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Lamprey Growth Griffiths and colleagues \(^{32}\) studied the larval sea lamprey in the Great Lakes, where these creatures represent a serious threat to the region's fisheries. They created a mathematical model given approximately by the equation \(G(T)=-1+0.3 T-0.02 T^{2},\) where \(T\) is the mean annual water temperature with \(6.2 \leq T \leq 9.8\) and \(G\) is the growth in millimeters per day. Graph. Find the instantaneous rate of change of growth with respect to water temperature when (a) \(T=7\), (b) \(T=9\). Give units and interpret your answer.

Short Answer

Expert verified
At \( T = 7 \), growth rate increases by 0.02 mm/day/°C; at \( T = 9 \), it decreases by 0.06 mm/day/°C.

Step by step solution

01

Understand the Function

The given function for the growth of larval sea lamprey is \( G(T) = -1 + 0.3T - 0.02T^2 \). This equation models how the growth in millimeters per day, \( G \), depends on the average water temperature, \( T \).
02

Find the Derivative

The instantaneous rate of change of a function with respect to a variable is given by its derivative. Derive the function \( G(T) = -1 + 0.3T - 0.02T^2 \) with respect to \( T \).\[ G'(T) = \frac{d}{dT}(-1 + 0.3T - 0.02T^2) = 0.3 - 0.04T \]
03

Calculate the Rate of Change at T = 7

Substitute \( T = 7 \) into the derived function to find the instantaneous rate of change.\[ G'(7) = 0.3 - 0.04 \times 7 = 0.3 - 0.28 = 0.02 \text{ millimeters per day per degree Celsius} \]
04

Calculate the Rate of Change at T = 9

Substitute \( T = 9 \) into the derivative to find the rate of change at this temperature.\[ G'(9) = 0.3 - 0.04 \times 9 = 0.3 - 0.36 = -0.06 \text{ millimeters per day per degree Celsius} \]
05

Interpret Results

For \( T = 7 \), the growth rate is increasing at a rate of 0.02 mm/day/°C, indicating slight sensitivity to temperature changes. For \( T = 9 \), the rate is negative, indicating a decrease in growth as temperature increases beyond this point.

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Key Concepts

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

Derivative
Derivatives are a fundamental concept in calculus that help us understand how a function changes at any given point. In simple terms, the derivative of a function is a formula that tells us the slope or steepness of the function at any point along its curve. In the context of the given problem, we used the derivative to determine how the growth rate of larval sea lamprey changes with changes in water temperature.
Derivatives are essential because they provide:
  • A way to calculate the instantaneous rate of change; this means how fast something changes at an exact moment.
  • An understanding of the behavior of curves, such as whether a curve is rising or falling at a given point.
To find the derivative of the function given in the problem, which is \( G(T) = -1 + 0.3T - 0.02T^2 \), we calculated:
  • \( G'(T) = 0.3 - 0.04T \)
This formula tells us the rate at which the growth function, \( G(T) \), changes as the temperature, \( T \), changes.
Instantaneous Rate of Change
The instantaneous rate of change is a concept used to describe the speed at which something is changing at any precise moment in time. For functions, this is equivalent to finding the derivative at a particular value. In our lamprey growth problem, we were tasked to find the instantaneous rate of change of growth concerning water temperature.
To better grasp this concept:
  • The instantaneous rate of change tells us the slope of the tangent line to a curve at a specific point. This gives a very accurate picture of what is happening exactly at that point.
  • When we calculated \( G'(7) \) and \( G'(9) \), these values gave us the rate at which the growth changes per degree Celsius at those specific temperatures.
So when we found that \( G'(7) = 0.02 \) mm/day/°C, it meant at \( T=7 \), the growth rate is increasing slightly. Conversely, \( G'(9) = -0.06 \) mm/day/°C showed a reduction in growth when \( T=9 \).
Mathematical Modeling
Mathematical modeling is a powerful tool that allows us to use equations to represent real-world phenomena. In the context of the lamprey growth, we used the model equation \( G(T) = -1 + 0.3T - 0.02T^2 \) to understand growth patterns based on water temperature.
There are several advantages to mathematical modeling:
  • It helps predict future trends and behaviors by providing a structured way to analyze complex systems.
  • Models allow scientists and researchers to simulate conditions and understand potential outcomes without intrusive or extensive field studies.
In this example, the model gives valuable insights into how water temperature affects lamprey growth. With \( T \) ranging from 6.2 to 9.8, the model helps determine conditions that either promote or inhibit growth. Understanding this is crucial for managing ecological systems, especially where species like the sea lamprey pose threats to fisheries.

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Most popular questions from this chapter

Cost Function Dean \(^{66}\) found that the cost function for indirect labor in a furniture factory was approximated by \(y=C(x)=0.4490 x-0.01563 x^{2}+0.000185 x^{3}\) for \(0 \leq\) \(x \leq 50 .\) Find the marginal cost for any \(x .\) Find \(C^{\prime}(10)\) \(C^{\prime}(30),\) and \(C^{\prime}(40) .\) Interpret what is happening. Graph marginal cost on a screen with dimensions [0,47] by [0,0.3]

Use the rules of limits to find the indicated limits if they exist. Support your answer using a computer or graphing calculator. $$ \lim _{x \rightarrow 1} \pi $$

Find the limits graphically. Then confirm algebraically. $$ \lim _{x \rightarrow 4} \frac{x-4}{\sqrt{x}-2} $$

Dean \(^{60}\) made a statistical estimation of the cost-output relationship for a shoe chain. The data for the firm is given in the following table. $$ \begin{array}{|l|lllllllll|} \hline x & 4.7 & 6.5 & 7.8 & 10 & 15 & 20 & 30 & 38 & 50 \\ \hline y & 4.7 & 5.5 & 6.4 & 6.8 & 8.5 & 12 & 20 & 16 & 30 \\ \hline \end{array} $$ Here \(x\) is the output in thousands of pairs of shoes, and \(y\) is the total cost in thousands of dollars. a. Use quadratic regression to find the best-fitting quadratic polynomial using least squares. b. Graph on your computer or graphing calculator using a screen with dimensions [0,56.4] by \([0,40] .\) Have your grapher find the derivative when \(x\) is \(6,18,24,30,\) and 36. Note the slope, and relate this to the rate of change. On the basis of this model, describe what happens to the rate of change of average cost as output increases.

On your computer or graphing calculator, graph \(y=\) \(f(x)=\cos x\) in radian mode, using a window with dimensions [-6.14,6.14] by [-1,1] to familiarize yourself with this function. As you see, this function moves back and forth between -1 and \(1 .\) We wish to estimate \(f^{\prime}(\pi / 2)\) where \(\pi / 2 \approx 1.57\). For this purpose, graph using a window with dimensions [1.07,2.07] by \([-0.5,0.5] .\) From the graph, estimate \(f^{\prime}(1.57)\).

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