Sugar dissolves in water at a rate proportional to the...

Sugar dissolves in water at a rate proportional to the amount still undissolved. If there were $50 \mathrm{kg}$ of sugar present initially, and at the end of $5 \mathrm{h}$ only $20 \mathrm{kg}$ are left, how much longer will it take until $90 \%$ of the sugar is dissolved?

Key Concepts

Exponential Decay

Exponential decay refers to a process in which the rate of change of a quantity is directly proportional to the current amount of that quantity. This results in a function that decreases rapidly at first and then more slowly over time, commonly modeled by an equation of the form A(t) = A0 * exp(-kt), where A0 is the initial amount, k is the decay constant, and t is time.

Differential Equations

A differential equation is a mathematical equation that relates a function with its derivatives, representing rates of change. In contexts such as exponential decay, the differential equation takes the form dA/dt = -kA, indicating that the derivative of the amount with respect to time is proportional to the amount itself. Solving such equations involves integration and can yield an explicit formula for the changing quantity over time.

Initial Conditions

Initial conditions specify the state of the system at the beginning of observation, which is essential for determining the particular solution to a differential equation. For instance, knowing the initial amount of a substance allows calculation of the constant in the decay equation, ensuring the solution correctly represents the system's behavior from time zero onward.

Solving for the Decay Constant

Determining the decay constant, k, is a crucial step in modeling exponential decay. This typically involves using known data points—such as the amount remaining at a certain time—to solve for k in the exponential decay equation. Once the decay constant is known, it can be used to predict the behavior of the system, including estimating how long it will take for a specified percentage of the substance to be dissolved.

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