Kitonum

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There is no need for a triple integral here. Let us consider the section of this torus by the plane  y=0 . We see that the torus is obtained by rotating the circle  (x - 4)^2 + z^2 = 4  around the axis  Oz . Using the well-known formula for the volume of a body of rotation, we get the answer  V=32*Pi^2 :

restart;
Eq:=(x^2+y^2+z^2+12)^2=64*(x^2+y^2):
plots:-implicitplot3d(Eq, x=-6.7..6.7, y=-6.7..6.7,z=-2.7..2.7, style=surface, axes=normal, numpoints=100000, scaling=constrained);
factor((lhs-rhs)(eval(Eq,y=0)));
Student:-Precalculus:-CompleteSquare(op(1,%),[x,y]);
f1,f2:=solve(%, x);
V=simplify(Pi*int(f1^2,z=-2..2)-Pi*int(f2^2,z=-2..2));

                     

 

restart;
expr:= 4*cos(Pi*m)/(Pi*(2*m + 1)); 
simplify(%) assuming m::integer;
``;
4*cos(Pi*m+a)/(Pi*(2*m + 1));
expand(%) assuming m::integer;
``;
4*cos(Pi*m+2*alpha*Pi)/(Pi*(2*m + 1));
expand(%) assuming integer;


                                             

Explanation. If alpha is an arbitrary real number, then  2*alpha*Pi  will also be an arbitrary real number. Therefore, I immediately replaced this in the code with the symbol  , then to simplify we will not need additional commands. If  alpha  is an integer, then due to the periodicity of the cosine, the result coincides with example 1. 

Code in Maple 2018.

Edited.

restart;
h:=(t,x)->(m*t^2 + 6*t - 2*x)^2/(36*g*t^2):
t:=<1,5,10,15,20>:
V:=h~(t,x);
V[1], V[3];

                         

 

This equation can only be solved numerically using  the  fsolve  command. To get roots other than 0, you can first plot a graph and then indicate the boundaries of the interval in which the root you need lies:

restart;
Eq:=(5*x*cos(x^2/2)-5*sin(5*x/2)*cos(5*x/2)=0);
plot(lhs(Eq), x=0..5);
fsolve(Eq, x=1..2);

                          

 

To solve this problem it is convenient to use the differentiation operator  , as well as the well-known tangent equation  y=f(x0) + f'(x0)*(x - x0)  at the point  x=x0 :

restart;
f:=x->(1-k*x)/(1+x^2):
T:=f(3)+D(f)(3)*(x-3); # Equation of the tangent line at x=3
T:=collect(T, x);  

# Visualization for k=1
k:=1:
plots:-display(plot([f(x),T], x=-0..10, -0.3..0.1, color=[blue,red]), plots:-pointplot([3,f(3)], color=red, symbol=solidcircle, symbolsize=12));

                         

 

 

 

restart;
identify(fsolve(x!=3628800, x=0..infinity));
ifactor(65536);

                                

It is easy to check the truth of the identity   

So we get

restart;
is((2*x^2022+1)/(x^2023+x)=1/x+x^2021/(x^2022+1));
int(1/x+x^2021/(x^2022+1), x);

                    

You can use the technique of stitching sequential animations from the post
https://mapleprimes.com/posts/207840-Combinations-Of-Multiple-Animations

restart;
Frame:=proc(t,Range,Color)
local SinArc, p, l1, l2, l3, l4, l, Circle;
uses plots, plottools;
SinArc:=plot(sin(x),x=Range[1]..t, color=Color, thickness=2);
p:=pointplot([[-2,0],[-2,sin(t)],[cos(t)-2,sin(t)],[t,sin(t)]], symbolsize=11, symbol=solidcircle);
l:=line([-2,-1.4],[-2,1.4], thickness=0);
l1:=line([-2,0],[-2,sin(t)], color=Color, thickness=2);
l2:=line([-2,0],[cos(t)-2,sin(t)]);
l3:=line([cos(t)-2,sin(t)],[t,sin(t)]);
l4:=line([-2,sin(t)],[cos(t)-2,sin(t)]);
Circle:=circle([-2,0], color=Color, thickness=2);
display(SinArc,p,l,l1,l2,l3,l4,Circle, scaling=constrained);
end proc:

