Since A=0, the equation relating the variables tells us:
0100This will probably be useful information. Since we’re also given the value of a derivative, let’s differentiate the equation relating the variables implicitly with respect to t. For ease of notation, we will write dtdA=A′, etc.A′B+AB′At t=10, A=C=D=0:A′B+0A′Bat t=10, A′=2 units per second:2BB=log(C2+D2+1)=C2+D2+1=C2+D2=C=D=C2+D2+12CC′+2DD′=0+0+10+0=0=0=0.