Question

Antibiotic decay The half-life of an antibiotic in the bloodstream is 10 hours. If an initial dose of 20 milligrams is administered, the quantity left after \(t\) hours is modeled by \(Q(t)=20 e^{-0.0693 t},\) for \(t \geq 0\) a. Find the instantaneous rate of change of the amount of antibiotic in the bloodstream, for \(t \geq 0\) b. How fast is the amount of antibiotic changing at \(t=0 ? \mathrm{At}\) \(t=2 ?\) c. Evaluate and interpret \(\lim _{t \rightarrow \infty} Q(t)\) and \(\lim _{t \rightarrow \infty} Q^{\prime}(t)\)

Short answer

Answer: As time goes to infinity, the amount of antibiotic in the bloodstream approaches 0, meaning it eventually leaves the body completely. Also, as time goes to infinity, the rate of change of the amount of antibiotic in the bloodstream approaches 0, meaning that the rate at which it decays becomes very slow as the amount becomes negligible.

Step-by-step solution

Find the derivative of Q(t)

To find the instantaneous rate of change of the antibiotic quantity, we need to find the derivative of \(Q(t)\). The function is given as \(Q(t) = 20e^{-0.0693t}\). We can use the chain rule (which states that the derivative of a composition of functions is the product of the derivative of the outer function times the derivative of the inner function) to find the derivative of this function. Let \(u(t) = -0.0693t\). Then, \(Q(t) = 20e^{u(t)}\), so: $$\frac{dQ}{dt} = \frac{dQ}{du} \cdot \frac{du}{dt}$$ Now compute each derivative: 1. \(\frac{dQ}{du}\): $$\frac{dQ}{du}(20e^{u}) = 20e^u$$ 2. \(\frac{du}{dt}\): $$\frac{du}{dt}(-0.0693t) = -0.0693$$ 3. Multiply the two derivatives together: $$\frac{dQ}{dt} = 20e^u \cdot (-0.0693) = -0.0693(20)e^{-0.0693t}$$ Therefore, the instantaneous rate of change of the quantity of antibiotic in the blood is: $$Q'(t) = -1.386 e^{-0.0693t}$$

Determine the rate of change at specific times

We are now asked to find the rate of change when \(t=0\) and \(t=2\). To do this, substitute each value of \(t\) into the derivative \(Q'(t)\) that we found in step 1. For \(t=0\): $$Q'(0) = -1.386 e^{-0.0693(0)} = -1.386$$ For \(t=2\): $$Q'(2) = -1.386 e^{-0.0693(2)} \approx -1.175$$ At time \(t=0\), the amount of antibiotic is changing at a rate of -1.386 mg/h, and at time \(t=2\), the rate is approximately -1.175 mg/h.

Evaluate and interpret the limits

We are asked to find and interpret the limits as \(t\) approaches infinity for both \(Q(t)\) and \(Q'(t)\). 1. \(\lim_{t\to\infty} Q(t)\): Since the exponential function approaches zero as its argument approaches negative infinity, we have: $$\lim_{t\to\infty} Q(t) = \lim_{t\to\infty} 20 e^{-0.0693t} = 20 \cdot \lim_{t\to\infty} e^{-0.0693t} = 20 \cdot 0 = 0$$ Interpretation: As time goes to infinity, the amount of antibiotic in the bloodstream approaches 0, meaning it eventually leaves the body completely. 2. \(\lim_{t\to\infty} Q'(t)\): Similarly, for the limit of \(Q'(t)\) as \(t\) goes to infinity: $$\lim_{t\to\infty} Q'(t) = \lim_{t\to\infty} -1.386 e^{-0.0693t} = -1.386 \cdot \lim_{t\to\infty} e^{-0.0693t} = -1.386 \cdot 0 = 0$$ Interpretation: As time goes to infinity, the rate of change of the amount of antibiotic in the bloodstream approaches 0, meaning that the rate at which it decays becomes very slow as the amount becomes negligible.

Key concepts

Instantaneous Rate of Change

The instantaneous rate of change is crucial when measuring how fast a quantity changes at a particular instant. In calculus, this concept is represented by the derivative of a function. Here, the focus is on understanding how the amount of antibiotic in the bloodstream changes over time.
To find the instantaneous rate of change, we calculate the derivative of the function that models the quantity of the antibiotic. In this case, the given function is \(Q(t)=20e^{-0.0693t}\).

• Finding the derivative indicates how the quantity of antibiotic changes with any small changes in time.
• The instantaneous rate of change can tell us whether the amount of antibiotic is increasing or decreasing.

The derivative provides a formula that lets us compute this rate of change for any time \(t\). By substituting specific values of \(t\), we gain insights into how the rate of change behaves at those moments.

Limit of a Function

A limit helps us understand the behavior of a function as the input approaches a certain value. When dealing with functions that model decay, such as our antibiotic problem, limits give us insights into what happens over an extended period.
The main task here is to determine

• what the quantity \(Q(t)\) approaches as \(t\) goes to infinity, and
• how the rate of change \(Q'(t)\) behaves as time continues indefinitely.

For \(Q(t)=20e^{-0.0693t}\), calculating these limits shows that as time passes, both the quantity of antibiotic and its rate of change become negligible. This illustrates the complete decay of the antibiotic, which is vital in understanding drug clearance from the body over time.

Exponential Decay

Exponential decay describes a process in which a quantity decreases at a rate proportional to its current value. This antibiotic problem is a classic example. In the equation \(Q(t)=20e^{-0.0693t}\), \(-0.0693\) is the decay constant, influencing the steepness of the decay.

• Higher decay constants result in faster decay.
• The half-life, in this case, is 10 hours, meaning it takes 10 hours for the amount to reduce to half its initial value.

This type of growth pattern is common in natural processes such as radioactive decay, population decline, and other similar phenomena. Understanding this helps us predict how the quantity of a substance changes over time, which is critical in fields such as pharmacology and environmental science.

Differentiation

Differentiation is a tool in calculus used to find the rate of change of a variable with respect to another. It allows us to compute the derivative of a function, like \(Q(t)=20e^{-0.0693t}\), to determine how one quantity changes as another one varies.
In our scenario, we've used the chain rule of differentiation, a fundamental method for finding derivatives of composite functions. The chain rule involves:

• Finding the derivative of outer function, \(20e^u\).
• Finding the derivative of inner function, \(-0.0693t\).
• Multiplying these two derivatives together to obtain \(Q'(t)\).

This process yields insight into how the amount of antibiotic changes with time, emphasizing the role of differentiation in analyzing dynamic systems. Understanding differentiation is crucial for interpreting changes in medical treatments, economic forecasts, and scientific experiments.