Maple 2018 Questions and Posts

These are Posts and Questions associated with the product, Maple 2018

I know how to contruct an isoscele ABCD trapeze knowing the 2L length of AB and BC=CD=CD=a. But I don't know answer to that question : L being fixed together with points A and B , show that the place of C when a varies is a branch of hyberbola. Here is my code.

restart; unprotect(O, D);
with(plots):
Vdot := proc (U, V) add(U[i]*V[i], i = 1 .. 2) end:
dist := proc (M, N) sqrt(Vdot(expand(M-N), expand(M-N))) end:
_EnvHorizontalName := x: _EnvVerticalName := y:
O:=[0,0]:A:=[-L,0]:alpha:=arccos((2*L-a)^2/(2*a*(2*L-a))):h:=tan(alpha)*(2*L-a)/2:
B:=[L,0]:C:=[L-(2*L-a)/2,h]:D:=[-L+(2*L-a)/2,h]:
L := 6; a := 7;
poly := [A, B, C, D, A];tp := textplot([[A[], "A"], [B[], "B"], [C[], "C"], [D[], "D"]], color = black, 'align' = {'above', 'right'});
trapeze := polygonplot(poly, axes = normal, color = "DarkGreen", transparency = .9);
display([tp, trapeze], scaling = constrained); Thank you foryour help.

We consider an ellipse  E defined by the equation x²/a²+y²/b²=1. Either F1 and F2 its foci. 
We consider a point M of E  such the angle F1,M,F2 is equal to a given value phi.
Find F1M et F2M in function of a, b and phi . Then find the coordiinates x0 and y0 of M in function of a, b, and phi. Here is the begining of my code;

restart;
_EnvHorizontalName := x: _EnvVerticalName := y:
with(plots):
line := proc (x1, y1, x2, y2) options operator, arrow; (x-x1)*(y2-y1)-(y-y1)*(x2-x1) end proc:
Vdot := proc (U, V) add(U[i]*V[i], i = 1 .. 2) end proc:
dist := proc (M, N) sqrt(Vdot(expand(M-N), expand(M-N))) end proc:

ell := x^2/a^2+y^2/b^2 = 1: 
c := sqrt(a^2-b^2):
F1 := [c, 0]: F2 := [-c, 0]:
M := [a*cos(t), b*sin(t)]:
MF2 := simplify(expand(sqrt((c+a*cos(t))^2+(b*sin(t))^2)));
MF1 := simplify(expand(sqrt((c-a*cos(t))^2+(b*sin(t))^2)));
varphi := arccos((MF1^2+MF2^2-4*c^2)/(2*MF1*MF2));
f := unapply(varphi, t);
`maxϕ` := f((1/2)*Pi);
diff(f(t), t);

a := 7: b := 5: t := (1/3)*Pi:
evalf(varphi);
evalf(MF1); evalf(dist(M, F1));
evalf(MF2); evalf(dist(M, F2));
MF1+MF2; evalf(%);
`maxϕ`; evalf(%);
ELL := implicitplot(ell, x = -a-2 .. a+2, y = -2-b .. b+2, color = blue):
tp := textplot([[F1[], "F1"],[F2[],"F2"],[M[],"M"]], 'align' = {'above', 'left'}):
po := plot([F1,F2, M], style = point, symbolsize = 15, symbol = solidcircle, color = red):

display([ELL, tp, po], scaling = constrained): Would  you help me to answer the questions ? Thank you.

Hi. i have a nonlinear integral equation. any idea for solving it? tnx in advance

integral-equation.mw

 

How to simplify this trigonometric expression with Maple? I only know the way with  identify command, which is difficult to call mathematically correct:

restart;
Expr:=arctan((1-tan(20*Pi/180))/(1-tan(25*Pi/180)));
evalf[15](Expr);
identify(%);

                           

        

 

Hello. Please help me find the value of the integral.

Integral.mw

How to write names on an drawing P1, P2, P3, Pn... I try
textplot({seq([(coordinates(point(P[i]))), cat("P", i)], i = 0 .. 5)}, 'align' = {'above', 'left'}):;
i got the message : wrong number of arguments that i don't how correct. Thank you.

