Using plot3d(..., style=surfacecontour, ...) or contourplot3d(...) displays wrong level curves when some axis are switched to a log mode.

Example:

Download WrongLevelCurves.mw

The problem is not dramatic because there is a workaround.
 

Download WrongLevelCurves_Workaround.mw

Even though this question is related to this one 
https://www.mapleprimes.com/questions/234781-How-Can-I-Get-The-Desired-Answer-From-solve
feel it is about a different issue. If any of you feel otherwise feel free to move it to the original one.

In this notional example  the name _Z1~ is created by RootOf: and here is an ad hoc way to catch it.

Download Example_1.mw

In this more complex example an assumption must be made on M to obtain ths desired solution g and the previous method no longer works.

Download Example_2.mw

I have tried using select to "capture" the name _Z2~ but I can't know how to distinguish M~ from _Z2~ (is there a type which could be used?).

Can you helpm fix this?
TIA

I don't understand why the solution of sys_2 isn't those of sys_1 when M__p=1 and M__a=0 ?

Traces of the computation seem to indicate that dsolve proceeds exactly the same for sys_2 and sys_1 .

Please note that sol_1 contains a term of the form t*cos(t) that sol_2 doesn't, thus the question: "Is sol_2 correct?"

Could you help me to fix this?
TIA

Download SomethingWrong.mw

PS: Already, in the following case, dsolve doesn't return the solution of sys_1.

sys_3 := {(A+B)*diff(x(t), t$2)=(A+B)*sin(t)-x(t), x(0)=0, D(x)(0)=0};
sol_3 := dsolve(sys_3)

If I do this

sys_4 := {(A+B)*diff(v(t), t)=(A+B)*sin(t)-x(t), diff(x(t), t)=v(t), x(0)=0, v(0)=0}:
sol_4 := dsolve(sys_4)

I get a very complex solution wich contains a piecewise function which separates the cases A+B=1 and A+B<>1.
Evaluating sol_4 for A+B=1 gives the same expression than sys_1:

simplify(eval(sol_4, A=1-B), trig)
       /       1                  1          1         \ 
      { v(t) = - sin(t) t, x(t) = - sin(t) - - cos(t) t }
       \       2                  2          2         / 

Here is a workaround to get the correct solution of sys_2:

sys_5 := {(M__P+M__A)*diff(v(t), t)=(M__P+C)*sin(t)-x(t), diff(x(t), t)=v(t), x(0)=0, v(0)=0}:
sol_5 := dsolve(sys_5):
simplify(eval(sol_5, [M__P=1, M__A=0, C=0]), trig)
       /       1                  1          1         \ 
      { v(t) = - sin(t) t, x(t) = - sin(t) - - cos(t) t }
       \       2                  2          2         / 

e

I compute the solution of this differential system

shock := piecewise(t <0, 0, t < 1, 10, 0):
sys   := {(M__p+M__a)*diff(x(t), t$2)=M__p*shock-x(t), x(0)=0, D(x)(0)=0}
sol   := unapply(rhs(dsolve(sys)), (M__p,M__a))

I'm interested in 3 quantities:

• the first time tend > 0 such that sol(tend) = 0,
• the time tmax in (0..tend) where sol(tmax) reaches its maximum value,
• the value xmax = sol(tmax).

Since sol has a relatively simple expression, I first attempted to use solve for calculating tend, but that didn't work.
The conclusion is still the same for tmax and xmax.

The values of these 3 quantities that I expect solve to provide, are those obtained using fsolve.

Can you explain me the failures I faced and show me how to force solve to get these values?
TIA

ToyProblem.mw

Given this "pattern" to transfporm an expression into an arrow procedure

e := 1/b:
f := unapply(e, b):
f(3)
                               1
                               -
                               3

I would like now to use this same "pattern" to transform an expression e which depends on 2 indeterminates a and b into a "double layered arrow procedure" f.
Here is a notional example

e := a/b:
f := unapply(unapply(e, b), a);
                     a
           a -> b -> -
                     b

 f(A)(B)
                              a  
                              -
                              B  

How can I proceed in order that f(A)(B) gives A/B ?

TIA