Problem 23
Question
If \(\frac{d B}{d t}\) represents the rate of change of biomass at time \(t\), explain what $$ \int_{1}^{6} \frac{d B}{d t} d t $$ means.
Step-by-Step Solution
Verified Answer
It represents the total change in biomass from time \( t = 1 \) to \( t = 6 \).
1Step 1: Understanding the Given Function
The function \( \frac{d B}{d t} \) represents the rate of change of biomass with respect to time. This means that it describes how quickly or slowly the biomass is changing at any given time \( t \).
2Step 2: Interpreting the Integral
The integral \( \int_{1}^{6} \frac{d B}{d t} \, d t \) represents the accumulation of the rate of change of biomass over the time interval from \( t = 1 \) to \( t = 6 \). Here, the definite integral is used to add up all the tiny changes in biomass from time \( t = 1 \) to time \( t = 6 \).
3Step 3: Applying the Fundamental Theorem of Calculus
According to the Fundamental Theorem of Calculus, if \( F(t) \) is an antiderivative of \( f(t) = \frac{d B}{d t} \), then the definite integral from 1 to 6 of \( f(t) \) is given by \( F(6) - F(1) \). Therefore, \( \int_{1}^{6} \frac{d B}{d t} \, d t = B(6) - B(1) \), which means it calculates the net change in biomass from \( t = 1 \) to \( t = 6 \).
4Step 4: Conclusion
The expression \( \int_{1}^{6} \frac{d B}{d t} \, d t \) finds the total change in biomass between the times \( t = 1 \) and \( t = 6 \).
Key Concepts
BiomassRate of ChangeFundamental Theorem of Calculus
Biomass
Biomass refers to the total amount of living matter present in a particular environment, measured by its mass. It encompasses the biological material derived from living, or recently living, organisms.
Biomass is crucial in various fields, including ecology, agriculture, and energy. Understanding its dynamics is essential for sustainability studies and natural resource management.
In the context of calculus, when we analyze biomass over time, we are interested in how biomass grows or decreases in a certain period. This leads us to explore the idea of its rate of change, which requires a deeper understanding by employing mathematical tools such as derivatives and integrals.
Biomass is crucial in various fields, including ecology, agriculture, and energy. Understanding its dynamics is essential for sustainability studies and natural resource management.
In the context of calculus, when we analyze biomass over time, we are interested in how biomass grows or decreases in a certain period. This leads us to explore the idea of its rate of change, which requires a deeper understanding by employing mathematical tools such as derivatives and integrals.
Rate of Change
The rate of change of a quantity determines how fast or slow it alters over time. In calculus, this concept is formally expressed through derivatives.
For biomass, the rate of change at a given time is represented by the derivative \( \frac{d B}{d t} \). This notation indicates how the biomass, denoted by \( B \), changes with respect to time.
Understanding \( \frac{d B}{d t} \) allows us to see whether the biomass is increasing or decreasing at any specific point in time.
For biomass, the rate of change at a given time is represented by the derivative \( \frac{d B}{d t} \). This notation indicates how the biomass, denoted by \( B \), changes with respect to time.
Understanding \( \frac{d B}{d t} \) allows us to see whether the biomass is increasing or decreasing at any specific point in time.
- If \( \frac{d B}{d t} > 0 \), it indicates an increase in biomass over time.
- If \( \frac{d B}{d t} < 0 \), it signals a decrease.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus bridges the concept of differentiation and integration, the two main operations in calculus. It fundamentally states that integration can be reversed by differentiation and vice versa.
For a function that measures the rate of change, the theorem provides a way to find the total change over an interval.
The result \( B(6) - B(1) \) is the total biomass change over this interval. This application exemplifies how powerful calculus can be in practical scenarios, enabling us to handle systems involving continuous growth or decay.
For a function that measures the rate of change, the theorem provides a way to find the total change over an interval.
- If a function \( f(t) \) represents the rate of change of another function, \( F(t) \), then the integral of \( f(t) \) from \( a \) to \( b \) gives \( F(b) - F(a) \).
The result \( B(6) - B(1) \) is the total biomass change over this interval. This application exemplifies how powerful calculus can be in practical scenarios, enabling us to handle systems involving continuous growth or decay.
Other exercises in this chapter
Problem 22
Write each sum in sigma notation. $$ 1-a+a^{2}-a^{3}+a^{4}-a^{5}+\cdots+(-1)^{n} a^{n} $$
View solution Problem 23
Use Leibniz's rule to find \(\frac{d y}{d x}\). $$ y=\int_{1}^{3 x^{2}+x}\left(1+t e^{t}\right) d t $$
View solution Problem 23
Use the algebraic rules for sums to evaluate each sum. Recall that $$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$$ and $$\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$$ $$
View solution Problem 24
Use Leibniz's rule to find \(\frac{d y}{d x}\). $$ y=\int_{2}^{\ln x} e^{-t} d t, x>0 $$
View solution