91Ó°ÊÓ

Since we are given that the rate of change is proportional to the amount present at any time, we can write that as a differential equation. Let x(t) be the amount of the first chemical present at time t, and k be the constant of proportionality. We can then write the differential equation as: \[\frac{dx}{dt} = -kx(t)\] We have a negative sign in the equation because the amount of the chemical is decreasing over time.

Integration can be used to solve the differential equation. Divide both sides of the equation by x(t) and integrate both sides with respect to t: \[\int \frac{1}{x(t)}dx(t) = -k \int dt\] After integrating, we obtain: \[\ln|x(t)| = -kt + C\] Exponentiating both sides: \[x(t) = e^{-kt + C}\] which, after rearranging, can be written as: \[x(t)= A e^{-kt}\] where \(A= e^C\).

Using the given data, we have two conditions: x(1)=50 and x(3)=25. Plugging these into our equation, we can solve for k and A: For \(x(1)=50\), we have: \[50 = A e^{-k}\] For \(x(3)=25\), we have: \[25 = A e^{-3k}\] Dividing the second equation by the first equation, and simplifying, we find: \[\frac{1}{2} = e^{-2k}\] Taking the natural logarithm of both sides: \[-\ln 2 = -2k\] Now, we can find the value of k: \[k = \frac{\ln 2}{2}\] Substituting the value of k into the first equation: \[50 = A e^{-\frac{\ln 2}{2}}\] Solving for A, we obtain: \[A = 50 e^{\frac{\ln 2}{2}}\]

(a) To find the initial amount x(0), we substitute t=0 into the equation and solve for x(0): \[x(0) = A e^{0} = 50 e^{\frac{\ln 2}{2}}\] So, initially, there were \(50 e^{\frac{\ln 2}{2}}\) grams of the first chemical. (b) To find the amount of chemical remaining after 5 hours, we substitute t=5 into the equation: \[x(5) = 50 e^{\frac{\ln 2}{2}} e^{-5\frac{\ln 2}{2}} = 50 \cdot \frac{1}{32}\] So, after 5 hours, only \(\frac{50}{32}\) grams of the first chemical will remain. (c) To find the time t when only 2 grams of the chemical remain, we substitute x(t)=2 into the equation and solve for t: \[2 = 50 e^{\frac{\ln 2}{2}} e^{-k t}\] Dividing both sides by 50, we get: \[\frac{1}{25} = e^{-\frac{\ln 2}{2} (1-t)}\] Taking the natural logarithm of both sides: \[\ln \frac{1}{25} = -\frac{\ln 2}{2} (1-t)\] Now, we solve for t: \[t = 1 + \frac{2\ln\frac{1}{5}}{\ln 2}\] When only 2 grams of the chemical remain, approximately \(\displaystyle 1 + \frac{2 \ln\frac{1}{5}}{\ln 2}\) hours will have elapsed.