Problem 10 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Exercise 4.1 .10 A nutritionist studying plasma epinephrine (EPI) kinetics with tritium labeled epinephrine, \(\left[{ }^{3} \mathrm{H}\right] \mathrm{EPI},\) observes that after a bolus injection of \(\left[{ }^{3} \mathrm{H}\right] \mathrm{EPI}\) into plasma, the time-dependence of \(\left[{ }^{3} \mathrm{H}\right]\) EPI level is well approximated by \(L(t)=4 e^{-2 t}+3 e^{-t}\) where \(L(t)\) is the level of \(\left[{ }^{3} \mathrm{H}\right] \mathrm{EPI}\) t hours after infusion. Sketch the graph of \(L\). Observe that \(L(0)=7\) and \(L(2)=0.479268 .\) The intermediate value property asserts that at some time between 0 and 2 hours the level of \(\left[{ }^{3} \mathrm{H}\right] \mathrm{EPI}\) will be \(1.0 .\) At what time, \(t_{1},\) will \(L\left(t_{1}\right)=1.0 ?\) (Let \(A=e^{-t}\) and observe that \(\left.A^{2}=e^{-2 t} .\right)\)

Step-by-Step Solution

Answer

The time \( t_1 \) when \( L(t_1) = 1.0 \) is approximately 1.386 hours.

Step 1: Understand the Function

The given function is \( L(t) = 4e^{-2t} + 3e^{-t} \). This function describes the level of tritium-labeled epinephrine \([^{3}H]EPI\) in the plasma over time \(t\) hours after the injection.

Step 2: Evaluate the Function at Given Points

We already know from the problem that at \(t = 0\), \(L(0) = 7\), and at \(t = 2\), \(L(2) = 0.479268\). To verify or explore these, we can substitute \(t = 0\) to get \(L(0) = 4e^{-2(0)} + 3e^{-0} = 4\times 1 + 3\times 1 = 7\) and for \(t = 2\), substitute to find \(L(2) = 4e^{-4} + 3e^{-2}\). Use a calculator to verify the value given.

Step 3: Rewriting the Function

Let \( A = e^{-t} \), then \( A^2 = e^{-2t} \). Rewriting the function in terms of \(A\), we have \( L(t) = 4A^2 + 3A \). This will simplify finding \(t\) when \(L(t) = 1\).

Step 4: Solve for Time \( t_1 \) when \( L(t_1) = 1 \)

Set \( L(t) = 4A^2 + 3A = 1 \). This gives us the quadratic equation \( 4A^2 + 3A - 1 = 0 \). Using the quadratic formula \( A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a=4, b=3, c=-1\):\[ A = \frac{-3 \pm \sqrt{3^2 - 4 \times 4 \times (-1)}}{2 \times 4} = \frac{-3 \pm \sqrt{9 + 16}}{8} = \frac{-3 \pm 5}{8} \]This provides two solutions for \(A\): \( A = \frac{2}{8} = \frac{1}{4} \) and \( A = \frac{-8}{8} = -1 \), but \( A = -1 \) is not valid as \( A = e^{-t} \) must be positive, thus, \( A = \frac{1}{4} \).

Step 5: Solve for \( t_1 \) using \( A = \frac{1}{4} \)

Since \( A = e^{-t} = \frac{1}{4} \), we take natural logarithm on both sides to solve for \( t \):\[ \ln(e^{-t}) = \ln(\frac{1}{4}) \]Which simplifies to \[ -t = \ln(\frac{1}{4}) \]Therefore, \[ t = -\ln(\frac{1}{4}) = \ln(4) \approx 1.386 \].

Step 6: Verify the Result for \( t_1 \)

Double-check the calculation: Since \( e^{-t_1} = \frac{1}{4} \), it matches our setup with \( L(t_1) = 1 \). So, when \( t \approx 1.386 \), the function \( L(t) \) is indeed 1, satisfying the intermediate value property.

Key Concepts

Exponential Functions

Exponential functions are a type of mathematical function that can be described in the form of \(f(x) = a \cdot e^{bx}\), where \(e\) is the base of the natural logarithms, approximately equal to 2.718, and \(a\) and \(b\) are constants. They are frequently used to model processes involving growth or decay, such as population growth, radioactive decay, or, as in this exercise, epinephrine kinetics in the plasma.
An exponential function can exhibit either rapid increase (exponential growth) or rapid decrease (exponential decay) depending on the sign and magnitude of \(b\).
In the context of the given exercise, the function \(L(t) = 4e^{-2t} + 3e^{-t}\) describes the decay of a substance, indicating that over time the levels of tritium-labeled epinephrine decrease. Both terms of our function reflect this exponentially decaying nature, highlighting how the substance diminishes in concentration over time, capturing the essence of drug kinetics.

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is a fundamental concept in calculus, establishing that for any continuous function \(f(x)\), if \(f(a)\) and \(f(b)\) have different signs, there exists at least one value \(c\) in the interval \([a, b]\) where \(f(c) = 0\).
This theorem ensures that no value is "skipped" over if the function transitions from above to below a particular point, or vice versa.
In this exercise, we applied the IVT to show that the tritium-labeled epinephrine level must reach an exact level (in this case, 1) between \(t = 0\) and \(t = 2\) because \(L(0) = 7\) and \(L(2) = 0.479268\). By applying this concept, we find it imperative that there exists a time \(t_1\) such that \(L(t_1) = 1\), highlighting the application of IVT in solving continuous functions.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree typically given in the form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\).
These equations can often be solved using factoring, completing the square, or more generally, the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In the step-by-step solution, the equation \(4A^2 + 3A - 1 = 0\) is a quadratic in form. We used the quadratic formula to find possible values for \(A\).
Understanding this mathematical form is vital for modeling and finding precise solutions in dynamic processes such as the kinetics highlighted in our exercise, enabling us to compute critical points, like when a specific substance level is reached, by considering these quadratic expressions.

Natural Logarithms

Natural logarithms are logarithms to the base \(e\). The natural log of a number \(x\) is represented as \(\ln(x)\), and is defined as the power to which \(e\) must be raised to obtain \(x\).
In mathematical modeling, natural logarithms are particularly useful when dealing with exponential growth and decay due to their inverse relationship with exponential functions.
In this exercise, once we determined that \(e^{-t} = \frac{1}{4}\), we used the natural logarithm to solve for \(t\), by converting \(e^{-t} = \frac{1}{4}\) into \(-t = \ln(\frac{1}{4})\). This leads us to the positive time solution \(t = \ln(4)\).
Knowing how to manipulate and solve expressions using natural logarithms is essential for reversing exponential equations, as seen in assessing the time at which a specific drug concentration level is achieved in our lab scenario.