Question

Verify that the function $$M(t)=K\left(\frac{M_{0}}{K}\right)^{\exp (-r t)}$$ satisfies the properties \(M(0)=M_{0}\) and \(\lim _{t \rightarrow \infty} M(t)=K\).

Short answer

Answer: Yes, the function satisfies both properties.

Step-by-step solution

Verify M(0) = M_{0}

Start by plugging t=0 into the function and simplifying. $$M(t)=K\left(\frac{M_{0}}{K}\right)^{\exp (-r t)}$$ $$M(0)=K\left(\frac{M_{0}}{K}\right)^{\exp (-r\cdot0)}$$ Since any number raised to the power of 0 is equal to 1: $$M(0)=K\left(\frac{M_{0}}{K}\right)^{1}$$ $$M(0)=KM_{0}/K$$ It is clear that: $$M(0)=M_{0}$$

Verify lim_(t->∞) M(t) = K

Now, we look at the limit as t approaches infinity. $$\lim _{t \rightarrow \infty} M(t)=K\left(\frac{M_{0}}{K}\right)^{\exp (-r t)}$$ As t approaches infinity, the expression inside the exponential function (-rt) approaches negative infinity, provided that r>0. $$\exp (-\infty)=0$$ Therefore: $$\lim _{t \rightarrow \infty} M(t)=K\left(\frac{M_{0}}{K}\right)^{0}$$ Since any number raised to the power of 0 is equal to 1: $$\lim _{t \rightarrow \infty} M(t)=K$$ Thus, we have shown that the given function satisfies both properties, M(0)=M_{0} and \(\lim _{t \rightarrow \infty} M(t)=K\).

Key concepts

Limit of a Function

The limit of a function explores how a function behaves as the input approaches a particular value, often infinity. For the function \(M(t)=K\left(\frac{M_{0}}{K}\right)^{\exp (-r t)}\), we are interested in the behavior of \(M(t)\) as \(t\) approaches infinity. This means we're determining what value \(M(t)\) tends towards as \(t\) grows larger.

• When we considered the limit \(\lim_{t \rightarrow \infty} M(t)\), it's essential to note that \(\exp(-rt)\) decreases towards 0 as \(t\) increases if \(r > 0\).
• Why? Because \(-rt\) becomes very negative, and the exponential of a large negative number is close to zero. So the term turns into \(\left(\frac{M_{0}}{K}\right)^{0} = 1\).

Finally, we find that \(\lim_{t \rightarrow \infty} M(t) = K\), indicating that the function levels off at \(K\) as time goes on. This behavior is typical in models where growth naturally caps at a maximum level, representing a stable state.

Initial Conditions

Initial conditions are critical in solving differential equations or verifying properties of a function. They provide values at specific points which help to determine parameters.

• In this problem, we verify the initial condition by substituting \(t=0\) into \(M(t)\) to see if itequals \(M_0\).
• Setting \(t=0\) gives us \(M(0)=K\left(\frac{M_{0}}{K}\right)^{\exp(-r\cdot0)}\).

Since \(\exp(0) = 1\), we simplify this to \(M(0) = K \cdot \frac{M_{0}}{K} = M_{0}\). Thus, the function satisfies the initial condition \(M(0) = M_{0}\). Initial conditions like these allow us to check the validity of model functions in representing real world scenarios.

Power Properties

Power properties are essential tools involving exponents and are widely used in simplifying expressions.

• The property we use often is that any number raised to the zero power is 1. In algebra, \(x^0 = 1\).
• This is crucial in the verification where \(\left(\frac{M_{0}}{K}\right)^{0} = 1\), which helps us confirm the limit condition \(\lim_{t \rightarrow \infty} M(t) = K\).

Understanding these power properties allows us to navigate through complex mathematical expressions and analyze their behavior. Having a solid grasp of these concepts reduces errors and enhances comprehension when handling exponential functions. Exponential equations frequently rely on these basic properties to evaluate limits or solve for unknowns.