HSC 2016 Past paper question on rate of change (1 Viewer)

[narges-ts]

This questions is from HSC paper 2016 Q.12 b)

In a chemical reaction, a compound X is formed from a compound Y. The mass
in grams of X and Y are x (t) and y (t) respectively, where t is the time in seconds
after the start of the chemical reaction.

Throughout the reaction the sum of the two masses is 500 g.
At any time t, the rate at which the mass of compound X is increasing is
proportional to the mass of compound Y.

At the start of the chemical reaction, x = 0 and dx/dt=2

(i) show that dx/dt = 0.004(500-x)

 

[jazz519]

X(t)+y(t) =500

dx/dt is proportional to mass of y so dx/dt = ky(t) where k is a constant
X(0) = 0, so y(0)=500

Sub that into the equation
2= k*500
K=0.004

Thus, as y(t)=500-x(t) and k=0.004

dx/dt = 0.004(500-x)