91Ó°ÊÓ

Let's denote the amount of U as a function of time, \(u(t)\) and the amount of W as a function of time as \(w(t)\). Since the rate of change of X being produced is proportional to the product of the amounts of U and W present, we can write the rate of change as: \(\frac{dx}{dt} = k \cdot u(t) \cdot w(t)\) where x(t) represents the amount of the compound X at any time t and k is the proportionality constant. Next, we need to find the rates of change of U and W as well. Since the compound X is formed by the combination of 2 parts of U and 3 parts of W, we can set up the following relationships: \(\frac{du}{dt} = -2 \cdot \frac{dx}{dt}\) \(\frac{dw}{dt} = -3 \cdot \frac{dx}{dt}\)

Now, we have a system of 3 differential equations: \(\frac{dx}{dt} = k \cdot u(t) \cdot w(t)\) \(\frac{du}{dt} = -2 \cdot \frac{dx}{dt}\) \(\frac{dw}{dt} = -3 \cdot \frac{dx}{dt}\) Let's substitute the second and third equations into the first equation: \(\frac{-1}{2} \cdot \frac{du}{dt} = k \cdot u(t) \cdot w(t) \) \(\frac{-1}{3} \cdot \frac{dw}{dt} = k \cdot u(t) \cdot w(t) \) Dividing the first equation by the second equation, we have: \(\frac{-1/2 \cdot \frac{du}{dt}}{-1/3 \cdot \frac{dw}{dt}} = 1 \) \(\frac{3}{2} \cdot \frac{du}{dt} = \frac{dw}{dt}\) Now, we can integrate this differential equation: \(\int \frac{3}{2} \cdot \frac{du}{dt} dt = \int \frac{dw}{dt} dt \) \(\frac{3}{2} \cdot \int du = \int dw \) \(\frac{3}{2} u(t) = w(t) + C\) where C is the constant of integration.

Using the initial conditions, at t = 0, u(0) = 10 lb. and w(0) = 8 lb., we can find the value of the constant C: \(\frac{3}{2} \cdot 10 = 8 + C \) \( 15 = 8 + C \) \( C = 7\) So, the relationship between u(t) and w(t) is: \(\frac{3}{2} u(t) = w(t) + 7\) Now, at t = 1, the amount of X produced is 2 lb. Therefore, at t = 1, we have u(1) = 8 lb. and w(1) = 4 lb. From the relationship we derived, we can find the value of k: \(\frac{dx}{dt} = k \cdot u(t) \cdot w(t)\) \( \frac{dx}{dt} = k \cdot 8 \cdot 4 \) \( \frac{2}{1} = k \cdot 8 \cdot 4 \) \( k = \frac{1}{16} \) Finally, the amount of X produced in time t is given by: \( x(t) = k \cdot \int u(t) \cdot w(t) dt \) Using the relationship between u(t) and w(t) we found earlier and the value of k, we can find x(t): \( x(t) = \frac{1}{16} \cdot \int u(t) ( \frac{3}{2}u(t) - 7) dt \) To find x(t), we would integrate this equation over the desired time interval (from t = 0 to t = T, where T is the chosen time). That will conclude the step-by-step solution to the given problem.