2 & p3: k2 ⊕ k'2 = k
Of course each of the parties cannot reconstruct the key by themselves.
The other possibilities are clearly wrong. Let's take the second possible answer and see if p2 & p3 can form a valid key. They cannot, because each have the same k'2 component, so that's not enough. Then p2 has a k'1, but p3 doesn't have a k1.
This is a solved problem, see Shamir's Secret Sharing:
Shamir's Secret Sharing is an algorithm in cryptography created by Adi Shamir. It is a form of secret sharing, where a secret is divided into parts, giving each participant its own unique part, where some of the parts or all of them are needed in order to reconstruct the secret.
Counting on all participants to combine the secret might be impractical, and therefore sometimes the threshold scheme is used where any
k of the parts are sufficient to reconstruct the original secret.
A possible weak 2 of 3 method might be possible using the same scheme that is used in three disk RAID5, any two can reconstitue the original.
This is an exercise that was suggested for my upcoming exam, the bottom is what I have gathered thus-far. All constructive input will be appreciated.
P1, P2 and P3 share three semaphores (x, y and z) each with 1 as initial value and three variables (a, b and c).
P1:
(1.1) wait(x);
(1.2) a = a + 1;
(1.3) wait(y);
(1.4) b = b - a;
(1.5) wait(z);
(1.6) c = a + 2*b -c;
(1.7) signal(z);
(1.8) signal(y);
(1.9) signal(x)
Code for P2:
(2.1) wait(y);
(2.2) b = b*2;
(2.3) wait(z);
(2.4) c = c - b;
(2.5) signal(y);
(2.6) wait(x);
(2.7) a = a + c;
(2.8) signal(x);
(2.9) signal(z)
Code for P3:
(3.1) wait(y);
(3.2) b = b*2;
(3.3) wait(z);
(3.4) c = c - b;
(3.5) signal(z);
(3.6) signal(y);
(3.7) wait(x);
(3.8) a = a / 10;
(3.9) signal(x)
A. If P1 and P2 run concurrently on a computer with only a single CPU, is it possible for these two processes to get into a deadlock? If so, show one execution sequence of the code that results in the deadlock, and show how to revise P2 only (P1 is not changed) to prevent deadlock.
B. If P1 and P3 are run concurrently on a computer with only a single CPU, is it possible for these two processes to get into a deadlock? If so, show one execution sequence of the code that results in the deadlock, and show how to revise P3 only (P1 is not changed) to prevent deadlock.
The changes you make should not violate the mutual exclusion requirement on shared variable access.
A) I'm not sure what it means by an example of when they would get into a deadlock? To me, it appears that y will cause a deadlock because line 1.3 will cause y to become -1 an would not be unlocked until 2.5 of P2.
To resolve this, 1.3 should be moved below 1.5 because that is when y is released in P2. It looks like there will be other conflicts though with x, but I don't know what a good way to rearrange P1 would be to resolve this without changing P2.
B) Here it appears 1.3 (wait(y)) causes a problem again since it is not signaled until 3.6. The resolution would then be to move it to after 1.6?
I'm trying to use Wiki's pages on Deadlock and Semaphore Programming to do this exercise.
Well, in the first case as an example;
Sequence Holds lock on
(1.1) wait(x); P1 x, P2 -
(1.2) a = a + 1; P1 x, P2 -
(2.1) wait(y); P1 x, P2 y
(2.2) b = b*2; P1 x, P2 y
(2.3) wait(z); P1 x, P2 yz
(2.4) c = c - b; P1 x, P2 yz
(2.5) signal(y); P1 x, P2 z
(2.6) wait(x); P1 x, P2 z - P2 locks waiting for x
(1.3) wait(y); P1 xy, P2 z
(1.4) b = b - a; P1 xy, P2 z
(1.5) wait(z); P1 xy, P2 z - P1 locks waiting for z
It could - for example - be fixed by locking P2 in the order same order as P1 (x, y, z instead of y, z, x). It may mean that you'll have to lock x before you really need to for mutual exclusion, but it will prevent deadlock.