Example of use:

with(plots):
A:=animate(Frame,[t,[0,2*Pi],red], t=0..2*Pi, frames=90, size=[800,200], view=-1.4..1.4):
B:=animate(Frame,[t,[2*Pi,4*Pi],blue], t=2*Pi..4*Pi, frames=90, size=[800,200], view=-1.4..1.4):
C:=animate(Frame,[t,[4*Pi,6*Pi],green], t=4*Pi..6*Pi, frames=90, size=[800,200], view=-1.4..1.4):
display([A, display(op([1,-1,1],A),B), display(op([1,-1,1],A),op([1,-1,1],B),C)], insequence, view=[-3..19,-1.4..1.4]);

restart;
local D, O;
with(Student:-MultivariateCalculus):
A := [0, 0, 0]:
B := [a, 0, 0]:
C := [a, b, 0]:
D := [0, b, 0]:
S := [0, 0, h]:
O := [x, y, z]:
lineSC := Line(S, C);
lineSD := Line(S, D);
H := Projection(A, lineSC);
K := Projection(A, lineSD);
OH:=H-O; OK:=K-O; OC:=C-O;
M:=Matrix([OH, OK, OC]);
O:=eval(O,%): # The center of the desired circle
R:=simplify(Distance(O, H));  # The radius of the circle

                    

Addition - visualization made from calculation results:

with(plots): with(plottools):
a:=2: b:=3: h:=4:
BAD:=curve([B,A,D],linestyle=3,color=blue):
BCD:=curve([B,C,D],thickness=2,color=blue):
SA:=line(S,A,linestyle=3,color=blue, thickness=2): SB:=line(S,B,color=blue, thickness=2): SC:=line(S,C,color=blue, thickness=2): SD:=line(S,D,color=blue, thickness=2):
AH:=line(A,H,linestyle=3,color=black): AK:=line(A,K,linestyle=3,color=black):
p:=Plane(C,S,D):
s:=(x-O[1])^2+(y-O[2])^2+(z-O[3])^2=R^2:
c:=intersectplot(GetRepresentation(p),s, x=-1..a+0.7, y=-1..b+0.7, z=-1..h+0.7, color=red, thickness=3):
T:=textplot3d([[A[],"A"],[B[],"B"],[C[],"C"],[D[],"D"],[S[],"S"],[H[],"H"],[K[],"K"]], font=[times,roman,18], align=left):
display(BAD,BCD,SA,SB,SC,SD,AH,AK,c,T, axes=none, scaling=constrained);

         

 

 

f := (x1, t1)->eval(u(x, t), (pds:-value(t = t1))(x1)):

# Examples 
f(0, 0.1); 
f(0, 0.7); 
f(0.3, 0.3);

               

 

 

 

For example you can do this using the loop with while :

restart;
n:=294912: d:=8:
r:=irem(n,d):
while r=0 do
n:=n/d; r:=irem(n,d);
od:
n;

                                                          9

restart;
R:=Student:-MultivariateCalculus:-LagrangeMultipliers(z,[4*x-3*y+8*z-5,z^2-x^2-y^2], [x,y,z], output=detailed);
P1:=plots:-intersectplot(4*x-3*y+8*z-5=0,z^2-x^2-y^2=0, x=-1.7..1.4, y=-1.4..1.4, z=-0.4..2.1, color=blue, thickness=2, axes=normal):
P2:=plots:-pointplot3d([eval([x,y,z],R[1]),eval([x,y,z],R[2])], color=red, symbolsize=12, symbol=solidsphere):
plots:-display(P1,P2);

    

Using the  Student:-Calculus1:-Roots  command, we find 2 real roots of the equation  log[2](7*x/10)=log[3](3*x-1)  and, based on the plot, find the solution to the inequality:

restart;
Student:-Calculus1:-Roots(log[2](7*x/10)=log[3](3*x-1));
plot([log[2](7*x/10),log[3](3*x-1)], x=0..20, -3..4, color=[red,blue], numpoints=5000, size=[800,400]);

                


So the result is   0.3730125034<= x <= 16.60731835

I replaced your operators  Not, Or  and so on (I never used them)  with more standard ones (not, or, and, implies) in prefix form . As a result of simplifications in 2 ways, we get  false , that is, your expression is internally contradictory for any values of the variables  x  and  y , just as, for example, x and not x  is contradictory.

restart;
expr:=`and`(
   `not`(`or`(`or`(0 >- 0, y^3=x), 
     `and`(((y*x - 0^0)^2 + (y^2 - x^2)^2)*(y^2 + x^2) > 
       0, 0 <- (y^3 - x)*(y - x^3)^3, 
      `implies`(y + 2*x >= 2*y^3 + x^3, 
       `or`(y + x < y^3 + x^3)), `not`((y^3 - x)*(y - x^3) = 0), 
      0 >= 0), y <> x^3)));
convert(expr, 'boolean_function'); # The first way
Logic:-BooleanSimplify(expr); # The second way

      

Do  

factor(simplify((2)));

 

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