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Growth of fish: Let \(w=w(t)\) denote the weight of a fish as a function of its age \(t\). For the North Sea cod, the equation of change $$ \frac{d w}{d t}=2.1 w^{2 / 3}-0.6 w $$ holds. Here \(w\) is measured in pounds and \(t\) in years. b. Make a graph of \(\frac{d w}{d t}\) against \(w\). Include weights up to 45 pounds. c. What is the weight of the cod when it is growing at the greatest rate? d. To what weight does the cod grow?

Short Answer

Expert verified
The weight for maximum growth is approximately 11.23 pounds, and potential maximum weight is approximately 45.8 pounds.

Step by step solution

01

Understanding the Equation of Change

The given equation \( \frac{d w}{d t} = 2.1 w^{2/3} - 0.6 w \) describes how the weight of the fish changes over time. The goal is to analyze this equation for different parts of the problem.
02

Graphing \( \frac{d w}{d t} \) Against \( w \)

To graph \( \frac{d w}{d t} \) against \( w \), calculate \( \frac{d w}{d t} \) for values of \( w \) ranging from 0 up to 45 pounds. Compute \( \frac{d w}{d t} = 2.1 w^{2/3} - 0.6 w \) for multiple points and plot them on the graph.
03

Finding Maximum Growth Rate

To find when the cod grows at the greatest rate, set the derivative of \( \frac{d w}{d t} \) with respect to \( w \) to zero: \( \frac{d}{dw}(2.1w^{2/3} - 0.6w) = 0 \). Solve this equation to find the optimal \( w \).
04

Solving for Maximum Growth

Take the derivative: \( \frac{d}{dw} (2.1w^{2/3}) = 1.4w^{-1/3} \) and \( \frac{d}{dw} (-0.6w) = -0.6 \). Set \( 1.4w^{-1/3} - 0.6 = 0 \) and solve for \( w \). This gives \( 1.4w^{-1/3} = 0.6 \). Solve for \( w \) to get \( w = \left(\frac{1.4}{0.6}\right)^3 \).
05

Finding Maximum Weight the Cod Reaches

The cod grows indefinitely towards the equilibrium weight where \( \frac{d w}{d t} = 0 \). Set \( 2.1w^{2/3} - 0.6w = 0 \) and solve for \( w \). Factor the equation to find \( w \) values.
06

Solving for Equilibrium Weight

Rewriting \( 0 = 2.1w^{2/3} - 0.6w \) gives \( w(2.1w^{-1/3} - 0.6) = 0 \). So, \( 2.1w^{-1/3} = 0.6 \). Solving gives \( w = \left(\frac{2.1}{0.6}\right)^3 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Growth Rate
The growth rate of the North Sea cod fish is represented by the differential equation \( \frac{d w}{d t} = 2.1 w^{2/3} - 0.6 w \). This equation consists of two terms contributing to changes in fish weight over time.
The first term, \( 2.1 w^{2/3} \), suggests a growth that depends on the current weight raised to the power of \( \frac{2}{3} \). As the weight increases, this term contributes to growth, but at a decreasing rate. This reflects a common biological phenomenon where organisms grow rapidly when young but slow down as they age.
The second term, \(-0.6 w \), represents a decay or loss factor that increases linearly with weight, counteracting the growth term. This could account for factors such as metabolic maintenance or environmental limits. Together, these terms model the overall growth rate, allowing us to predict growth patterns and understand how factors like weight affect the fish's development over time.
Exploring Equilibrium
The concept of equilibrium in this context refers to a steady state where the fish's weight no longer changes, meaning the derivative \( \frac{d w}{d t} \) equals zero. This occurs when the forces driving weight gain balance those causing weight loss.
For the given equation, finding equilibrium involves solving \( 2.1 w^{2/3} - 0.6 w = 0 \). Factoring the equation and solving gives the weight \( w \) at this state. The calculation results in \( w = \left(\frac{2.1}{0.6}\right)^3 \), approximately equalling the maximum possible weight the fish will naturally reach.
Reaching equilibrium is significant since it helps biologists determine the potential maximum size in optimal conditions, providing insights into the lifecycle and growth limitations of the species.
Understanding Weight Function
A weight function, like \( w(t) \), represents how a variable—in this case, a fish’s weight—changes over time. This function is critical for modeling biological phenomena and predicting future scenarios.
In this exercise, the focus is on how \( w(t) \) evolves due to the growth rate equation provided. By applying mathematical techniques, such as taking derivatives, we analyze how much and at what rate the fish's weight changes with age.
Successfully determining the weight function allows us to make important conclusions about life expectancy, needed resources for sustaining populations, and effective strategies for fisheries management. It highlights the interplay of multiple factors influencing growth, emphasizing the importance of mathematical models in understanding biological systems.
Graphing Functions Effectively
Graphing the derivative \( \frac{d w}{d t} \) against \( w \) provides a visual representation of the growth rate at different weights. A graph reveals critical points where growth is maximized or ceased, like the maximum growth rate point and equilibrium weight.
Creating such a graph involves calculating \( \frac{d w}{d t} \) for various weight values up to 45 pounds and plotting these on a coordinate plane.
  • A negative \( \frac{d w}{d t} \) indicates weight decrease or negative growth.
  • A positive \( \frac{d w}{d t} \) signifies positive growth.
  • A zero value shows equilibrium where growth stops.
Graphing helps in understanding complex relations visually, making abstract concepts more accessible and aiding in strategic decision-making related to ecological and conservation practices.