As far as I can see, P1 and P3 can't mutually deadlock, since P3 is locked in the sequence y, z (which follows the x, y, z sequence, just skips the lock on x), then releases the locks and only locks/unlocks x.
Before I begin the problem, I use P0, P1, P2, and P3 for the four cubic Bezier points, and 't' since it's parametric. Also, I have searched for a similar problem in this site, as well as Google, and couldn't find one. I apologize if this is a common question.
The problem: I am getting a slope of 0 for both dx/dt and dy/dt for cubic Beziers in these two cases
1: t = 0 and P0 == P1
2: t = 1 and P2 == P3
Here's an example to illustrate (1), where t = 0 and P0 == P1.
Find the tangent (i.e dx/dt and dy/dt) of the following cubic Bezier at t = 0:
(100, 100) (100, 100) (150, 150) (200, 100)
To find the tangent, we'd want the first derivative of the cubic Bezier:
Cubic Bezier definition
B(t) = (1-t)^3P0 + 3t(1-t)^2P1 + 3t^2(1-t)P2 + t^3P3
First derivative of a bezier curve (if you'd like to see the steps I used to get here, let me know)
B'(t) = (-3P0 + 9P1 - 9P2 + 3P3)t^2 + (6P0 - 12P1 + 6P2)t + (-3P0 + 3P1)
Plugging in t = 0 to the first derivative equation, we get
B'(0) = -3P0 + 3P1
And finally, recall that P0 = P1 = (100, 100), so dx/dt and dy/dt are:
dx/dt = dy/dt = -3*(100) + 3*(100) = 0
This tells me...there is no tangent at t = 0 for this cubic Bezier. Which makes no sense if you were to graph and look at it.
What I'm doing to get a non-zero slope is to:
Treat the points P1, P2, and P3 like a quadratic Bezier, convert them into the equivalent cubic Bezier, THEN find the first derivative at t = 0.
Is there any way I can avoid doing that? I'm finding it difficult to accept a tangent that has 0 for dx/dt and dy/dt.
Thanks for your help.
The derivative B'(t) at t = 0 is indeed undefined for case 1 (and at t = 1 for case 2).
To see why this is the case, we can run the de Casteljau algorithm "backwards" on your example to double the parameter range of the curve from t = 0 ... 1 to t = -1 ... 1. This results in the following cubic Bezier curve control points:
(300,400) (0,-100) (100,200) (200,100)
If you plot this curve, you'll see your original curve from t = 0.5 ... 1. You'll also see that there is a cusp at t = 0.5 on this extended curve, right at the beginning of your original. This cusp is why your curve is not differentiable at its starting point.
However, the tangent of the curve is not quite the same thing as the derivative. So if all you need is the tangent, you're in luck. (The derivative is tangent to the curve, but so is any other vector perpendicular to the curve's normal.)
It turns out that the tangents at the ends of the curve are generally equivalent to:
P1 - P0 at t = 0
P3 - P2 at t = 1
However, if (and only if) P0 = P1 and/or P2 = P3, then the tangent at the degenerate point (that is, at t = 0 if P0 = P1 and/or t = 1 if P2 = P3) is equivalent to:
P2 - P1
You can verify that this is the case by evaluating B'(t) as t->0.
In fact, if you split the extended curve in two at t = 0.5 and then apply the P2 - P1 equation to each side, you'll see that there are two different tangents at the cusp. The tangent for each half of the curve point in the exact opposite directions. This is another illustration of why the derivative is undefined at this point.
One final note: your trick of treating the points P1, P2, and P3 like a quadratic Bezier will also give you a correct tangent. However, this will not give you the correct derivative.