restart; with(plots):unprotect(gamma):
_EnvHorizontalName := 'x';_EnvVerticalName := 'y';
line := proc (x1, y1, x2, y2) options operator, arrow; (x-x1)*(y2-y1)-(y-y1)*(x2-x1) end proc:
R := 3:
ang := [0, (1/3)*Pi, 3*Pi*(1/4)+.2, 7*Pi*(1/6)+.4, 8*Pi*(1/5), 13*Pi*(1/7)]:
for i to 6 do P || i := [R*cos(ang[i]), R*sin(ang[i])] end do:
pts := [seq(P || i, i = 1 .. 6)]:
for i to 6 do tang || i := x*P || i[1]+y*P || i[2] = R^2 end do:
sol:=solve({tang1,tang3},{x,y}):Q13:=[subs(sol,x),subs(sol,y)]:
sol := solve({tang1, tang2}, {x, y}): Q1 := [subs(sol, x), subs(sol, y)]:
sol := solve({tang2, tang3}, {x, y}): Q2 := [subs(sol, x), subs(sol, y)]:
sol := solve({tang3, tang4}, {x, y}): Q3 := [subs(sol, x), subs(sol, y)]:
sol := solve({tang4, tang5}, {x, y}): Q4 := [subs(sol, x), subs(sol, y)]:
sol := solve({tang5, tang6}, {x, y}): Q5 := [subs(sol, x), subs(sol, y)]:
sol := solve({tang1, tang6}, {x, y}): Q6 := [subs(sol, x), subs(sol, y)]:
ptQ := [seq(Q || i, i = 1 .. 6)]:
line14 := line(Q1[1], Q1[2], Q4[1], Q4[2]): L14 := implicitplot(line14, x = -10 .. 10, y = -10 .. 10, color = red):
line25 := line(Q2[1], Q2[2], Q5[1], Q5[2]): L25 := implicitplot(line25, x = -10 .. 10, y = -10 .. 10, color = red):
line36 := line(Q3[1], Q3[2], Q6[1], Q6[2]): L36 := implicitplot(line36, x = -10 .. 10, y = -10 .. 10, color = red):
sol := solve({line14, line25}, {x, y}): I1 := [subs(sol, x), subs(sol, y)]:

lineP23 := line(P2[1], P2[2], P3[1], P3[2]): lineP56 := line(P5[1], P5[2], P6[1], P6[2]):
sol := solve({lineP23, lineP56}, {x, y}): gamma := [subs(sol, x), subs(sol, y)]:
lineP12 := line(P1[1], P1[2], P2[1], P2[2]): lineP45 := line(P4[1], P4[2], P5[1], P5[2]):
sol := solve({lineP12, lineP45}, {x, y}): beta := [subs(sol, x), subs(sol, y)]:
lineP34 := line(P3[1], P3[2], P4[1], P4[2]): lineP16 := line(P1[1], P1[2], P6[1], P6[2]):
sol := solve({lineP16, lineP34}, {x, y}): alpha := [subs(sol, x), subs(sol, y)]:
pl:= line(alpha[1], alpha[2], gamma[1], gamma[2]):
hexa := seq(implicitplot(tang||i, x = -20 .. 20, y = -20 .. 20, linestyle=3,color = blue),i=1..6):
#hexa:=plot([seq([P||i,P||(i mod 6)+1],i=1..6),color=green): 
#hexa:=plot([seq([P||i,P||(i+1)],i=1..6)],thickness=4,color=green): 
hex:=plot([[Q1,Q2],[Q2,Q3],[Q3,Q4],[Q4,Q5],[Q5,Q6],[Q6,Q1]],thickness=4,color=green):
tp := textplot({seq([op(pts[i]), cat("P", i)], i = 1 .. 6)}, 'align' = {'above', 'left'}):
tpq := textplot({seq([op(ptQ[i]), cat("Q", i)], i = 1 .. 6)}, 'align' = {'above', 'left'}):
TP:=textplot([[I1[],"I"],[alpha[],"alpha"],[beta[],"beta"],[gamma[],"gamma"]],'align' = {'above', 'left'}):
slopes:=[seq(((dx,dy)->dy/dx)((pts[i]-pts[(i mod 6)+1])[]),i=1..6)]:
lines:=zip((pt,slope)->y=slope*(x-pt[1])+pt[2],pts,slopes):
plotpts:=plot(pts,style=point,colour=red,symbol=solidcircle,symbolsize=5):
plotptQ:=plot(ptQ,style=point,colour=black,symbol=solidcircle,symbolsize=5):
plotlines:=plot(rhs~(lines),style=line,linestyle=3,colour=magenta):
cir:=implicitplot(x^2+y^2=R^2,x=-R..R,y=-R..R,color=black):
PL:=implicitplot(pl,x=-1..15,y=-15..6,color=blue,thickness=3):
display(plotpts,plotptQ,plotlines,hex,cir,L14,L25,L36,PL,tp,tpq,TP,axis = [gridlines = [4, color = blue]],
view=[-6..10,-15..6],scaling=constrained,axes=none,size=[800,800]):
How to simplify this program ? Thank you.

Hello. I have certain requirements for drawings. Please tell me, is it possible to implement them when plotting in Maple? Thanks.
Drawing width 15 cm
Be sure to frame and mesh. The thickness of the frame and serifs on the axes is 0.5 pt. Mesh thickness 0.25 pt. The length of the serifs is 1.2 mm.
The thickness of the graph itself is 1 pt
Axis names 10 pt

How to show that the function z0(t)=t²+exp(-t)  is solution of z"(t)+2z'(t)+z(t)=2exp(-t); Thank you.

