@Robert Israel I have often enough suggested that the couple of occurences of subs in the animate procedure be replaced by eval's. I never got any response. Is it harmful in any way?

What I do to remedy the situation is the following and illustrated with Erik's example:

restart;

`plots/animate`:=subs(subs=((x,y)->eval(y,x)),eval(`plots/animate`)):

with(plots);
f := proc(t) if 5 < t then sin(t) else 0 end if end proc;
ball := proc (x, y) plots[pointplot]([[x, y]], color = blue, symbol = solidcircle, symbolsize = 40) end proc;
animate(ball, [4, 'f'(t)], t = 0 .. 10, frames = 200);

In Maple 14 the result doesn't simplify so easily. However, doing simplify(xxx, symbolic); on the real looking solution does give the same result as Maple 13 does with just simplify.

In Maple 14 the result doesn't simplify so easily. However, doing simplify(xxx, symbolic); on the real looking solution does give the same result as Maple 13 does with just simplify.

Use 'numframes', not 'frames'. See the help page for DEplot (not DEplot3d).

Use 'numframes', not 'frames'. See the help page for DEplot (not DEplot3d).

@hirnyk Yes, that was my point. It was intended as a counterexample to Axel Vogt's argument.

@hirnyk Yes, that was my point. It was intended as a counterexample to Axel Vogt's argument.

@Axel Vogt It was for you.

@Axel Vogt It was for you.

Take F(x) = x^2*sin(1/x) for x<>0 and F(0) = 0. Then F is  differentiable everywhere. The derivative at zero is zero by a direct application of the definition of differentiability.

However, F'(x) = 2*x*sin(1/x) -cos(1/x) doesn't have a limit at 0. 

Edited: This a counterexample to Axel Vogt's argument.

Take F(x) = x^2*sin(1/x) for x<>0 and F(0) = 0. Then F is  differentiable everywhere. The derivative at zero is zero by a direct application of the definition of differentiability.

However, F'(x) = 2*x*sin(1/x) -cos(1/x) doesn't have a limit at 0. 

Edited: This a counterexample to Axel Vogt's argument.

@Axel Vogt It seems that you are right. I'm glad that I said, "If Maple is right .. ".

eq:=(-t)^5=abs((t^2+t+1)/(-t));
eq2:=eval(eq,t=x+I*y) assuming x::real,y::real;
expand(%);
sys:={Re(lhs(%))=rhs(%),Im(lhs(%))=0} assuming real;
sol:=solve(sys,{x,y}):
evalf(sol):
L:=remove(has,[%],I);

Here Maple finds 3 solutions, but clearly if t is a solution then conjugate(t) is a solution, thus 5 solutions.

@Axel Vogt It seems that you are right. I'm glad that I said, "If Maple is right .. ".

eq:=(-t)^5=abs((t^2+t+1)/(-t));
eq2:=eval(eq,t=x+I*y) assuming x::real,y::real;
expand(%);
sys:={Re(lhs(%))=rhs(%),Im(lhs(%))=0} assuming real;
sol:=solve(sys,{x,y}):
evalf(sol):
L:=remove(has,[%],I);

Here Maple finds 3 solutions, but clearly if t is a solution then conjugate(t) is a solution, thus 5 solutions.

Since we are in the complex plane, abs(z) is not just either z or -z.

Since we are in the complex plane, abs(z) is not just either z or -z.