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Most popular questions from this chapter

Looking up: The constant \(g=32\) feet per second per second is the downward acceleration due to gravity near the surface of the Earth. If we stand on the surface of the Earth and locate objects using their distance up from the ground, then the positive direction is up, so down is the negative direction. With this perspective, the equation of change in velocity for a freely falling object would be expressed as $$ \frac{d V}{d t}=-g $$ (We measure upward velocity \(V\) in feet per second and time \(t\) in seconds.) Consider a rock tossed up- ward from the surface of the Earth with an initial velocity of 40 feet per second upward. a. Use a formula to express the velocity \(V=V(t)\) as a linear function. (Hint: You get the slope of \(V\) from the equation of change. The vertical intercept is the initial value.) b. How many seconds after the toss does the rock reach the peak of its flight? (Hint: What is the velocity of the rock when it reaches its peak?) c. How many seconds after the toss does the rock strike the ground? (Hint: How does the time it takes for the rock to rise to its peak compare with the time it takes for it to fall back to the ground?)

Estimating rates of change: Use your calculator to make the graph of \(f(x)=x^{3}-5 x\). a. Is \(\frac{d f}{d x}\) positive or negative at \(x=2\) ? b. Identify a point on the graph of \(f\) where \(\frac{d f}{d x}\) is negative.

Logistic growth with a threshold: Most species have a survival threshold level, and populations of fewer individuals than the threshold cannot sustain themselves. If the carrying capacity is \(K\) and the threshold level is \(S\), then the logistic equation of change for the population \(N=N(t)\) is $$ \frac{d N}{d t}=-r N\left(1-\frac{N}{S}\right)\left(1-\frac{N}{K}\right) . $$ For Pacific sardines, we may use \(K=2.4\) million tons and \(r=0.338\) per year, as in Example 6.10. Suppose we also know that the survival threshold level for the sardines is \(S=0.8\) million tons. a. Write the equation of change for Pacific sardines under these conditions. b. Make a graph of \(\frac{d N}{d t}\) versus \(N\) and use it to find the equilibrium solutions. How do the equilibrium solutions correspond with \(S\) and \(K\) ? c. For what values of \(N\) is the graph of \(N\) versus \(t\) increasing, and for what values is it decreasing? d. Explain what can be expected to happen to a population of \(0.7\) million tons of sardines. e. At what population level will the population be growing at its fastest?

Chemical reactions: In a second-order reaction, one molecule of a substance \(A\) collides with one molecule of a substance \(B\) to produce a new substance, the product. If \(t\) denotes time and \(x=x(t)\) denotes the concentration of the product, then its rate of change \(\frac{d x}{d t}\) is called the rate of reaction. Suppose the initial concentration of \(A\) is \(a\) and the initial concentration of \(B\) is \(b\). Then, assuming a constant temperature, \(x\) satisfies the equation of change $$ \frac{d x}{d t}=k(a-x)(b-x) $$ for some constant \(k\). This is because the rate of reaction is proportional both to the amount of \(A\) that remains untransformed and to the amount of \(B\) that remains untransformed. Here we study a reaction between isobutyl bromide and sodium ethoxide in which \(k=0.0055, a=51\), and \(b=76\). The concentrations are in moles per cubic meter, and time is in seconds. 10 a. Write the equation of change for the reaction between isobutyl bromide and sodium ethoxide. b. Make a graph of \(\frac{d x}{d t}\) versus \(x\). Include a span of \(x=0\) to \(x=100\). c. Explain what can be expected to happen to the concentration of the product if the initial concentration of the product is 0 .

7\. A population of bighorn sheep: A certain group of bighorn sheep live in an area where food is plentiful and conditions are generally favorable to bighorn sheep. Consequently, the population is thriving. There are initially 30 sheep in this group. Let \(N=N(t)\) be the population \(t\) years later. The population changes each year because of births and deaths. The rate of change in the population is proportional to the number of sheep currently in the population. For this particular group of sheep, the constant of proportionality is \(0.04\). a. Express the sentence "The rate of change in the population is proportional to the number of sheep currently in the population" as an equation of change. (Incorporate in your answer the fact that the constant of proportionality is \(0.04\).) b. Find a formula for \(N\). c. How long will it take this group of sheep to grow to a level of 50 individuals?

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