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While obtaining the answer for A3 and A3 in a set of equations, the answer is in vector-format(??). I want to use these answer further on. How can I pick these values? Right now I just copy and paste the answer but that's not ideal.

IN: eq1=-A3-A4Kl+A3cos(Kl)+A4sin(Kl)+((qzcl^2)/(2K^2))==0

eq2=-A3K^2cos(Kl)-A4K^2sin(Kl)+(qzc/(K^2))==0

solve([eq1,eq2],A3,A4)

OUT: ​[[A3 == 1/2(2Kl - (K^2l^2 + 2)sin(Kl))qzc/(K^5lcos(Kl) - K^4sin(Kl)), A4 == 1/2((K^2l^2 + 2)cos(Kl) - 2)qzc/(K^5lcos(Kl) - K^4sin(Kl))]]

How to get A3=... and A4=...

While obtaining the answer for A3 and A3 in a set of equations, the answer is in vector-format(??). I want to use these answer further on. How can I pick these values? Right now I just copy and paste the answer but that's not ideal.

IN: eq1=-A3-A4Kl+A3cos(Kl)+A4sin(Kl)+((qzcl^2)/(2K^2))==0IN:

eq1=-A3-A4*K*l+A3*cos(K*l)+A4*sin(K*l)+((qzc*l^2)/(2*K^2))==0
 eq2=-A3*K^2*cos(K*l)-A4*K^2*sin(K*l)+(qzc/(K^2))==0
 
solve([eq1,eq2],A3,A4)solve([eq1,eq2],A3,A4)

OUT: ​[[A3 OUT:

​ [[A3 == 1/2(2Kl 1/2*(2*K*l - (K^2l^2 (K^2*l^2 + 2)sin(Kl))qzc/(K^5lcos(Kl) 2)*sin(K*l))*qzc/(K^5*l*cos(K*l) - K^4sin(Kl)), K^4*sin(K*l)), A4 == 1/2((K^2l^2 1/2*((K^2*l^2 + 2)cos(Kl) 2)*cos(K*l) - 2)qzc/(K^5lcos(Kl) 2)*qzc/(K^5*l*cos(K*l) - K^4sin(Kl))]]K^4*sin(K*l))]]

How to get A3=... and A4=...

While obtaining the answer for A3 and A3 in a set of equations, the answer is in vector-format(??). I want to use these answer further on. How can I pick these values? Right now I just copy and paste the answer but that's not ideal.

IN:

eq1=-A3-A4*K*l+A3*cos(K*l)+A4*sin(K*l)+((qzc*l^2)/(2*K^2))==0

eq2=-A3*K^2*cos(K*l)-A4*K^2*sin(K*l)+(qzc/(K^2))==0

solve([eq1,eq2],A3,A4)

OUT:

​ ​[[A3 == 1/2*(2*K*l - (K^2*l^2 + 2)*sin(K*l))*qzc/(K^5*l*cos(K*l) - K^4*sin(K*l)), A4 == 1/2*((K^2*l^2 + 2)*cos(K*l) - 2)*qzc/(K^5*l*cos(K*l) - K^4*sin(K*l))]]

How to get A3=... and A4=...

While obtaining the answer for A3 and A3 in a set of equations, the answer is in vector-format(??). I want to use these answer further on. How can I pick these values? Right now I just copy and paste the answer but that's not ideal.

IN:

eq1=-A3-A4*K*l+A3*cos(K*l)+A4*sin(K*l)+((qzc*l^2)/(2*K^2))==0

eq2=-A3*K^2*cos(K*l)-A4*K^2*sin(K*l)+(qzc/(K^2))==0

solve([eq1,eq2],A3,A4)

OUT:

​[[A3 == 1/2*(2*K*l - (K^2*l^2 + 2)*sin(K*l))*qzc/(K^5*l*cos(K*l) - K^4*sin(K*l)), A4 == 1/2*((K^2*l^2 + 2)*cos(K*l) - 2)*qzc/(K^5*l*cos(K*l) - K^4*sin(K*l))]]

How to get A3=... and A4=...