Mechoption

24 Reputation

2 Badges

15 years, 130 days

MaplePrimes Activity


These are questions asked by Mechoption

Let y1=exp(a*t), and y2=ln(y1)=a*t The error in y1 and y2 is found from sigma=sqrt(diff(y1,a)^2*siga^2) where siga is the uncertainty in a. Therefore, sigy1 = exp(a*t)*t*siga sigy2 = t*siga Therefore, y1 = y1 +/- y1*t*siga y2 = y2 +/- t*siga What I don't get, is as exp(y2)=y1 , then this gives me: y1 = y1 +/- exp(t*siga) What is the correct errror in y1? y1*t*siga or exp(t*siga) ??

How do I calculate the derivative of C(r,t) with respect to r evaluated at r=a? I can do it in multiple lines like this:

temp:=diff(C(r,t),r);

r:=a;

R:=temp;

r:='r';

But this is a pretty ghetto way of doing this, how do I solve for R in one line? I thought diff or D could do it but I can't seem to get the syntax right :(

Thanks.

I'm trying to solve a diffusion PDE with initial conditions, but can't seem to get the syntax right? Here's what I have: pde:= diff(C(r, t), t) = Di*(diff(r^2*(diff(C(r, t), r)), r))/r^2-lambda*C(r, t)+Beta Initial conditions (which I'm not sure how to define) C(r,0)=0 C(0,t)=finite C(a,t)=0 r=0 to a t>0 I can use pdsolve(pde,build) to get a reasonable solution, but when I try: pdsolve(eq, C(a, t) = 0, build) I get: Error, (in pdsolve/info) wrong extra arguments: {C(a, t) = 0}
Page 1 of 1