This question is already correctly answered but I thought you'd like to know the underlying mathematics:
You are looking to find the end-slopes of a cubic bezier. Since the curve (i.e. its x and y values) is parametric in t, you will differentiate x and y w.r.t. t separately. The pair that you arrive at may be conceived of as the instantaneous "velocity" of a point travelling along the curve. So the point's initial velocity is zero (or more precisely a null-vector) in this case, but (most probably) the acceleration (or failing that, at least the rate of change of acceleration) will be non-zero, and hence the point's velocity will become non-zero (a non-null vector) and hence it will will move from those coordinates and trace out the curve.
But the slope as you visually perceive it is not parametric i.e. it does not depend on the time. I.O.W. what you are looking for is dy/dx, and not the pair (dx/dt, dy/dt), and given that dx/dt and dy/dt both are zero at t=0 for your curve, dy/dx = (dy/dt)/(dx/dt) = 0/0 which is indeterminate. To evaluate this, one must apply L'Hopital's rule. You can get the detailed treatment of the rule from the Wikipedia article but basically it means that to evaluate such indeterminate limits, we can differentiate the numerator f and denominator g separately to get f' and g' and then limit(f/g) is equal to limit(f'/g'). Now if p0, p1, p2 and p3 are the points defining your cubic, then:
dy / dt = ypart ( 3 * ( p1 - p0 ) + 6 * t * ( p2 - 2 * p1 + p0 ) + 3 * t ** 2 * ( p3 - 3 * p2 + 3 * p1 - p0 ) )
dx / dt = xpart ( 3 * ( p1 - p0 ) + 6 * t * ( p2 - 2 * p1 + p0 ) + 3 * t ** 2 * ( p3 - 3 * p2 + 3 * p1 - p0 ) )
-> (dy/dt) / (dx/dt) = ypart ( p1 - p0 ) / xpart ( p1 - p0 )
But this becomes indeterminate when p0 == p1. Now by L'Hopital's rule,
limit(t->0) [ (dy/dt) / (dx/dt) ] = limit(t->0) [ (d2y/dt2) / (d2x/dt2) ]
Now:
d2y/dt2 = ypart ( 6 * ( p2 - 2 * p1 + p0 ) + 6 * t * ( p3 - 3 * p2 + 3 * p1 - p0 ) )
d2x/dt2 = xpart ( 6 * ( p2 - 2 * p1 + p0 ) + 6 * t * ( p3 - 3 * p2 + 3 * p1 - p0 ) )
and at t = 0, (d2y/dt2) / (d2x/dt2) = ypart ( p2 - 2 * p1 + p0 ) / xpart ( p2 - 2 * p1 + p0 )
But since p1 == p0, this becomes: ypart ( p2 - p0 ) / xpart ( p2 - p0 ), which is exactly the result what Naaff told you. Note that if even p2 == p0 (a really degenerate curve, especially if cubic, in which case it will be just a straight line!), then even this will be indeterminate, and you can once more differentiate the numerator and denominator to get:
limit(dy/dx) = limit(t->0) [ (d3y/dt3) / (d3x/dt3) ] = ypart ( p3 - p0 ) / xpart ( p3 - p0 )
I hope it was useful for you... (BTW it doesn't seem as if TeX-like notation works here unlike on math.stackexchange, else I would provide math markup.)
I have a method to calculate the exponent, but it doesnt like the c := c * a. If i do something like c := a. it works and im unsure why its behaving this way when i try and do c := c *a.. Im new to Smalltalk, so maybe its just something im missing. Im using Pharo as my implementaton.
testPow: i1 exp: i2
"Testing exponent. i1 and i2 are integers, we calculate and then return the value as a Church numeral"
| a b c |
a := i1.
b := i2.
1 to: b do: [ :i |
c:= c*a.
].
^c.
You need to initialise c before using it. With your statement:
c := a
it works because a has a value. With:
c := c * a
it won't work because the first time through the loop, c will not be initialised to anything useful.
Just change your initialisation section to:
a := i1.
b := i2.
c := 1.
In the sample c is pointing to nil because c was not initialized. And nil (which is an object) doesn't understand the message *. Look at the values in the debugger when the exception is thrown.