Problem 70 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

NET CHANGE IN BIOMASS A protein with mass \(m\) (grams) disintegrates into amino acids at a rate given by $$ \frac{d m}{d t}=\frac{-15 t}{t^{2}+5} $$ What is the net change in mass of the protein during the first 4 hours?

Step-by-Step Solution

Answer

Approximately -10.76 grams.

Step 1: Identify the Given Information

The rate of mass change of the protein is given by \ \( \frac{d m}{d t} = \frac{-15 t}{t^2 + 5} \). The time interval of interest is from \( t = 0 \) to \( t = 4 \) hours.

Step 2: Set Up the Integral

To find the net change in mass, integrate the rate of change from \( t = 0 \) to \( t = 4 \). This gives us: \ \( \int_{0}^{4} \frac{d m}{d t} \ d t \).

Step 3: Substitute the Rate of Change

Substitute \( \frac{d m}{d t} = \frac{-15 t}{t^2+5} \) into the integral: \ \( \int_{0}^{4} \frac{-15 t}{t^2 + 5} \ d t \).

Step 4: Solve the Integral using Substitution

Use substitution where \( u = t^2 + 5 \) and \( du = 2t \, dt \). Then \( t \, dt = \frac{du}{2} \). The limits of integration change accordingly, at \( t = 0, u = 5 \) and at \( t = 4, u = 21 \). Substitute and integrate: \ \( \begin{align*} \begin{aligned} \ & = \int_{5}^{21} \frac{-15}{2} \, \frac{1}{u} \, du \ & = -\frac{15}{2} \int_{5}^{21} \frac{1}{u} \, du \ & = -\frac{15}{2} \left[ \ln | u | \right ]_{5}^{21} \ & = -\frac{15}{2} \left( \ln 21 - \ln 5 \right) \ & = -\frac{15}{2} \ln \left( \frac{21}{5} \right) \ & = -\frac{15}{2} \ln \left( 4.2 \right) \ \end{aligned} \end{align*} \).

Step 5: Compute the Net Change in Mass

Evaluate the final expression: \ \( -\frac{15}{2} \ln(4.2) = -\frac{15}{2} \cdot 1.4351 \approx -10.76 \) grams. Thus, the net change in mass is approximately -10.76 grams.

Key Concepts

integral calculus

Integral calculus is a fundamental branch of mathematics focused on concepts of accumulation and the area under curves. When dealing with rates of change, integral calculus allows us to summarize changes over a specific interval by integrating the rate function. For our exercise, the integral of the rate of change of the protein mass over time tells us the total net change in biomass over that interval. Integrating the given rate function, \(\frac{d m}{d t} = \frac{-15 t}{t^2 + 5}\), over the 4-hour period from \(t = 0\) to \(t = 4\) results in the net change in mass.

rate of change

The rate of change describes how a quantity changes over time. In our problem, we are given the rate at which the protein mass decreases, expressed as \(\frac{d m}{d t} = \frac{-15 t}{t^2 + 5}\). This function indicates how fast and in what manner the protein's mass is decreasing at any given time \(t\). Understanding the rate of change is crucial for setting up the integral that will help determine the total change in mass over a specified time frame.

integration by substitution

Integration by substitution is a technique used to simplify the process of finding integrals, especially when the integral contains complex expressions. In our exercise, substitution makes the integration more manageable. We start by letting \(u = t^2 + 5\) and finding the differential \(du = 2t \, dt\). This simplifies \(\int_{0}^{4} \frac{-15 t}{t^2 + 5} \, d t\) into a more straightforward integral of \(\int_{5}^{21} \frac{-15}{2} \, \frac{1}{u} \, du\). This step transforms our original integral, allowing for simpler evaluation and ultimately finding the net change in mass.

applied calculus

Applied calculus takes the principles and techniques of calculus and uses them to solve real-world problems. In this exercise, we use integral calculus to address a biological problem: calculating the net change in biomass of a disintegrating protein. By understanding how rates of change and integration apply practically, such as in measuring changes over time, students can appreciate the power of calculus in solving tangible problems in fields like biology, economics, engineering, and more. Using the given rate function and integral calculus, we determine that the protein’s mass decreases by approximately 10.76 grams over the first 4 hours.