Kitonum

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MaplePrimes Activity


These are answers submitted by Kitonum

Your decision will be correct if the line MN does not intersect the sphere S. Otherwise, the answer is 0. If the point M(1, 2, -1) then at the end of your code, add the command allvalues(%)

Let O be a center of the sphere.

1) Find the equation of the plane passing through the points M, N and O.

2) Find the equation of the line of intersection of this plane and the sphere.

3) The points for which the distance from the sphere to the line MN the maximum and minimum are on the line.

If n is known, then your equation is reduced to finding the roots of a polynomial of degree n. It is easy to prove that, for different  a_i, this polynomial will have n different real roots. But finding them in a closed form for large n seems impossible. See the example for n = 5:

restart:

f:=unapply(sum(1/(x+a[i]),i=1..n), (x,n));

n:=5;

a:=[seq(k,k=1..n)];

f(x,n);

solve(f(x,n)=1);

allvalues([%]);

evalf(%);

plot([f(x,n),1],x=-18..8,-3..5, color=[blue,red], thickness=2, discont=true);

Try this variant:

restart;

 k := 0; B[1] := 0.001/365; B[2] :=0.002/365; B[3] :=0.005/365:

ode := diff(U(t), t) = -(A[1]+A[2]*U(t))*U(t);

ic[0] := U(365*k) = 1000;

sol[0] := dsolve({ic[0], subs(A[1] = B[1], A[2] = B[2], ode)}, U(t), numeric);

sigma := .5;

for k to 4 do 

if k = 3 then A[1] := B[3]:

ode1 := diff(U(t), t) = -(A[1]+A[2]*U(t))*U(t):  end if;  tk := 365*k; 

V := rhs(sol[k-1](tk)[2]);  ic[k] := U(tk) = sigma*V;

if k=3 then sol[k] := dsolve({ic[k], subs(A[2] = B[2], ode1)}, U(t), numeric); else 

sol[k] := dsolve({ic[k], subs(A[1] = B[1], A[2] = B[2], ode)}, U(t), numeric); fi;

end do;

Maple already has such a procedure:

ceil(3.5);

     4

If i understand your problem, you're trying to experiment to find a straight line passing through the point (4, 4/3) and to be tangent to semicircle  y=-sqrt(1-x^2)+3 .  This problem can be solved exactly, as follows:  

1) At any point of the semicircle find the tangent line to this semicircle. You get the equation of a straight line, depending on the parameter x that specifies the position of the point of tangency.  

2) Using the condition that the line passes through the point  (4, 4/3) , find the point of tangency, and then your desired angle.

 

I got the angle phi=arctan((20-4*sqrt(10))/45)=0.1619229414

Sorry, this is slip of the pen. In the code, replace everywhere 2^(u*v) by 2*u*v.

Make it as follows:

A:= `2*(a+b)`:

B:= `2*a+2*b`:

evalb(A = B);

    false

Blue color- x = constgreen colory = const:

plots[implicitplot]([seq(u^2-v^2=C, C=-10..10), u^2-v^2=-1/3, u^2-v^2=1/3, seq(2^u*v=C, C=-10..10), 2^u*v=-1/3, 2^u*v=1/3], u=-10..10, v=-10..10, color=[blue$23, green$23],numpoints=10000, view=[-1..3, -3..3], thickness=3,scaling=constrained);

If the parameters C1 and C2 are unknown, then use the command  Statistics[NonlinearFit]

Why is m not zero? It is natural to assume that m is any none-negative number:

int((x-m^2)/(x+1),x=0..m^2)  assuming m >= 0;

-ln(m^2+1)+m^2-ln(m^2+1)*m^2

add(add(A[i]*B[j], i=1..piecewise(6<8-j, 6, 8-j)), j=1..6);

 

It can be generalized to the continuous case (integration), but the previous two approaches can't.

int( int(x*y, y=1..piecewise(8-x>6, 6, 8-x)), x=1..6);

1627/12

 

Compare:

int( int(x*y, y=1..min(6, 8-x)), x=1..6);

A:=plot([seq(50+C*exp(-0.196*t), C=-40..40,20)], t=0..10,0..100, thickness=2, tickmarks=[default, [seq(10*k, k=0..9)]]):

B:=plots[textplot]([seq([0.5,14+20*i, C=-40+20*i], i=0..4)], font=[TIMES,ROMAN,14]):

plots[display](A, B);

 

L:=[ ]:

for t from 10 by 2 to 100 do

h := solve({x*y*z = 6*t^3, x-y-z = 0, x+y+z = 6*t}, {x, y, z}):

L:= [op(L), rhs(h[1][2])]:

end do:

A:=matrix([L]);

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