Hello. Tell me, please, is it possible to somehow remove the labels on the axes in Maple when plotting graphs? I mean labels 1, 2, 3 and so on. Thank you for your help.

If the equation with homogeous coordinates in the base (A,B,C) is pYZ+qZX+rXY=0, an affine  equation in the base (A,AB,AC) is
pxy+qy(1-x-y)+r(1-x-y)=0, it is also written rx²+(q+r-p)xy+qy²=0. The discrim of  trinom is rx²+(q+r-p)xy+qy² is of the sign of 
delta=(q+r-p)²-4qr; We get a hyperbole, a parabole, an ellipse according to delta is >0, =0 or <0 respectvely. How implement
this property. Thank you very much.

The problem: to simplify the expression

for any negative  x  and  y .

Below we see that Maple copes with the task brilliantly (example 1). For example, it presents  sqrt(x*y)  as  sqrt(-x)*sqrt(-y)  and so on. But the same technique, applied only to the numerator of this expression does not give the desired presentation in the form of a square (example 2 and example 3).

restart;
# Example 1
A:=(x+y-2*sqrt(x*y))/(sqrt(-x)+sqrt(-y));
simplify(A) assuming negative;
factor(%,{sqrt(-x),sqrt(-y)});

                                         

restart;
# Example 2
B:=x+y-2*sqrt(x*y);
simplify(B) assuming negative;
factor(%,{sqrt(-x),sqrt(-y)});

                                           

 

restart;
# Example 3
B:=x+y-2*sqrt(x*y);
R:=simplify(B) assuming positive;
combine(R) assuming positive;
factor(R,{sqrt(x),sqrt(y)});

                                                   

Two questions:

1. Does anyone know the reasons for this behavior.

2. Does anyone know an easy way to simplify in examples 2 and 3 (without  substitutions  like  x=+-u^2  and  y=+-v^2 and so on,  of course)

 

Download sqrt.mw

 

An exercise of 1948 commits me to form the equation of conics passing by 3 points. Let P=0, Q=0, R=0 be the equations of the sides of the triangle ABC; if we associate these sides 2 to 2 we obtain 3 conics passing through points A, B, C having for equations QR=0, RP=0, PQ=0. As a result, the general equation of conics around the triangle ABC is: aQR+bRP+cPQ=0. P, Q, R being equations of the form mx+ny+pz=0 (so-called homogeneous coordinates). Then change to refined coordinates with x+y+z=1 (formula found on the internet and surely misinterpreted). Is it necessary to change of base ? Thank you for your help.

In the two examples below (in the second example, the range for the roots is simply expanded), we see bugs in both examples (Maple 2018.2). I wonder if these errors are fixed in Maple 2020?
 

restart;

solve({log[1/3](2*sin(x)^2-3*cos(2*x)+6)=-2,x>=-7*Pi/2,x<=-2*Pi}, explicit, allsolutions); # Example 1 - strange error message
solve({log[1/3](2*sin(x)^2-3*cos(2*x)+6)=-2,x>=-4*Pi,x<=-2*Pi}, explicit, allsolutions);  # Example 2 - two roots missing

Error, (in assume) contradictory assumptions

 

{x = -(11/3)*Pi}, {x = -(10/3)*Pi}

(1)

plot(log[1/3](2*sin(x)^2-3*cos(2*x)+6)+2, x=-7*Pi/2..-2*Pi);
plot(log[1/3](2*sin(x)^2-3*cos(2*x)+6)+2, x=-4*Pi..-2*Pi);

 

 

Student:-Calculus1:-Roots(log[1/3](2*sin(x)^2-3*cos(2*x)+6)=-2, x=-7*Pi/2..-2*Pi);  # OK
Student:-Calculus1:-Roots(log[1/3](2*sin(x)^2-3*cos(2*x)+6)=-2, x=-4*Pi..-2*Pi);  # OK

[-(10/3)*Pi, -(8/3)*Pi, -(7/3)*Pi]

 

[-(11/3)*Pi, -(10/3)*Pi, -(8/3)*Pi, -(7/3)*Pi]

(2)

 


I am glad that  Student:-Calculus1:-Roots  command successfully handles both examples.

 

Download bugs-in-solve.mw

At the moment t=0, we place a body at 100 ° C in a room at 25°C; we designate by q(t) the temperature at the moment t. The differential equation is.q'(t)+k*(q(t)-25)=t, with k cooling coefficient equal to 2.
Determine the solution that checks the initial condition.
What is the body temperature after 30 minutes.
After how long the temperature drops to 50°C. Thank you for the help.

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