In Purdue's Chemistry department, the chemists have found that in a water based solution containing 24 grams of certain undissolved chemicals, the rate of change of the amount of chemicals dissolved in the solution is proportional to the amount of the undissolved chemicals. Let Q(t) (in grams) be the amount of dissolved chemicals at time t and let k be the positive proportionality constant. The differential equation describing the given situation is:
To derive the differential equation that describes the situation in Purdue's Chemistry department, we need to analyze the information provided.
Given:
- The amount of undissolved chemicals in the water-based solution is 24 grams.
- The rate of change of the amount of chemicals dissolved in the solution is proportional to the amount of undissolved chemicals.
- Let Q(t) (in grams) represent the amount of dissolved chemicals at time t.
- k is the positive proportionality constant.
We can express the rate of change of the amount of chemicals dissolved, dQ(t)/dt, as:
dQ(t)/dt = k * f(t)
where f(t) represents the amount of undissolved chemicals at time t.
Since the rate of change of the amount of chemicals dissolved is proportional to the amount of undissolved chemicals, we can conclude that f(t) is Q(t) itself (as the amount dissolved is proportional to the amount undissolved). Therefore, we can write:
f(t) = Q(t)
Substituting this into our equation, we have:
dQ(t)/dt = k * Q(t)
So, the differential equation that describes the given situation is:
dQ(t)/dt = k * Q(t)
The differential equation describing the given situation is:
dQ/dt = k(24 - Q)
where dQ/dt represents the rate of change of the amount of dissolved chemicals with respect to time t, k is the positive proportionality constant, and (24 - Q) represents the amount of undissolved chemicals remaining in the solution at time t.
if there are Q grams of dissolved chemicals, then there are 24-Q undissolved. So,
dQ/dt = k(24-Q)
Note that the question was kind of loosely worded.