@goli In my example x(t) = tan(t) is the solution satisfying the initial condition x(0)=0 and having as its maximal interval of definition (-Pi/2, Pi/2) using the usual requirement that the solution has to be (at least) defined and continuous on the interval. If you give up only continuity then

x(t) = piecewise(t<Pi/2,tan(t),tan(t-Pi/4))

is a solution satisfying x(0) = 0 and defined on the open interval (-Pi/2, 3*Pi/4), but it is not continuous at t = Pi/2.

Instead of tan(t-Pi/4) you could have used tan(t-c) for infinitely many other values of c and still have a solution defined on an interval containing (-Pi/2,Pi/2) as a proper subset.

That is why I say that it doesn't make sense to continue the solution on the other side of the singularity. If you  give up the requirement that the solution be defined at all points of the interval then the situation is of course no better.

@goli In my example x(t) = tan(t) is the solution satisfying the initial condition x(0)=0 and having as its maximal interval of definition (-Pi/2, Pi/2) using the usual requirement that the solution has to be (at least) defined and continuous on the interval. If you give up only continuity then

x(t) = piecewise(t<Pi/2,tan(t),tan(t-Pi/4))

is a solution satisfying x(0) = 0 and defined on the open interval (-Pi/2, 3*Pi/4), but it is not continuous at t = Pi/2.

Instead of tan(t-Pi/4) you could have used tan(t-c) for infinitely many other values of c and still have a solution defined on an interval containing (-Pi/2,Pi/2) as a proper subset.

That is why I say that it doesn't make sense to continue the solution on the other side of the singularity. If you  give up the requirement that the solution be defined at all points of the interval then the situation is of course no better.

I just installed Maple 15 and checked that indeed the subs-problem in animate is corrected in Maple 15, meaning that the two occurrences of subs have been replaced with eval's. 

I just installed Maple 15 and checked that indeed the subs-problem in animate is corrected in Maple 15, meaning that the two occurrences of subs have been replaced with eval's. 

@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);

@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.