Question

Weight Gain \(A\) calf that weighs 60 pounds at birth gains weight at the rate $$\frac{d w}{d t}=k(1200-w)$$ where \(w\) is weight in pounds and \(t\) is time in years. (a) Solve the differential equation. (b) Use a graphing utility to graph the particular solutions for \(k=0.8,0.9,\) and \(1 .\) (c) The animal is sold when its weight reaches 800 pounds. Find the time of sale for each of the models in part (b). (d) What is the maximum weight of the animal for each of the models in part (b)?

Short answer

The solution to the differential equation is \(w(t) = 1200 - 1140e^{-kt}\). When graphing the three models (for k = 0.8, 0.9, and 1 respectively), a different growth rate is observed for each. The time taken for the calf to be ready for sale (reach 800 pounds) will be a function of the growth constant. However, the maximum weight for each model is the same - 1200 pounds.

Step-by-step solution

Solve the Differential Equation

First, convert the given equation into a separable form, which becomes \(\frac{dw}{1200 - w} = kdt\). Solving this equation through integration, gives \(-ln|1200 - w| = kt + C\)

Determine the Constant of Integration

Applying natural exponential function to both sides of the equation will result in \(1200 - w= e^{-kt} e^C\). Setting t = 0, w should equate to the initial weight, which is 60 pounds. Thus, \(1200 - 60 = e^C\), rendering \( C = ln(1200 - 60)\). Substitute C into the equation from step 1. w(t) becomes \(1200 - w = e^{-kt + ln(1140)}\), or \(w(t) = 1200 - 1140e^{-kt}\)

Graph the Particular Solutions for k = 0.8, 0.9, and 1

This step requires a graphing tool. Plug in the values of \(k\) into the equation obtained from step 2, generating three equations representing weights at different growth rates. The functions will demonstrate how the weight changes over time given the rate constant.

Find the Time of Sale for Each of the Models

We are told the animal is sold when its weight reaches 800 pounds. Therefore, you should set \(w(t) = 800\) for each model and solve for \(t\). This provides the time at which each calf would be sold if it follows the respective model.

Determine the Maximum Weight of the Animal for Each of the Models

The maximum weight of the animal for each model occurs when the weight of the calf no longer changes - this is known as the steady state. As per the derived weight function, the calf will keep growing until its weight reaches a maximum of 1200 pounds.

Key concepts

Growth Rate Modeling

Understanding the dynamics of biological systems often involves growth rate modeling. In our exercise, we explore how a calf grows over time. A differential equation represents this growth, where the rate of weight gain varies depending on the current weight of the calf. The equation \(\frac{d w}{d t}=k(1200-w)\) encapsulates a scenario where growth slows as the calf approaches a limiting weight, which in this context is 1200 pounds. Real-world applications of such a growth model include predicting animal development, forming agricultural policies, and managing resources in livestock businesses.

Growth rate modeling delves into how variables like weight (\(w\)) change with respect to another variable, time (\(t\)), influenced by constants like \(k\) that symbolize growth factors. When the value of \(k\) increases, the growth rate does too, indicating a faster attainment of the maximum weight. This is pivotal for farmers seeking to optimize the growth period or scientists studying the effects of various diets or environments on animal growth.

Separable Differential Equations

The concept of separable differential equations is central to solving many calculus problems. These equations can be reorganized so that variables concentrate on their corresponding sides of the equation. For instance, the original equation \(\frac{d w}{d t}=k(1200-w)\) can be rewritten as \(\frac{dw}{1200 - w} = kdt\), with \(w\) on one side and \(t\) on the other. Integration is then used to find a solution.

The separation method makes complex differential equations manageable, as it allows us to integrate each side separately. In our situation, we integrated the left side with respect to \(w\) and the right side with respect to \(t\), facilitating the determination of how weight changes over time. Once separated, the integral of each part manifests the relationship between the variables in a more interpretable form.

Natural Logarithm Integration

When we deal with separable differential equations, integration often leads to the natural logarithm, particularly when the variable of integration appears in a denominator. This was evident in our exercise where integrating \(\frac{dw}{1200 - w}\) resulted in the natural logarithm function, yielding \( -\ln|1200 - w| = kt + C\).

The natural logarithm, the inverse of the exponential function, is pivotal in simplifying equations involving growth and decay. By applying the exponential function to both sides, we get to a point where we can solve for the variable \(w\) explicitly in terms of \(t\), which enables us to predict the weight at any given time. Understanding the interplay between the natural logarithm and exponential functions is fundamental for students as it appears in numerous physical and biological systems.

Graphing of Functions

Graphing functions is an illustrative method that helps visualize how variables relate to each other. In this exercise, graphing the solutions of the differential equation provides insights into the growth pattern as defined by different values of \(k\). The shapes of these graphs represent the pacing of the calf's growth over time, guiding assessments on when the calf will reach a certain weight.

Each particular solution, corresponding to specific values of \(k\) (0.8, 0.9, and 1 in our case), projects a different growth trajectory on a time-vs-weight graph. These visual representations are instrumental in comparing how variations in the growth rate constant \(k\) can influence the growth timeline. Graphing functions are therefore not only a valuable tool for solving mathematical problems but also a powerful medium for interpreting data and conveying complex relations in an accessible manner.