## 26 Badges

16 years, 25 days

## Answer...

Ax:=t->cos(5*t): Ay:=t->0: Bx:=t->-cos(5*t)/2: By:=t->3*cos(5*t)/2:

S:=seq(plots[arrow]([Ax(i*Pi/72),Ay(i*Pi/72)],color=blue),i=0..72):

T:=seq(plots[arrow]([Bx(i*Pi/72),By(i*Pi/72)],color=green),i=0..72):

U:=seq(plots[display]([S[i],T[i]]),i=1..73):

plots[display](U,view=[-2..2,-2..2],insequence=true);

## Answer...

You must assign  z  as a function, such as in an example

restart:

dsolve({diff(z(t),t) = 2*t-1, z(0) = 1});

z:= unapply(rhs(%), t);

z(60);

## Answer...

a:= 1: b:= 1/sqrt(3):

for n to 2010 do

c:= simplify((a + b)/(1-a*b)):  a:=b:  b:=c:

od:

c;

## Head-on solution...

Brute force computation shows that there are no solutions. Explanation of code: firstly founded a list of matrices  H  of 0 and 1, satisfying the first condition. The number of such matrices will be 3190. As an example displayed 4 such matrices. Then, all of these 3190 matrices tested for the second condition.

st:= time():

It:=proc(K)

[seq([0,op(K[i])],i=1..nops(K)),seq([1,op(K[i])],i=1..nops(K))];

end proc:

M:=(It@@5)([[ ]]):

H:=[]:

for L[1] in M do

for L[2] in M do

for L[3] in M do

for L[4] in M do

for L[5] in M do

if add(add(L[i][j],i=1..3),j=1..3)=5 and add(add(L[i][j],i=1..3),j=2..4)=5 and add(add(L[i][j],i=1..3),j=3..5)=5 and

add(add(L[i][j],i=2..4),j=1..3)=5 and add(add(L[i][j],i=2..4),j=2..4)=5 and add(add(L[i][j],i=2..4),j=3..5)=5 and

add(add(L[i][j],i=3..5),j=1..3)=5 and add(add(L[i][j],i=3..5),j=2..4)=5 and add(add(L[i][j],i=3..5),j=3..5)=5 then

H:=[op(H), Matrix([L[1],L[2],L[3],L[4],L[5]])]: fi:

od: od: od: od: od:

'nops(H)'=nops(H);  seq(H[n],n=1001..1004);

P:=[ ]:

for k in H do

if add(add(k[i,j],i=1..2),j=1..4)=4 and add(add(k[i,j],i=1..2),j=2..5)=4 and add(add(k[i,j],i=2..3),j=1..4)=4 and

add(add(k[i,j],i=2..3),j=2..5)=4 and add(add(k[i,j],i=3..4),j=1..4)=4 and add(add(k[i,j],i=3..4),j=2..5)=4 and

add(add(k[i,j],i=4..5),j=1..4)=4 and add(add(k[i,j],i=4..5),j=2..5)=4 and add(add(k[i,j],j=1..2),i=1..4)=4 and

add(add(k[i,j],j=1..2),i=1..4)=4 and add(add(k[i,j],j=1..2),i=1..4)=4 and add(add(k[i,j],j=1..2),i=1..4)=4 and

add(add(k[i,j],j=1..2),i=1..4)=4 and add(add(k[i,j],j=1..2),i=1..4)=4 and add(add(k[i,j],j=1..2),i=1..4)=4 and

add(add(k[i,j],j=1..2),i=1..4)=4

then P:=[op(P), k]: fi:

od:

P;

time() - st;

## Solution of the problem...

The evalf command after solve command for systems as well as fsolve gives only 1 solution.

Possible decision of this problem:

f:=x->x^3-3*x^2+2:

g:=x->k*(x+1)+3:

RealDomain[solve]([f(x) = g(x), diff(f(x),x) = diff(g(x),x)]);

evalf(allvalues([%]));

Got all the solutions in both symbolic and numerical form!

## Answer...

Your solution is right.

