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Replace a variable with a function

asked 2014-07-10 07:42:14 -0500

tetraeder gravatar image

updated 2014-07-11 15:04:49 -0500

kcrisman gravatar image

Hello i hope anyone can help me with the following problem.

i have the following code

T_w,y,T_m,k_f,rho_f,U_0,h_mstar,R, alpha, delta, Phi = var('T_w y T_m k_f rho_f U_0 h_mstar R alpha delta Phi')

T=T_w+y*(-2*(T_w-T_m)/delta+rho_f*U_0*h_mstar*cos(Phi)/k_f)+y^2*((T_w-T_m)/delta^2-rho_f*U_0*h_mstar*cos(Phi)/(k_f*delta))

u=-6*U_0*R*y*(y-delta)*sin(Phi)/delta^3

integralTu=integrate(T*u,y,0,delta)

now i would like to differentiate integralTu with respect to x (diff(integralTu,x)), but delta must be a function of x (i first defined it as a variable because of the integration).

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@tetraeder, @kcrisman: add tags?

slelievre gravatar imageslelievre ( 2014-07-11 05:50:45 -0500 )edit

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answered 2014-07-10 08:42:17 -0500

kcrisman gravatar image

You are right that Maxima doesn't want delta to be a function, since it's a limit of integration. That was slightly surprising to me.

Anyway, it turns out you can fix this post-hoc.

sage: integralTu.subs(delta=function('delta',x))
1/10*(2*U_0*h_mstar*rho_f*cos(Phi)*delta(x)^4 + 7*T_m*k_f*delta(x)^3 + 3*T_w*k_f*delta(x)^3)*R*U_0*sin(Phi)/(k_f*delta(x)^3)

That might seem a little hackish, but is just fine.

sage: IT = integralTu.subs(delta=function('delta',x))
sage: diff(IT,x)
1/10*(8*U_0*h_mstar*rho_f*cos(Phi)*delta(x)^3*D[0](delta)(x) + 21*T_m*k_f*delta(x)^2*D[0](delta)(x) + 9*T_w*k_f*delta(x)^2*D[0](delta)(x))*R*U_0*sin(Phi)/(k_f*delta(x)^3) - 3/10*(2*U_0*h_mstar*rho_f*cos(Phi)*delta(x)^4 + 7*T_m*k_f*delta(x)^3 + 3*T_w*k_f*delta(x)^3)*R*U_0*sin(Phi)*D[0](delta)(x)/(k_f*delta(x)^4)

By the way, I've updated your question because you needed one more variable defined.

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Thanks, it works!

tetraeder gravatar imagetetraeder ( 2014-07-10 08:53:46 -0500 )edit

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Asked: 2014-07-10 07:42:14 -0500

Seen: 46 times

Last updated: Jul 10 '14