Hi, could someone help me please with these?
Problem 1) solve the system equations: dx/dt=4x+3y dy/dt=-x+t with x(0)=2, y(0)=-1 as initial conditions.
Problem 2) Find the surface area of that portion of the surface z=1- root-square of x^2+y^2 that lies in the first octant.
Thanks in advance.

Question

Hi, could someone help me please with these?
Problem 1) solve the system equations: dx/dt=4x+3y  dy/dt=-x+t with x(0)=2, y(0)=-1 as initial conditions.
Problem 2) Find the surface area of that portion of the surface z=1- root-square of x^2+y^2 that lies in the first octant.
Thanks in advance.

anonymous · Answer

1)try to integrate the two equations

anonymous · Answer

p.1: from equation 1, we get x''=4x'+3y', substituting y'=-x+t, the problem is now a differential equation: x''-4x'+3x=3t with i.c. x(0)=2, x'(0)=5. if you've done 18.03, you can solve it easily.
p.2: area of the portion is $$\iint_{S}dS=\iint_{R_xy}\sqrt{(z'_x)^2+(z'_y)^2+1}dxdy$$. S is lying over area Rxy.