Problem 45 - Biocalculus Calculus for the Life Sciences

Question

Measles pathogenesis The function $$f(t)=-t(t-21)(t+1)$$ has been used to model the measles virus concentration in an infected individual. The area under the graph of \(f\) represents the total amount of infection. We saw in Section 5.1 that at \(t=12\) days this total amount of infection reaction 5.1 threshold beyond which symptoms appear. Use the Evaluation Theorem to calculate this threshold value.

Step-by-Step Solution

Answer

The threshold infection value at \(t=12\) is 7848.

Step 1: Identify Relevant Integral

To find the total amount of infection, we need to find the area under the curve of the function \(f(t) = -t(t-21)(t+1)\). This is done by evaluating the definite integral from \(t = 0\) to \(t = 12\). Therefore, our integral is \[ \int_0^{12} f(t) \, dt = \int_0^{12} -t(t-21)(t+1) \, dt. \]

Step 2: Expand the Function Inside the Integral

First, expand \(-t(t-21)(t+1)\): \[-t(t^2 - 20t - 21) = -t^3 + 20t^2 + 21t.\]Now our integral becomes \(\int_0^{12} (-t^3 + 20t^2 + 21t) \, dt\).

Step 3: Integrate Term by Term

Integrate each term of the polynomial separately: \[\int_0^{12} (-t^3 + 20t^2 + 21t) \, dt = \left[ \frac{-t^4}{4} + \frac{20t^3}{3} + \frac{21t^2}{2} \right]_0^{12}.\]

Step 4: Evaluate the Integral at the Bounds

Substitute the upper limit of 12 into the integral: \[ \frac{-12^4}{4} + \frac{20 \times 12^3}{3} + \frac{21 \times 12^2}{2} \].Calculate this expression.

Step 5: Calculate Substitution at Upper Limit

Calculate each term:- \(\frac{-12^4}{4} = -5184\),- \(\frac{20 \times 12^3}{3} = 11520\),- \(\frac{21 \times 12^2}{2} = 1512\).Sum them up: \(-5184 + 11520 + 1512 = 7848\).

Step 6: Evaluate at Lower Bound

Substitution of the lower limit 0 into the integral gives 0, since every term involves \(t\) which becomes zero when \(t = 0\). The lower bound contribution is \(0\).

Step 7: Calculate the Total Integration Value

The value of the integral, and hence the area under the curve, is the difference between the values evaluated at the upper and lower limits, which is \(7848 - 0 = 7848\). This represents the total infection amount at \(t=12\).

Key Concepts

Definite Integral

A definite integral is a powerful concept in calculus used to find the area under a curve between two limits. In this exercise, we are calculating the total amount of a measles virus infection in an individual over a specific time period. The function given, \(f(t)=-t(t-21)(t+1)\), models the virus concentration in an individual's body. By finding the definite integral of this function from \(t=0\) to \(t=12\), we determine the area under the curve, which corresponds to the total infection.

To compute the definite integral, we must calculate the antiderivative of the function and then evaluate it at these limits. This approach provides an accurate measurement of the accumulated amount of infection, as it considers all variations in concentration over time.

Understanding definite integrals helps in many fields beyond calculus, such as physics and engineering, where one often needs to calculate physical quantities like distance or force over an interval.

Polynomial Integration

Polynomial integration involves finding the antiderivative of a polynomial expression, which gives us a new function whose derivative is the original polynomial. In this exercise, the function \(-t(t-21)(t+1)\) is expanded into a polynomial \(-t^3 + 20t^2 + 21t\) for integration purposes.

Each term of the polynomial is integrated separately:

• \(\int -t^3 \, dt = \frac{-t^4}{4}\)
• \(\int 20t^2 \, dt = \frac{20t^3}{3}\)
• \(\int 21t \, dt = \frac{21t^2}{2}\)

Combining these integral results gives the full antiderivative, which allows us to evaluate the integral at specific upper and lower limits. Polynomial integration simplifies the process of finding the area under complex polynomial curves, a common task in solving real-world problems.

Measles Virus Concentration

In this particular context, the function \(f(t)=-t(t-21)(t+1)\) serves as a model for the concentration of measles virus in a patient's body over time. The behavior of this function provides insights into how the infection evolves. At \(t=12\) days, it's critical to recognize that the total area under the curve has significant meaning; it represents the threshold at which symptoms are likely to manifest.

The ability to model virus concentration with mathematical functions such as this can inform healthcare professionals about infection dynamics. With accurate models, interventions can be better timed, leading to improved outcomes and understanding of disease progressions.

Through calculus, we can transform abstract data about virus behavior into actionable insights for managing public health concerns.

Evaluation Theorem

The Evaluation Theorem is a fundamental tool in calculus used to compute the value of a definite integral once the antiderivative is known. It states that if \(F\) is an antiderivative of \(f\) on \([a, b]\), then the definite integral \(\int_a^b f(x) \, dx = F(b) - F(a)\).

In this exercise, the theorem is applied by substituting the bounds \(t=0\) and \(t=12\) into the antiderivative of \(-t^3 + 20t^2 + 21t\). This results in:

• Evaluating at \(t=12\): \(\frac{-12^4}{4} + \frac{20 \, 12^3}{3} + \frac{21 \, 12^2}{2} = 7848\).
• Evaluating at \(t=0\): the expression equals zero since all terms include \(t\).

Subtracting these results gives the total value 7848 at \(t=12\), marking the threshold beyond which measles symptoms appear.

Knowing how to apply the Evaluation Theorem enables students and professionals to solve integrals efficiently, providing important quantitative results in scientific research and various applications.