I suggest another variant to solve your problem without solve command by direct computation (firstly find the points of tangancy):

restart:  with(geom3d):  with(linalg):

line(d1,[-5+2*t,-3*t+1,-13+2*t],t):  line(d2,[-7+3*t,-1-2*t,8],t):

a:=ParallelVector(d1):  b:=ParallelVector(d2):

c:=crossprod(a,b):

sphere(S, x^2+  y^2 + z^2 -10*x +2*y +26*z -113=0, [x,y,z]):

T:=coordinates(center(S)):  R:=radius(S):

H:=[T + [seq(c[i]/norm(c,2)*R,i=1..3)], T - [seq(c[i]/norm(c,2)*R,i=1..3)]]:

for i to 2 do

sort(Equation(plane(P,c[1]*(x-H[i,1]) +c[2]*(y-H[i,2]) +c[3]*(z-H[i,3]) =0,[x,y,z])), {x,y,z}):

od;

## Answer...

I have not found a suitable way to sort, so I suggest a way to manually isolate complete squares.

The code:

restart:  with(geom3d):

a:=[-t, -t, t]: point(T, a): point(A, 1, -1, -1): point(B, 2, 1, 2): point(C, 1, 3, 1): plane(ABC, [A,B,C], [x, y, z]):

d:= distance(T, ABC):  R:= distance(T, A):

r:= R^2 - d^2;  simplify(r) assuming real;

sol:= minimize(r, location);  Sol:= op(sol[2])[1];

point(T, subs(Sol, a)):  R:= subs(Sol, R):

Eq:=lhs(Equation(sphere(S, [T,R], [x, y, z])));

(x+coeff(Eq,x)/2)^2+(y+coeff(Eq,y)/2)^2+(z+coeff(Eq,z)/2)^2=sqrt((coeff(Eq,x)^2+coeff(Eq,y)^2+coeff(Eq,z)^2)/4-tcoeff(Eq))^`2`;  # Canonical equation of S

## Answer...

You can use the parametric equations of the surface.

The example:

theta:=Pi/3:

plot3d( [u*cos(t)*cos(theta), u*cos(t)*sin(theta), u*sin(t)], u=0..1, t=0..2*Pi, color=red, axes=normal, scaling=constrained, style=surface, view=[-1..1, -1..1, -1.2..1.2] );

## Answer...

For example:

Sol:= 9.356887917*10^5 , { [{P = 330.1448618, W = 3.339008977}, 9.356887917*10^5 ] }:

P:=rhs(Sol[2,1,1,1]);  W:=rhs(Sol[2,1,1,2]);

P:=330.1448618

W:= 3.339008977

## Another version of your code...

I would write your code so:

restart:  with(geom3d):

a:=[-t, -t, t]: point(T, a): point(A, 1, -1, -1): point(B, 2, 1, 2): point(C, 1, 3, 1):

plane(ABC, [A,B,C], [x, y, z]):

d:= distance(T, ABC):  R:= distance(T, A):

r:= R^2 - d^2;  simplify(r) assuming real;

sol:= minimize(r, location);  Sol:= op(sol[2])[1];

point(T, subs(Sol, a)):  R:= subs(Sol, R):

Equation(sphere(S, [T,R], [x, y, z]));

Eq:=Student[Precalculus][CompleteSquare](Equation(S));

op(1, Eq) - op(4, op(1, Eq)) = sqrt(-op(4, op(1, Eq)))^`2`;  # Canonical equation of S

Here you will find answers to your questions!

## Slip...

Sorry! I made a slip of the pen. Instead  2  must be  d .

## Answer...

After  v:=crossprod(w,a)  insert two lines into your code:

d:=igcd(v[1],v[2],v[3]):

v:=map(x->x/2,v):

## Answer...

The lower boundary your region is a common segment  [-1, 1] . So look for  maximum and minimum of the function of one variable  f(x,0)=x^2+x  on this segment by the commands  maximize  and  minimize . In this decision, the derivatives are not needed.

## Explanation...

Computed plane passes through the line AB so that AB and its projection onto given plane  are perpendicular to the line of intersection of computed plane with given plane. Draw a picture and everything will be clear!

## Code...

restart:

with(LinearAlgebra):

A:=<1, -2, 4>:  B:=<3, 5, -1>:  n1:=<1, 1, 1>:  M:=<x, y, z>:

n2:=CrossProduct(n1, B - A):  n:=CrossProduct(n2, B - A):  d:=igcd(seq(n[i],i=1..3)):

sort((n/d).(M - A)) = 0;   # Equation of the required plane

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