# Why does differentiating Boyle's Law to find chance in volume lead to positive change in V as well as P?

here's my work:

V1*P1=V2*P2 where V1=1000 m^3 P1=5 N/m^2 P2=10 N/m^2 and dP/dT=.05 N/m^2

if V1(dP/dT) + P1(dV/dT)=V2(dP/dT) + P2(dV/dT)

then dV/dT=(V2-V1)*(dP/dT)/(P1-P2)

so the increase in pressure gave decrease in volume but dV/dT comes out positive...where did I go wrong?

V1*P1=V2*P2 where V1=1000 m^3 P1=5 N/m^2 P2=10 N/m^2 and dP/dT=.05 N/m^2

if V1(dP/dT) + P1(dV/dT)=V2(dP/dT) + P2(dV/dT)

then dV/dT=(V2-V1)*(dP/dT)/(P1-P2)

so the increase in pressure gave decrease in volume but dV/dT comes out positive...where did I go wrong?

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