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8 years, 131 days

MaplePrimes Activity

These are answers submitted by nm

in numerical integration, you can't have parameter with no numerical value in the integrand. (how is Maple going to find numerical value if a has no value?)

Assign a some value before calling int, then it will work

f := (x, y) ->(1/2)*a*(sinh(y-x^2)+tanh(x-y^3)); 
sol := evalf(int(int(f(x, y), x = -6 .. 5), y = -5 .. 5));



   evalf(int(int((1/2)*a*(sinh(y-x^2)+tanh(x-y^3)), x = -6 .. 5), y = -5 .. 5)),

You can also use Explore, and change the raduis using the slider with the mouse.


Explore(plots:-display( plottools:-sphere([0,0,0],r),view=[-3 .. 3,-3..3,-3..3]),

always use exact numbers, unless there is a reason not to

a := 12/100;
kn := n -> (n + 1/2)*Pi/a;
g := (n, x) -> cos(kn(n)*x);
g(1, a);


You could try collect

p:= (1+x+y)*(2*x-y*x+z)*(y^2-z*y);

Just tell it the full specs of y(x) which is option after the ode and bc. This is needed when you have y(L) and y(x) in the input, so it knows which is the actual dependent variable to solve for.

And I is complex number by default So changed it to I__0 for the  second moment of area. You can use different symbol if you want.

ode := diff(y(x), x$2) = -P*x/(I__0*E);
bc  := y(L)=0,D(y)(L)=0;
sol := dsolve([ode,bc],y(x))


              [0, 0, 0]

30 tons pollutant per year coming in? this means the concetration of pollution coming in, from the river, is 1/2 ton per km^3? Since there is 60 km^3 per year coming in? 

This is very strange way to state these type of problems. Normally pollutant is given in the problem as mass/volume directly and not mass per year.  but given the above, you could do (can't type latex, so here is screen shot)